NCERT Exemplar Class 10 Maths Solutions Chapter 7 Coordinate Geometry 2021

Q: Which theorem is used to derive distance formula?

The distance formula enables us to find out the distance between two points of coordinates. This formula is derived by the Pythagoras theorem.

Q: What is external division of a line segment?

If the point which divides the line segment lies outside the line segment and the ratio of its distance from the given two points of line segment are in the given ratio.

Q: Can we find the area of any polygon if the coordinate of all vertices is given?

Yes, we can find out the area of any polygon. To calculate the area of any polygon we have to divide this area into different triangles and we can use the formula of the area of the triangle.

Q: Is the chapter Coordinate Geometry important for Board examinations?

The chapter Coordinate Geometry holds around 6-8% weightage of the whole paper.

Q: Will these problems of NCERT exemplar require any additional knowledge of Coordinate Geometry?

NCERT exemplar Class 10 Maths solutions chapter 7 help students with a multidimensional approach. It will help them to understand the concept of Coordinate Geometry quite easily.

NCERT Exemplar Class 10 Maths Solutions Chapter 7 Coordinate Geometry

Edited By Ravindra Pindel | Updated on Sep 05, 2022 04:10 PM IST | #CBSE Class 10th

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NCERT exemplar Class 10 Maths solutions chapter 7 provides with the understanding for coordinate geometry in continuation to the basics studied in Class 9. The students attempting exemplar can use these NCERT exemplar Class 10 Maths chapter 7 solutions and build a strong base for the NCERT Class 10 Maths. These NCERT exemplar Class 10 Maths solutions chapter 7 generate a better understanding of concepts of coordinate geometry as they are detailed and expressive. The Class 10 Maths NCERT exemplar chapter 7 solutions are in accord with the CBSE Syllabus for Class 10.

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Features of NCERT Exemplar Class 10 Maths Solutions Chapter 7:

Question:1

The distance of the point P (2, 3) from the x-axis is
(A) 2
(B) 3
(C) 1
(D) 5

Answer:

$\\We know that at x-axis y = 0\\ Hence we have to find distance of P(2, 3) from Q(2, 0)\\ (x\textsubscript{1}, y\textsubscript{1}) = (2,3) (x\textsubscript{2}, y\textsubscript{2}) = (2,0)\\ Let the distance between them is PQ\\ Distance formula = \sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ PQ=\sqrt{{{(2-2)}^{2}}+{{(0-3)}^{2}}} \\ PQ=\sqrt{0+9}=\sqrt{9}=3 \\ PQ = 3\\$

Question:2

The distance between the points A (0, 6) and B (0, –2) is
(A) 6
(B) 8
(C) 4
(D) 2

Answer:

$\\The given points are A(0, 6), B(0, -2)\\ (x\textsubscript{1}, y\textsubscript{1}) = (0, 6) (x\textsubscript{2}, y\textsubscript{2}) = (0, -2)\\ Let the distance between them is AB\\ AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ AB=\sqrt{{{(0-0)}^{2}}+{{(-2-6)}^{2}}} \\ \begin{aligned} & AB=\sqrt{{{(-8)}^{2}}} \\ & AB =\sqrt{64} \\ & AB =8 \\ \end{aligned} \\ AB = 8\\$

Question:3

The distance of the point P (–6, 8) from the origin is
(A) 8
(B) $2\sqrt7$
(C) 10
(D) 6

Answer:

$\\The given point is P(-6, 8) origin (0,0)\\ \ (x, y) = (-6, 8)\\ Let the distance from origin is OP\\ OP=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ OP=\sqrt{{{(-6)}^{2}}+{{(8)}^{2}}}=\sqrt{36+64}=\sqrt{100}=10 \\ OP = 10\\$

Question:4

The distance between the points (0, 5) and (–5, 0) is
(A) 5
(B) $5\sqrt2$
(C) $2\sqrt5$
(D) 10

Answer:

$\\The given points are A (0, 5),B (-5, 0)\\ (x\textsubscript{1}, y\textsubscript{1}) = (0, 5) (x\textsubscript{2}, y\textsubscript{2}) = (-5, 0)\\ Let the distance between the points is AB \\ AB=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ AB=\sqrt{{{(-5-0)}^{2}}+{{(0-5)}^{2}}} \\ AB=\sqrt{25+25}=\sqrt{50}=5\sqrt{2} \\ AB=5\sqrt{2} \\$

Question:5

AOBC is a rectangle whose three vertices are vertices A (0, 3), O (0, 0) and B (5, 0). The length of its diagonal is
(A) 5
(B) 3
(C) $\sqrt{34}$
(D) 4

Answer:

Given : Vertices of rectangle A(0, 3), O(0, 0), B(5, 0)

To find the length of diagonal we have to find the distance between A and B
$\\Let the length of diagonal is AB\\ \ \ x\textsubscript{1}=0 \ ,\ x\textsubscript{2}=5\ ,y\textsubscript{1}=3\ \ ,y\textsubscript{2}=0 \\ Hence, the length of diagonal is \sqrt{34} .\\$

Question:6

The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is
(A) 5
(B) 12
(C) 11
(D) $7+ \sqrt5$

Answer:

Given: Vertices of the triangle are A (0, 4), B(0, 0) and C (3, 0)
screenshot-2021-01-23-224757
$\\ \mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}} \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,4) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,0) \\ \mathrm{AB}=\sqrt{(0-0)^{2}+(0-4)^{2}}=\sqrt{(4)^{2}}=4 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(3,0) \\ \mathrm{BC}=\sqrt{(3-0)^{2}+(0-0)^{2}}=\sqrt{(3)^{2}}=3 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(3,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,4) \\ \mathrm{CA}=\sqrt{(0-3)^{2}+(4-0)^{2}}=\sqrt{9+16}$
$\\ \mathrm{CA}=\sqrt{25}=5$
Perimeter of $\Delta$ ABC = AB + BC + AC = 4 + 3 + 5 = 12
Hence perimeter of $\Delta$ ABC is 12

Question:7

The area of a triangle with vertices A (3, 0), B (7, 0) and C (8, 4) is
(A) 14
(B) 28
(C) 8
(D) 6

Answer:

The given vertices are A(3, 0), B(7, 0) and C(8, 4)
$\\$ Area of $\Delta ABC = \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\ = \frac{1}{2} [3(0 - 4) + 7(4 - 0) + 8(0 - 0)]\\ = \frac{1}{2} [-12 + 28 + 0]\\ = \frac{1}{2} [16] = 8\\ $Area of $\Delta ABC = 8\\$

Question:8

The points (–4, 0), (4, 0), (0, 3) are the vertices of a
(A) right triangle
(B) isosceles triangle
(C) equilateral triangle
(D) scalene triangle

Answer:

The given points are (–4, 0), (4, 0), (0, 3)
screenshot-2021-01-23-225459
$\\ \mathrm{AB}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}} \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(-4,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(4,0) \\ \mathrm{AB}=\sqrt{\left(4-(-4)^{2}+(0-0)^{2}\right.} \\ \mathrm{AB}=\sqrt{(8)^{2}} \\ \mathrm{AB}=8 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(4,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,3) \\ \mathrm{BC}=\sqrt{(0-4)^{2}+(3-0)^{2}} \\ \mathrm{BC}=\sqrt{16+9}=\sqrt{25}=5 \\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0,3) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(-4,0) \\ \mathrm{AC}=\sqrt{(0+4)^{2}+(3-0)^{2}} \\ \mathrm{AC}=\sqrt{16+9}=\sqrt{25}=5$
In this triangle BC = AC.
Hence it is an isosceles triangle.

Question:9

The point which divides the line segment joining the points (7, –6) and (3, 4) in ratio 1: 2 internally lies in the
(A) I quadrant
(B) II quadrant
(C) III quadrant
(D) IV quadrant

Answer:

Here the points are A(7, -6), B(3, 4) and the ratio is 1: 2.
$(x\textsubscript{1}, y\textsubscript{1}) = (7, -6) (x\textsubscript{2}, y\textsubscript{2}) = (3, 4)\\$
Let point which divides the line is (x, y)
$\\ (\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{m}_{1} \mathrm{x}_{2}+\mathrm{m}_{2} \mathrm{x}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}, \frac{\mathrm{m}_{1} \mathrm{y}_{2}+\mathrm{m}_{2} \mathrm{y}_{1}}{\mathrm{~m}_{1}+\mathrm{m}_{2}}\right) \\ (\mathrm{x}, \mathrm{y})=\left(\frac{1 \times 3+2 \times 7}{1+2}, \frac{14+2 \times-6}{1+2}\right) \\ (\mathrm{x}, \mathrm{y})=\left(\frac{3+14}{3}, \frac{4-12}{3}\right) \\ (\mathrm{x}, \mathrm{y})=\left(\frac{17}{3}, \frac{-8}{3}\right)$
Here x is positive and y is negative.
Hence the point lies in IV quadrant.

Question:10

The point which lies on the perpendicular bisector of the line segment joining the points A (–2, –5) and B (2, 5) is
(A) (0, 0)
(B) (0, 2)
(C) (2, 0)
(D) (–2, 0)

Answer:

The point which lies on the perpendicular bisector is the co-ordinate of mid-point on the line joining the points A(–2, –5), B(2, 5).
Let this point is (x, y)
$\\ \left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(-2,-5) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(2,5) \\ (\mathrm{x}, \mathrm{y})=\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}, \frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right) \\ (\mathrm{x}, \mathrm{y})=\left(\frac{-2+2}{2}, \frac{-5+5}{2}\right) \\ (\mathrm{x}, \mathrm{y})=(0,0)$
Hence, the required point is (0, 0)

Question:11

The fourth vertex D of a parallelogram ABCD whose three vertices areA (–2, 3), B (6, 7) and C (8, 3) is
(A) (0, 1)
(B) (0, –1)
(C) (–1, 0)
(D) (- 2 , 0)

Answer:

Let D(x, y)
A(–2, 3), B(6, 7), C(8, 3) (given)
We know that in parallelogram diagonals are equal
mid-point of AC = mid-point of BD
$\\ \left(\frac{-2+8}{2}, \frac{3+3}{2}\right)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ \left(\frac{6}{2}, \frac{6}{2}\right)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ (3,3)=\left(\frac{6+x}{2}, \frac{7+y}{2}\right) \\ \frac{6+x}{2}=3 \quad \frac{7+y}{2}=3 \\ 6+x=6 \quad 7+y=6 \\ x=0 \quad y=-1 \\ \text { Hence, } D=(0,-1)$
option B is correct.

Question:12

If the point P (2, 1) lies on the line segment joining points A (4, 2) and B (8, 4),then
$\\(A) \mathrm{AP}=\frac{1}{3} \mathrm{AB} \\(B) \mathrm{AP}=1 / 3 \mathrm{~PB} \\(C) \mathrm{PB}=\frac{1}{3} \mathrm{AB} \\(D) \mathrm{AP}=\frac{1}{2} \mathrm{AB}$

Answer:

Let P divide A B in ratio k: 1
$\\\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(4,2) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(8,4)\\P=\left(\frac{k(8)+1(4)}{k+1}, \frac{k(4)+1(2)}{k+1}\right)$
By using section formula
Here,
$\mathrm{P}=(2,1)$
$(2,1)=\left(\frac{8 k+4}{k+1}, \frac{4 k+12}{k+1}\right)$
Compare x co-ordinate
$\\\frac{8 k+4}{k+1}=2 \\8 k+4=2 k+2 \\6 k=-2 \\k=\frac{-2}{6}=\frac{-1}{3} \\Here, k is negative. Hence P divides A B in ratio 1: 3 externally. \\\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{3}$ \\$\mathrm{AP}=\frac{1}{3} \mathrm{~PB}$

Question:13

If P(a/3 , 4) is the mid-point of the line segment joining the points Q (– 6, 5) and R (– 2, 3), then the value of a is
(A) – 4
(B) – 12
(C) 12
(D) – 6

Answer:

$\\ \mathrm{Q}(-6,5) \quad \mathrm{R}(-2,3)\\\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)$
By using mid-point formula
$\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)$
$P=\left(\frac{-6-2}{2}, \frac{5+3}{2}\right)$
$\\ \left(\frac{a}{3}, 4\right)=\left(\frac{-8}{2}, \frac{8}{2}\right) \\\left(\frac{a}{3}, 4\right)=(-4,4)$
Compare x co-ordinate
$\\\frac{a}{3}=-4$ \\$a=-4 \times 3$ \\$a=-12$
Hence, option B is correct.

Question:14

The perpendicular bisector of the line segment joining the points A (1, 5) andB (4, 6) cuts the y-axis at
(A) (0, 13)
(B) (0, –13)
(C) (0, 12)
(D) (13, 0)

Answer:

$\\At y-axis, x = 0 \therefore point P is (0, y).\\ A (1, 5) and B (4, 6)\\ \therefore AP = BP\\$
Squaring both sides we get
$\\AP\textsuperscript{2} = BP\textsuperscript{2}\\ (x\textsubscript{1} - 0)\textsuperscript{2} + (y\textsubscript{1} + y)\textsuperscript{2} = (x\textsubscript{2}- 0)\textsuperscript{2} + (x\textsubscript{2} - y)\textsuperscript{2 }\\ \textsuperscript{ }(Because distance formula =\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} )\\ (1 - 0)\textsuperscript{2} + (5 - y)\textsuperscript{2} = (4 - 0)\textsuperscript{2} + (6 - y)\textsuperscript{2}\\ (1)\textsuperscript{2} + (5)\textsuperscript{2} + (y)\textsuperscript{2} - 2 \times 5 \times y = (4)\textsuperscript{2} + (6)\textsuperscript{2} + (y)\textsuperscript{2} - 2 \times 6 \times y\\ \{ using (a - b)\textsuperscript{2} = a\textsuperscript{2} + b\textsuperscript{2} - 2ab \} \\ 1 + 25 + y\textsuperscript{2} - 10y = 16 + 36 + y\textsuperscript{2} - 12y\\ 26 - 10y + 12y = 52\\ 2y = 52 - 26\\ 2y = 26\\ y = 13\\$
Hence, point P is (0, 13)
Therefore, option A is correct.

Question:15

The coordinates of the point whichis equidistant from the three verticesof the ΔAOB as shown in the figure is
screenshot-2021-01-23-232200
$(A) (x, y)$
$(B) (y, x)$
$(C) \frac{x}{2},\frac{y}{2}$
$(D)\frac{y}{2},\frac{x}{2}$

Answer:

screenshot-2021-01-23-232200
In the given figure, it is clear that DAOB is a right-angle triangle.
And in a right-angle triangle, the mid-point of the hypotenuse is equidistant from the three vertices. Thus, co-ordinates must be mid-point of AB
$\\ A (0, 2y), B(2x, 0)\\ (x\textsubscript{1}, y\textsubscript{1}) = (0, 2y) ,(x\textsubscript{2}, y\textsubscript{2}) = (2x, 0)\\$
Now find mid-point of AB using mid-point formula
$\left( \frac{{{x}_{1}}+{{x}_{2}}}{2},\frac{{{y}_{1}}+{{y}_{2}}}{2} \right) \\$
$\left( \frac{0+2x}{2},\frac{{2y}+{0}}{2} \right)= \left ( x,y \right ) \\$
Hence, option A is correct.

Question:16

A circle drawn with origin as the Centre passes through(13/2, 0) . The point which does not lie in the interior of the circle is :
$\\(A) \frac{-3}{4}, 1 \\(B) 2, \frac{7}{3} \\(C) 5, \frac{-1}{2} \\(D) \left(-6, \frac{5}{2}\right)$

Answer:

A) Distance of the point (-3/4, 1) from (0,0) is
$=\sqrt{\left(\frac{-3}{4}-0\right)^{2}+(1-0)^{2}}=\sqrt{\frac{9}{16}+1}=\sqrt{\frac{25}{16}}=\frac{5}{4}=1.25 \mathrm{units}$
The distance 1.25<6.5 . so the point (-3/4, 1) is lie
B) Distance point (2, 7/3) from (0,0) is
$=\sqrt{\left(\frac{7}{3}-0\right)^{2}+(2-0)^{2}}=\sqrt{\frac{49}{9}+4}=\sqrt{\frac{85}{9}}=\frac{9.2195}{3}=3.0731<6.25$
So the point is lie
C) Distance point (5,-1/2) from (0,0) is
$\\=\sqrt{\left(-\frac{1}{2}-0\right)^{2}+(5-0)^{2}}=\sqrt{\frac{1}{4}+25}=\sqrt{\frac{101}{4}}=\frac{10.0498}{2}=5.0249<6.25$
So the point is lie in the interior of the circle
D)
The circle passes through (13/2, 0) having a centre (0,0)
$\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(13 / 2,0) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(0,0)$
$\text{Apply distance formula}=\sqrt{\left(\mathrm{x}_{2}-\mathrm{x}_{1}\right)^{2}+\left(\mathrm{y}_{2}-\mathrm{y}_{1}\right)^{2}}$
$\text{Radius}=\sqrt{(6.5-0)^{2}+(0-0)^{2}}$
$\\=\sqrt{(6.5)^{2}}=6.5$ \\$=5^{2}+0.5^{2}-6.5^{2} \quad (Negative) \\\left(-6, \frac{5}{2}\right)=(-6)^{2}+(2.5)^{2}-6.5$ \\$=6^{2}+2.5^{2}-6.5 \quad(positive)$
Hence, point D is the correct answers.

Question:17

A line intersects the y-axis and x-axis at the points P and Q, respectively. If(2, –5) is the mid-point of PQ, then the coordinates of P and Q are respectively
(A) (0, – 5) and (2, 0)
(B) (0, 10) and (– 4, 0)
(C) (0, 4) and (– 10, 0)
(D) (0, – 10) and (4, 0)

Answer:

Point
$P=(0, y)\{ at\ y -axis x=0\}$
$\\\mathrm{Q}=(\mathrm{x}, 0) \quad\{ at \mathrm{x} axis \mathrm{y}=0\} \\\left(\mathrm{x}_{1}, \mathrm{y}_{1}\right)=(0, \mathrm{y}) \quad\left(\mathrm{x}_{2}, \mathrm{y}_{2}\right)=(\mathrm{x}, 0)$
$Mid-point =(2,-5)$
$\\\left[\left(\frac{\mathrm{x}_{1}+\mathrm{x}_{2}}{2}\right),\left(\frac{\mathrm{y}_{1}+\mathrm{y}_{2}}{2}\right)\right]=(2,-5)\\\left(\frac{0+x}{2}, \frac{y+0}{2}\right)=2,-5 \\\frac{x}{2}=2 \quad \frac{y}{2}=-5$
$\\ x=4 \: \: \: \: \: \: \: \: y=-10 \\ \text { Point } P(0,-10), \ and\ Q(4,0)$
Hence, option D is correct.

Question:18

The area of a triangle with vertices (a, b + c), (b, c + a) and (c, a + b) is
(A) (a+b+c)²
(B) 0
(C)( a + b + c)
(D) abc

Answer:

Vertices are (a, b + c), (b, c + a), (c, a + b)
Here $A(x\textsubscript{1}, y\textsubscript{1}) = (a , b + c)\\$
$\\B\left( {{x}_{2}},{{y}_{2}} \right)= (b , c + a)\\ C\left( {{x}_{3}},{{y}_{3}} \right)= (c , a + b)\\$
We know that
$$Area of triangle $= \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\$
$\\ = \frac{1}{2} [a(c + a - a - b) + b(a + b - b - c) + c(b + c - c -a)]\\ = \frac{1}{2} [a(c - b) + b(a - c) + c(b - a)]\\ = \frac{1}{2} [ac - ab + ab - bc + bc - ac] = 0\\$
Hence, option B is correct.

Question:19

If the distance between the points (4, p) and (1, 0) is 5, then the value of p is
(A) 4 only
(B) ± 4
(C) – 4 only
(D) 0

Answer:

Points are A(4, p) and B(1, 0)
Distance
$\ (AB)\ = 5 (given )\\$
Here
$\\{{x}_{1}} =4\ \ {{x}_{2}} =1\\ {{y}_{1}} =p \: \: \: \: \: {{y}_{2}} =0\\$
Using distance formula
$\\=\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ AB =\sqrt{{{({{x}_{2}}-{{x}_{1}})}^{2}}+{{({{y}_{2}}-{{y}_{1}})}^{2}}} \\ Squaring both sides\\ (1 - 4)\textsuperscript{2} + (0 - p)\textsuperscript{2} = (5)\textsuperscript{2}\\ (-3)\textsuperscript{2} + (p)\textsuperscript{2} = 25\\ 9 + p\textsuperscript{2} = 25\\ p\textsuperscript{2} = 25 - 9\\ p\textsuperscript{2} = 16\\ p = \pm 4\\$
Hence, option B is correct.

Question:20

If the points A (1, 2), O (0, 0) and C (a, b) are collinear, then
(A) a = b
(B) a = 2b
(C) 2a = b
(D) a = –b

Answer:

If points A(1, 2), O(0, 0),C (a, b) are collinear then area of triangle
formed by these points must be zero.
$\\A(1, 2)=A(x\textsubscript{1} ,y\textsubscript{1})\\ O(0, 0)=O(x\textsubscript{2} ,y\textsubscript{2})\\ C (a, b)=C(x\textsubscript{3} ,y\textsubscript{3})\\ $area of triangle$= \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})]\\ \Rightarrow \frac{1}{2} [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})] = 0\\ \Rightarrow [x\textsubscript{1}(y\textsubscript{2} - y\textsubscript{3}) + x\textsubscript{2}(y\textsubscript{3} - y\textsubscript{1}) + x\textsubscript{3}(y\textsubscript{1} - y\textsubscript{2})] = 0\\ \Rightarrow [1(0 - b) + 0(b - 2) + a(2 - 0)] = 0\\ \Rightarrow [-b + 0 + 2a] = 0\\ \Rightarrow -b + 2a = 0\\ \Rightarrow 2a = b\\$
Hence, option C is correct.

Question:1

$\bigtriangleup$ ABC with vertices A(–2, 0), B(2, 0) and C(0, 2) is similar to $\bigtriangleup$ DEF with vertices D(–4, 0), E(4, 0) and F(0, 4).

Answer:

Answer [False]
Solution.
59066
$Distance\, Formula= \sqrt{\left ( {x_{2}-x_{1}} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
In $\bigtriangleup$ ABC,
$AB= \sqrt{\left ( 2+2 \right )^{2}+\left ( 0-0 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}}= 4$
$BC= \sqrt{\left ( 0-2 \right )^{2}+\left ( 2-0 \right )^{2}}= \sqrt{\left ( 4+4 \right )}= \sqrt{8}= 2\sqrt{2}$
$AC= \sqrt{\left ( 0+2 \right )^{2}+\left ( 2-0 \right )^{2}}= \sqrt{\left ( 4+4 \right )}= 2\sqrt{2}$
In $\bigtriangleup$ DEF,
$DE= \sqrt{\left ( 4+4 \right )^{2}+\left ( 0-0 \right )^{2}}= \sqrt{\left ( 8 \right )^{2}}= 8$
$EF= \sqrt{\left ( 0-4\right )^{2}+\left ( 4-0 \right )^{2}}= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}$
$DF= \sqrt{\left ( 0+4\right )^{2}+\left ( 4-0 \right )^{2}}= \sqrt{16+16}= 4\sqrt{2}$
$Here,\frac{AB}{DE}= \frac{BC}{EF}= \frac{AC}{DF}$
$\frac{4}{8}= \frac{2\sqrt{2}}{4\sqrt{2}}= \frac{1}{2}$ ,
Hence, $\bigtriangleup$ ABC is similar to $\bigtriangleup$ DEF i.e $\bigtriangleup ABC\sim \bigtriangleup DEF$

Question:2

Point P (– 4, 2) lies on the line segment joining the points A (– 4, 6) and B (– 4, – 6).

Answer:

Answer [true]
Solution.
The given points A (–4, 6) and B (–4, –6)
(x₁, y₁) = (-4, 6) (x₂, y₂) = (-4, -6) (x₃,y₃)=(-4,2)
Point p(-4,2) lie on line AB if the area of triangle ABP =0
$\frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{2}+ \right )+x_{3}\left ({y_{1}-{y_{2}}} \right ) \right ]= 0$
$\frac{1}{2}\left [ -4\left ( -6-2 \right )-4\left ( 2-6 \right ) -4\left ( 6+6 \right )\right ]= 0$
$\frac{1}{2}\left [ -4\left ( -8 \right )-4\left ( -4 \right ) -4\left ( 12 \right )\right ]= 0$
$\left [ +32+16-48 \right ]= 0$
so we can say that p(-4,2) must lie on line joining, AB

Question:3

The points (0, 5), (0, –9) and (3, 6) are collinear.

Answer:

Answer. [False]
The given points area (0, 5), (0, –9) and (3, 6)
If the point area collinear then the area of triangle is 0.
x₁ =0 x₂ =0 x₃ =3
y₁=5 y₂=9 y₃=6
$We\, know \, that\, area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 0\left ( -9-6 \right ) +0\left ( 6-5 \right )+3\left ( 5+9 \right )\right ]$
$= \frac{1}{2}\left [ 0+0+42 \right ]$
$= \frac{42}{2}=21$
Area of triangle = 21
Here the area of the triangle is not equal to zero.
Hence, the point is not collinear.

Question:4

Point P (0, 2) is the point of intersection of y–axis and perpendicular bisector of line segment joining the points A (–1, 1) and B (3, 3).

Answer:

Solution. If the point P is a perpendicular bisector of the line joining the point A(–1, 1) and B(3, 3) then it must be mid-point of AB.
$Mid -point \, of AB = \left ( \frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2} \right )$
(x₁, y₁) = (-1, 1) (x₂, y₂) = (3, 3)
$= \left ( \frac{-1+3}{2},\frac{1+3}{2} \right )= \left ( \frac{2}{2} ,\frac{4}{2}\right )= \left ( 1,2 \right )$
Which is not point P.
Hence, the given statement is false.

Question:5

Points A (3, 1), B(12, –2) and C(0, 2) cannot be the vertices of a triangle.

Answer:

Answer. [True]
Solution. If they are not the vertices of a triangle then
Area of $\triangle$ ABC = 0
x₁ =3 x₂ =12 x₃ =0
y₁=1 y₂=-2 y₃=2
Let us find the area of $\triangle$ ABC
$Area \, of \triangle ABC = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 3\left ( -2-2 \right )+12\left ( 2-1 \right ) +0\left ( 1+2 \right )\right ]$
$= \frac{1}{2}\left [ 3\left ( -4 \right ) +12\left ( 1 \right )+0\right ]$
$= \frac{1}{2}\left [ -12+12 \right ]$
$= \frac{0}{2}$
= 0
Area of $\triangle$ ABC = 0
Hence, they are collinear or not the vertices of a triangle.

Question:6

Points A (4, 3), B (6, 4), C (5, –6) and D (–3, 5) are the vertices of a parallelogram.

Answer:

Answer. [False]
Solution.
The given points area A(4, 3), B(6, 4), C(5, –6), D(–3, 5)
$Distance \: between\: AB = \sqrt{\left ( 6-4 \right )^{2}+\left ( 4-3 \right )^{2}}$
$AB = \sqrt{\left ( 2 \right )^{2}+\left ( 1 \right )^{2}}= \sqrt{5}$
$Distance \: between\: BC = \sqrt{\left ( 5-6 \right )^{2}+\left ( -6-4 \right )^{2}}$
$BC= \sqrt{1+100}= \sqrt{101}$
$Distance \: between\: CD = \sqrt{\left ( -3-5 \right )^{2}+\left ( 5+6 \right )^{2}}$
$CD = \sqrt{64+121}= \sqrt{185}$
$Distance \: between\: DA = \sqrt{\left ( 4+3 \right )^{2}+\left ( 3-5 \right )^{2}}$
$DA =\sqrt{49+4}= \sqrt{53}$
Here, opposite sides are not equal i.e.
$AB\neq CD,BC\neq DA$
Hence, it is not a parallelogram

Question:7

A circle has its Centre at the origin and a point P (5, 0) lies on it. The point Q (6, 8) lies outside the circle.

Answer:

Answer. [True]
Solution. The centre of the circle is O (0, 0).
If point P(5, 0) lies on the circle then the distance between O(0, 0) and P(5, 0) is the radius of the circle
$OP= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$OP= \sqrt{\left ( 5-0 \right )^{2}+\left ( 0-0 \right )^{2}}$
$OP= \sqrt{\left ( 5 \right )^{2}}= 5$
OP = 5
Radius of circle = 5
If point Q (6, 8) is outside the circle then the distance between O(0, 0) and Q(6, 8) is grater then the radius of the circle
(x₁, y₁) = (0, 0) (x₂, y₂) = (6, 8)
$OQ= \sqrt{\left ( 6-0 \right )^{2}+\left ( 8-0 \right )^{2}}$
$OQ= \sqrt{36+64}$
$= \sqrt{100}=10$
Here point OQ is greater than the radius of circle
Hence, point Q(6, 8) lies outside the circle

Question:8

The point A (2, 7) lies on the perpendicular bisector of line segment joining the points P (6, 5) and Q (0, – 4).

Answer:

Answer. [False]
Solution. If point A(2, 7) is bisector then it must be mid-point of the line joining the points P(6, 5) and Q(0, –4)
$Mid-point \, of \, PQ = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$
(x₁, y₁) = (6, 5) (x₂, y₂) = (0, -4)
$= \left ( \frac{6+0}{2},\frac{5-4}{2} \right )= \left ( 3,\frac{1}{2} \right )$
Hence, A not lies on bisector.

Question:9

Point P (5, –3) is one of the two points of trisection of the line segment joining the points A (7, – 2) and B (1, – 5).

Answer:

Answer. [True]
Solution.
FireShot%20Capture%20093%20-%20Careers360%20CMS%20-%20cms-article%20-%20learn Let the two point of trisection are C, D $Point \, \, C= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} ,\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}}\right )$
(x₁, y₁) = (7, -2) (x₂, y₂) = (1, -5)
m₁ = 1 , m₂ = 2 (Because C divide AB in ratio 1:2 )
$C=\left ( \frac{1\times 1+2\times 7}{1+2} ,\frac{1\times -5+2\times -2}{1+2}\right )$
$C= \left ( \frac{1+14}{3} ,\frac{-5-4}{3}\right )$
$C= \left ( 5,- 3 \right )$
$Point \, \, D= \left ( \frac{n_{1}x_{2}+n_{2}x_{1}}{n_{1}+n_{2}} ,\frac{n_{1}y_{2}+n_{2}y_{1}}{n_{1}+n_{2}}\right )$
n₁ = 2 , n₂ = 1 (Because D divide AB in ratio 2:1)
$D=\left ( \frac{2\times 1+1\times 7}{2+1} ,\frac{2\times -5+1\times -2}{2+1}\right )$
$D=\left ( \frac{2+7}{3},\frac{-10-2}{3} \right )$
$D=\left ( 3,-4 \right )$
Hence, the given statement is true.

Question:10

Points A (–6, 10), B (–4, 6) and C (3, –8) are collinear such that AB =2/9 AC.

Answer:

Answer. [Ture]
Solution. If the points A (–6, 10), B(–4, 6) and C(3, –8) are collinear then area of $\triangle$ ABC = 0
$Area \, of \, \triangle ABC= \frac{1}{2}\left [ -6\left ( 6+8 \right ) +\left ( -4 \right )\left ( -8-10 \right )+3\left ( 10-6 \right )\right ]$
$= \frac{1}{2}\left [ -84+72+12 \right ]$
Area of $\triangle$ ABC = 0
Hence, A, B and C are collinear
$Distance \, between \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$AB= \sqrt{\left ( -4+6 \right )^{2}+\left ( 6-10 \right )^{2}}$
$AB=\sqrt{4+16}= \sqrt{20}= 2\sqrt{5}$
$Distance \, between \, AC= \sqrt{\left ( 3+6 \right )^{2}+\left ( -8-10 \right )^{2}}$
$= \sqrt{81+324}$
$= \sqrt{405}$
$AC= 9\sqrt{5}$
Hence,
$\frac{AB}{AC}= \frac{2\sqrt{5}}{9\sqrt{5}}$
$AB= \frac{2}{9}AC$

Question:11

The point P (–2, 4) lies on a circle of radius 6 and centre C (3, 5).

Answer:

Answer. [False]
Solution.
The radius of the circle is 6 and centre C(3, 5)
If point P(–2, 4) lies on the circle then the distance between the centre and point P is equal to the radius of the circle.
$Distance\, between\, PC = \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2} -y_{1}\right )^{2}}$
(x₁, y₁) = (-2, 4) (x₂, y₂) = (3, 5)
$PC= \sqrt{\left ( 3+2 \right )^{2}+\left ( 5-4 \right )^{2}}$
$PC=\sqrt{25+1}= \sqrt{26}$
$PC\neq radius\left ( 6 \right )$
Hence, point P(–2, 4) not lies on the circle with centre C(3, 5).

Question:12

The points A (–1, –2), B (4, 3), C (2, 5) and D (–3, 0) in that order form a rectangle.

Answer:

Answer. [True]
Solution. The given points are A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0)
$Length \, of \, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$AB= \sqrt{\left ( 4+1 \right )^{2}+\left ( 3+2 \right )^{2}}$
$AB= \sqrt{50}= 5\sqrt{2}$
$Length \, of \, BC= \sqrt{\left ( 2-4 \right )^{2}+\left ( 5-3 \right )^{2}}$
$BC=\sqrt{4+4}= \sqrt{8}= 2\sqrt{2}$
$Length \, of \, CD= \sqrt{\left ( -3-2 \right )^{2}+\left ( 0-5 \right )^{2}}$
$CD= \sqrt{25+25}= \sqrt{50}= 5\sqrt{2}$
$Length \, of \, DA= \sqrt{\left ( -1+3 \right )^{2}+\left ( -2-0 \right )^{2}}$
$DA= \sqrt{4+4}= \sqrt{8}= 2\sqrt{2}$
AB = CD , BC = DA
$Length \, of \, AC= \sqrt{\left ( 2+1 \right )^{2}+\left ( 5+2 \right )^{2}}$
$AC= \sqrt{9+49}= \sqrt{58}$
$Length \, of \, BD= \sqrt{\left ( -3-4 \right )^{2}+\left ( 0-3 \right )^{2}}$
$BD= \sqrt{49+9}= \sqrt{58}$
AC = BD (Diagonals)
Hence, ABCD is a rectangle because
AB = CD, BC = DA, AC = BD

Question:1

Name the type of triangle formed by the points A (–5, 6), B (–4, –2) and C (7, 5).

Answer:

Solution.
Given vertices are A(–5, 6), B(–4, –2), C(7, 5)
$Distance \, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$Distance \, of\, AB= \sqrt{\left ( -4-\left ( -5 \right ) \right )^{2}+\left ( -2-6 \right )^{2}}$
$= \sqrt{\left ( 1 \right )^{2}+\left ( -8 \right )^{2}}$
$= \sqrt{1+64}$
$= \sqrt{65}$
$Distance \, of\, BC= \sqrt{\left ( 7-\left ( -4 \right ) \right )^{2}+\left ( 5-\left ( -2 \right ) \right )^{2}}$
$= \sqrt{\left ( 11 \right )^{2}+\left ( 7 \right )^{2}}$
$= \sqrt{121+49}$
$= \sqrt{170}$
$Distance \, of\, AC= \sqrt{\left ( 7-\left ( -5 \right ) \right )^{2}+\left ( 5-6 \right )^{2}}$
$= \sqrt{\left ( 12 \right )^{2}+\left ( -1 \right )^{2}}$
$=\sqrt{144+1}$
$=\sqrt{145}$
$AB\neq BC\neq AC$
Therefore, ABC is a scalene triangle.

Question:2

Find the points on the x-axis which are at a distance of $2\sqrt{5}$ from the point(7, –4). How many such points are there

Answer:

Solution. Let point on x-axis is (x, 0) { $\because$ y is zero on x-axis}
Given point (7, –4)
$Distance = 2\sqrt{5}$
$2\sqrt{5}= \sqrt{\left ( 7-x \right )^{2}+\left ( -4-0 \right )^{2}}$
Squaring both sides
$\left ( 2\sqrt{5} \right )^{2}= \left ( 7-x \right )^{2}+\left ( -4 \right )^{2}$
$20= \left ( 7 \right )^{2}+\left ( x \right )^{2}-2\times 7\times x+16\left \{ \because \left ( a-b \right )^{2}= a^{2}+b^{2}-2ab \right \}$
$20=49+x^{2}-14x+16$
$x^{2}-14x+65-20= 0$
$x^{2}-14x+45= 0$
$x^{2}-9x-5x+45= 0$
$x\left ( x-9 \right )-5\left ( x-9 \right )= 0$
$\left ( x-9 \right )\left ( x-5 \right )= 0$
x = 9, x =5
Point are (9, 0) and (5, 0)
Hence, two points are there.

Question:3

What type of a quadrilateral do the points A(2, –2), B(7, 3), C(11, –1) and D(6,–6)taken in that order, form?

Answer:

Solution.
Let the points be A(2, –2), B(7, 3), C(11, –1), D(6, –6) of a quadrilateral ABCD
$AB= \sqrt{\left ( 7-2 \right )^{2}+\left ( 3+2 \right )^{2}}= \sqrt{5^{2}+5^{2}}= \sqrt{25+25}= \sqrt{50}$
$BC= \sqrt{\left ( 7-11 \right )^{2}+\left ( -1-3 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}+\left ( -4 \right )^{2}}= \sqrt{16+16}= \sqrt{32}$
$CD= \sqrt{\left ( 6-11 \right )^{2}+\left ( -6+1 \right )^{2}}= \sqrt{\left ( -5 \right )^{2}+\left ( -5 \right )^{2}}= \sqrt{25+25}= \sqrt{50}$
$DA= \sqrt{\left ( 6-2 \right )^{2}+\left ( -6+2 \right )^{2}}= \sqrt{\left ( 4 \right )^{2}+\left ( -4 \right )^{2}}= \sqrt{16+16}= \sqrt{32}$
$AC= \sqrt{\left ( 11-2 \right )^{2}+\left ( -1+2 \right )^{2}}= \sqrt{\left ( 9 \right )^{2}+\left ( 1 \right )^{2}}= \sqrt{81+1}= \sqrt{82}$
$BD= \sqrt{\left ( 6-7 \right )^{2}+\left ( -6-3 \right )^{2}}= \sqrt{\left ( 1 \right )^{2}+\left ( -9 \right )^{2}}= \sqrt{1+81}= \sqrt{82}$
As, AB = CD and BC = DA and AC = BD
Hence, the quadrilateral is a rectangle.

Question:4

Find the value of a, if the distance between the points A (–3, –14) and B (a, –5) is 9 units.

Answer:

Solution. Here, points are A(–3, –14) and B(a, –5)
Distance = 9
(x₁, y₁) = (-3, -14) (x₂, y₂) = (a, -5)
$9= \sqrt{\left ( a+3 \right )^{2}+\left ( -5+14 \right )^{2}}$
Squaring both sides we get
$\left ( 9 \right )^{2}= \left ( a+3 \right )^{2}+\left ( 9 \right )^{2}$
$\left ( a+3 \right )^{2}= 0$
$\Rightarrow a+3= 0$
a = -3
Value of a is -3

Question:5

Find a point which is equidistant from the points A (–5, 4) and B (–1, 6)? How many such points are there?

Answer:

Solution. Let P(x, y) is a point which is equidistant from point A(–5, 4) and B(–1, 6) i.e. PA = PB
Squaring both sides we get
$PA^{2}= PB^{2}$
$\left ( -5-x \right )^{2}+\left ( 4-y \right )^{2}= \left ( -1-x \right )^{2}+\left ( 6-y \right )^{2}$
25 + x² + 10x + 16 + y² – 8y = 1 + x² + 2x + 36 + y² – 12y
{Using : (a + b)² = a² + b² + 2ab; (a – b)² = a² + b² – 2ab}
25 + 10x + 16 – 8y = 1 + 2x + 36 – 12y
10x – 8y + 41 – 2x + 12y – 37 = 0
8x + 4y + 4 = 0
Dividing by 4 se get
2x + y + 1 = 0 ……..(1)
$Mid-point\, of\, AB= \left ( \frac{-5-1}{2},\frac{4+6}{2} \right )$
= (-3,5)
$\because Mid-point= \left \{ \left ( \frac{x_{1}+x_{2}}{2} \right ) ,\left ( \frac{y_{1}+y_{2}}{2} \right )\right \}$
Put point (–3, 5) in eqn. (1)
2(–3) + 5 + 1 = 0
– 6 + 6 = 0
0 = 0
Mid-point of AB satisfy equation (1)
Hence, infinite numbers of points are there.

Question:6

Find the coordinates of the point Q on the x-axis which lies on the perpendicular bisector of the line segment joining the points A (–5, –2) and B(4, –2). Name the type of triangle formed by the points Q, A and B.

Answer:

Solution.
$\text{Use distance formula}= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
Let Q(x, 0) { $\because$ on x-axis y coordinate is zero}
Q lies on the perpendicular bisector of the line AB i.e.
AQ = BQ
Squaring both sides
AQ² = BQ²
(x – 5)² + (0 + 2)² = (x – 4)² + (0 + 2)²
x² + 25 – 10x + 4 = x² + 16 – 8x + 4
29 – 10x = 20 – 8x
29 – 20 = –8x + 10x
9 = 2x
$\frac{9}{2}= x$
x = 4.5
$\therefore$ The co-ordinate of Q is (4.5, 0)
$\because$ AQ = BQ and Q lies on the perpendicular bisector of the line AB
$\therefore$ $\triangle$ ABQ is an isosceles triangle.

Question:7

Find the value of m if the points (5, 1), (–2, –3) and (8, 2m ) are collinear.

Answer:

Solution.
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
Given points are A(5, 1), B(–2, –3), C(8, 2m)
If points are collinear then the area of triangle = 0
$= \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]= 0$
$\Rightarrow \left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]= 0$
$\Rightarrow$ 5(–3 – 2m) – 2(2m – 1) + 8(1 + 3) = 0
$\Rightarrow$ – 15 – 10m – 4m + 2 + 32 = 0
$\Rightarrow$ – 14m + 19 = 0
$\Rightarrow$ 14m = 19
$\Rightarrow m= \frac{19}{14}$

Question:8

If the point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the values of y. Also find distance PQ.

Answer:

Solution. Given: Point A(2, –4) is equidistant from P(3, 8) and Q(–10, y)
AP = AQ
Square both sides
AP² = AQ²
(3 – 2)2 + (8 + 4)2 = (– 10 – 2)2 + (y + 4)2 ( using distance formula)
(1)² + (12)² = (12)² + (y)² + (4)² + 2 × y × 4 { $\because$ (a+ b)² = a² + b² + 2ab}
1 + 144 = 144 + y² + 16 + 8y
y² + 8y + 15 = 0
y² + 3y + 5y + 15 = 0
y.(y + 3) + 5.(y + 3) = 0
(y + 5) (y + 3) = 0
y = –5, y = –3
Case-I when y = –5
$PQ= \sqrt{\left ( -10-3 \right )^{2}+\left ( -5-8 \right )^{2}}= \sqrt{169+169}= 13\sqrt{2}\, units$
Case-I when y = –3
$PQ= \sqrt{\left ( -10-3 \right )^{2}+\left ( -3-8 \right )^{2}}= \sqrt{169+121}= \sqrt{290} \, units$

Question:9

Find the area of the triangle whose vertices are (–8, 4), (–6, 6) and (–3, 9).

Answer:

Solution.
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
Given vertices are (–8, 4), (–6, 6), (–3, 9)
x₁ =-8 x₂ =-6 x₃ = -3
y₁=4 y₂= 6 y₃= 9
We know that area of triangle is
$= \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ \left ( -8 \right )\left ( 6-9 \right )+\left ( -6 \right )\left ( 9-4 \right )+\left ( -3 \right ) \left ( 4-6 \right )\right ]$
$= \frac{1}{2} \left [ 24+30+6 \right ]$
$= \frac{1}{2} \left [ 60 \right ]$
$= 30 sq.units$

Question:10

In what ratio does the x–axis divide the line segment joining the points (– 4, – 6) and (–1, 7)? Find the coordinates of the point of division.

Answer:

Solution.
59125
$\left ( X,Y \right )= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Let the point on x-axis (x, 0) when divides the given points (–4, –6) and (–1, 7) in the ration k : 1.
(x₁, y₁) =A (-4, -6) (x₂, y₂) = B(-1, 7)
m₁ = k, m₂ = 1
Using section formula, we have
$p\left ( x,0 \right )= \left [ \frac{-k-4}{k+1},\frac{7k-6}{k+1} \right ]$
By comparing the left-hand side and the right-hand side we get
$\frac{7k-6}{k+1}= 0$
7k-6 = 0
$k= \frac{6}{7}$
$x= \frac{-k-4}{k+1}= \frac{-\frac{6}{7}-4}{\frac{6}{7}+1}= \frac{-6-28}{6+7}= \frac{-34}{13}$
The required ratio is 6: 7.
$Co-ordinate \, of\, P\left ( \frac{-34}{13},0 \right )$

Question:11

Find the ratio in which the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$ divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$ and $B (2, -5)$ .

Answer:

Solution
59127
$P\left ( X,Y \right )= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Let the point $P\left ( \frac{3}{4},\frac{5}{12} \right )$ divides the line segment joining the points $A\left ( \frac{1}{2},\frac{3}{2} \right )$ and B(2, –5) in the ration k : 1.
(x₁, y₁) = (1/2, 3/2) (x₂, y₂) = (2, -5)
m₁ = k, m₂ = 1
Using section formula we have
$\left ( \frac{3}{4},\frac{5}{12} \right )= \left [ \frac{2k+1/2}{k+1} ,\frac{-5k+3/2}{k+1}\right ]$
$\left ( \frac{3}{4},\frac{5}{12} \right )= \left [ \frac{4k+1}{2k+1},\frac{-10k+3}{2k+2} \right ]$
$\frac{4k+1}{2k+1}= \frac{3}{4}$ $\frac{-10k+3}{2k+2}= \frac{5}{12}$
16k + 4 = 6k + 6 –120k + 36 = 10k + 10
16k – 6k = 6 – 4 –130k = –26
$k= \frac{2}{10}= \frac{1}{5}$
Therefore the required ratio is 1 : 5
i.e. 1 : 5

Question:12

If P(9a – 2, –b) divides line segment joining A(3a + 1, –3) and B(8a, 5)in the ratio 3: 1, find the values of a and b.

Answer:

Solution
59128
Point P(9a – 2, –b) divides line segment joining the points A(3a + 1, –3) and B(8a, 5) in ration 3 : 1.
(x₁, y₁) = (3a+1, -3) (x₂, y₂) = (8a, 5)
m₁ = 3, m₂ = 1
Using section formula we have
$\left ( 9a-2,-b \right )= \left [ \frac{3\left ( 8a \right )+1\left ( 3a+1 \right )}{3+1} ,\frac{3\left ( 5 \right )+1\left ( -3 \right )}{3+1}\right ]$
$\left ( 9a-2,-b \right )= \left [ \frac{24a+3a+1}{4},\frac{15-3}{4}\right ]$
$\left ( 9a-2,-b \right )= \left [ \frac{27a+1}{4},\frac{12}{4}\right ]$
Equate left-hand side and the right-hand side we get
$9a-2= \frac{27a+1}{4}$ $-b= \frac{12}{4}$
36a – 8 = 27a + 1 –b = 3
9a = 9 b = –3
$a= \frac{9}{9}$
a = 1

Question:13

If (a, b) is the mid-point of the line segment joining the points A (10, –6) and B (k, 4) and a – 2b = 18, find the value of k and the distance AB.

Answer:

Solution

$mid-point\, formula\, x= \frac{x_{1}+x_{2}}{2},y= \frac{y_{1}+y_{2}}{2}$
Point P (a, b) divide A(10, –6) and B(k, 4) in two equal parts.
$a= \frac{10+k}{2}$ $b= \frac{-6+4}{2}= \frac{-2}{2}= -1$
Given : a – 2b = 18
Put b = –1
a – 2(–1) = 18
a = 18 – 12
a = 16
$Now, a= \frac{10+k}{2}$
$16= \frac{10+k}{2}$
32 = 10 + k
32 – 10 = k
22 = k
$\therefore$ A (10, –6), B(22, 4)
$AB= \sqrt{\left ( 22-10 \right )^{2}+\left ( 4+6 \right )^{2}}$
$= \sqrt{\left ( 12 \right )^{2}+\left ( 10 \right )^{2}}$
$= \sqrt{144+100}$
$= \sqrt{244}$
$=2 \sqrt{61}$

Question:14

The centre of a circle is (2a, a – 7). Find the values of a if the circle passes through the point (11, –9) and has diameter $10\sqrt{2}$ units.

Answer:

Solution

$distance\: formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
Given points area A(2a, a –7) and B(11, –9)
$Diameter= 10\sqrt{2}$
$Radius= \frac{10\sqrt{2}}{2}= 5\sqrt{2}$
$Distance= 5\sqrt{2}$
(x₁, y₁) = (2a, a - 7) (x₂, y₂) = (11, -9)
$= \sqrt{\left ( 11-2a \right )^{2}+\left ( -9-a+7 \right )^{2}}= \left ( 5\sqrt{2} \right )$
Squaring both sides
$\left ( 11-2a \right )^{2}+\left ( -2-a \right )^{2}= \left ( 5\sqrt{2} \right )^{2}$
121 + 4a² – 44a + 4 + a² + 4a = 50
502 – 40a + 125 – 50 = 0
5a² – 40a + 75 = 0
Dividing by 5 we get
a² – 8a + 15 = 0
a² – 5a - 3a + 15 = 0
a(a – 5) – 3(a – 5) = 0
(a – 5) (a – 3) = 0
a = 5, 3

Question:15

The line segment joining the points A (3, 2) and B (5,1) is divided at the point P in the ratio 1:2 and it lies on the line 3x – 18y + k = 0. Find the value of k.

Answer:

Solution:

Here points A(3, 2) and B(5, 1) is divided at the point P in the ratio 1 : 2
(x₁, y₁) = (3, 2) (x₂, y₂) = (5, 1)
m₁ = 1, m₂ = 2
By section formula we have
$P\left ( x,y \right )= \left [ \frac{1\times 5+2\times 3}{1+2},\frac{1\times 1+2\times 2}{1+2} \right ]$
$= \left [ \frac{5+6}{3},\frac{1+4}{3} \right ]$
$= \left [ \frac{11}{3},\frac{5}{3} \right ]$
Line is 3x – 18y + k = 0 ......(1)
$Put\, point\, P\left ( \frac{11}{3},\frac{5}{3} \right ) in \left ( 1 \right )$
$Put\, x= \frac{11}{3},y= \frac{5}{3}$
$3\left ( \frac{11}{3} \right )-18\left ( \frac{5}{3} \right )+k= 0$
11 – 30 + k = 0
–19 + k = 0
k = 19

Question:16

$\text{If }$ $D\left ( \frac{-1}{2},\frac{5}{2} \right )$ $, E (7, 3) \ and$ $F\left ( \frac{7}{2},\frac{7}{2} \right )$ are the midpoints of sides of $\Delta ABC$ find the area of the $\Delta ABC$ .

Answer:

Solution.

D is mid-point of AB using mid-point formula we get:
$-\frac{1}{2}= \frac{x_{1}+x_{2}}{2}$ $\frac{5}{2}= \frac{y_{1}+y_{2}}{2}$
x₁ + x₂ = –1 ……(1) 5 = y₁ + y₂ .…(2)
E is mid-point of BC using mid-point formula we get:
$\frac{x_{2}+x_{3}}{2}= 7$ $\frac{y_{2}+y_{3}}{2}= 3$
x₂ + x₃ = 14 …..(3) y₂ + y₃ = 6 ….(4)
F is mid-point of AB using mid-point formula we get:
$\frac{x_{1}+x_{3}}{2}= \frac{7}{2}$ $\frac{y_{1}+y_{3}}{2}= \frac{7}{2}$
x₁ + x₃ = 7 ….(5) y₁ + y₃ = 7 …..(6)
Simplifying the above equations for values of x₁, y₁, x₂, y₂, x₃ and y₃
x₁ + x₂ = –1
x₃ + x₃ = 14 {using (1) and (3)}
– – –
x₁ – x3 = –15 ….(7)
Use (5) and (7) we get
x₁ + x₃ = 7
x₁– x₃ = –15
2x₁ = –8
x₁ = –4
put x₁ = 4 in (1) we get
–4 + x₂ = –1
X₂ = –1 + 4 = 3
Punt x₂ = 3 in (3) we get
3 + x₃ = 14
x₃ = 11
Using equation (2) and (4) we get
y₁ + y₂ = 5
y₃ + y₃ = 6
- - –
y₁ - y₃ = –1 ….(8)
Adding equation (6) and (8) we get
y₁ + y₃ = 7
y₁ – y₃ = –1
-------------------
2y₁ = 6
y₁ = 3
Put y₁ = 3 in eqn. (2)
5 – 3 = y₂
2 = y₂
Put y₂ = 2 in eqn. (4)
2 + y₃ = 6
y₃ = 6 – 2
y₃ = 4
Hence, x₁ = –4 y1 = 3
x₂ = 3 y₂ = 2
x₃ = 11 y₃ = 4
A = (x1, y₁) = (–4, 3), B = (x₂, y₂) = (3, 2), C = (x₃, y₃) = (11, 4)
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ \left ( -4 \right )\left ( 2-4 \right )+\left ( 3 \right )\left ( 4-3 \right ) +11\left ( 3-2 \right )\right ]$
$= \frac{1}{2}\left [ \left ( -4 \right )\left ( -2 \right )+3\left ( 1 \right )+11\left ( 1 \right ) \right ]$
$= \frac{1}{2}\left [ 8+3+11 \right ]$
$= \frac{1}{2}\left [ 22 \right ]$
= 11 sq. units

Question:17

The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of $\triangle$ ABC.

Answer:

Solution.

$distance\, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$\triangle$ ABC is a right angle triangle by using Pythagoras theorem we have
(AC)² = (BC)² + (AB)²
[(5 – 2)² + (5 - 9)²] = [(2 – a)²+ (9 – 5)²] + [(a – 5)² + (5 + 5)²]
(3)² + (–4)² = 4 + a² – 4a + (4)² + a² + 25 – 10a
9 + 16 = 4 + a² – 4a + 16 + a² + 25 – 10a
25 = 2a² – 14a + 45
2a² – 14a + 45 – 25 = 0
2a² – 14a + 20 = 0
Dividing by 2 we have
a² – 7a + 10 = 0
a² – 5a - 2a + 10 = 0
a(a – 5) – 2(a – 5) = 0
(a – 5) (a – 2) = 0
a = 5, a = 2
a = 5 is not possible because if a = 5 then point B and C coincide.
$\therefore$ a = 2
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 2\left ( 5-5 \right )+2\left ( 5-9 \right )+5\left ( 9-5 \right )\right ]$
$= \frac{1}{2}\left [ 0-8+20\right ]$
$= \frac{1}{2}\left [ 16\right ]$
= 8 sq. units.

Question:18

Find the coordinates of the point R on the line segment joining the points P(–1, 3) and Q(2, 5) such that $PR= \frac{3}{5}PQ$

Answer:

Solution

$section\, formula \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
According to question let R = (x, y) and PR = $\frac{3}{5}$ PQ
$\frac{PQ}{PR}= \frac{3}{5}$
R lies on PQ $\therefore$ PQ = PR +RQ
$\frac{PR+PQ}{PR}= \frac{5}{3}$
On dividing separately we get
$1+\frac{RQ}{PR}= \frac{5}{3}$
$\frac{RQ}{PR}= \frac{5}{3}-1= \frac{2}{3}$
$\Rightarrow PR:RQ= 3:2$
Hence, R divides PQ in ratio 3 : 2 using section formula we have
(x₁, y₁) = (-1, 3) (x₂, y₂) = (2, 5)
m₁ = 3, m₂ = 2
$R\left ( x,y \right )= \left ( \frac{3\left ( 2 \right )+2\left ( -1 \right )}{3+2} ,\frac{3\left ( 5 \right )+2\left ( 3 \right )}{3+2}\right )$
$R\left ( x,y \right )= \left ( \frac{6-2}{5},\frac{15+6}{5} \right )$
$R\left ( x,y \right )= \left ( \frac{4}{5},\frac{21}{5} \right )$
Here co- ordinates of R is $\left ( \frac{4}{5},\frac{21}{5} \right )$

Question:19

Find the values of k if the points A(k + 1, 2k), B(3k, 2k + 3) and C (5k –1, 5k)are collinear.

Answer:

Solution. If points A(k + 1, 2k), B(3k, 2k + 3) and C(5k – 1, 5k) are collinear then area of triangle is equal to zero
$\frac{1}{2}\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right )\right ]= 0$
$\left [ x_{1}\left ( y_{2}-y_{3} \right )+x_{2}\left ( y_{3}-y_{1} \right ) +x_{3}\left ( y_{1}-y_{2} \right )\right ]= 0$
[(k + 1) (2k + 3 – 5k) + 3k(5k – 2k) + (5k – 1)(2k – 2k – 3)] = 0
[(k + 1) (3 – 3k) + 3k(3k) + 5(k – 1) (–3)] = 0
[3k – 3k² + 3 – 3k + 9k² – 15k + 3] = 0
6k² – 15k + 6 = 0
6k² – 12k – 3k + 6 = 0
6k(k – 2) – 3(k – 2) = 0
(k – 2)(6k – 3 )
$k= 2,k= \frac{3}{6}$
$= \frac{1}{2}$
Hence, values of k are 2, 1/2

Question:20

Find the ratio in which the line 2x + 3y – 5 = 0 divides the line segment joining the points (8, –9) and (2, 1). Also find the coordinates of the point of division.

Answer:

Solution

$Section\, formula\left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Let point p(x, y) divides the line segment joining the points A(8, –9) and B(2, 1) in ratio k : 1.
(x₁, y₁) = (8, -9) (x₂, y₂) = (2, 1)
m₁ : m2 = k:1
using section formula we have
$P\left ( x,y \right )= \left [ \frac{2k+8}{k+1},\frac{k-9}{k+1} \right ]$
Given equation is 2x + 3y – 5 = 0 …(2)
Put values of x and y in eqn. (2)
$2\left [ \frac{2k+8}{k+1} \right ]+3\left [ \frac{k-9}{k+1} \right ]-5= 0$
2(2k + 3) + 3(k – 9) – 5(k + 1) = 0
4k + 16 + 3k – 27 – 5k – 5 = 0
2k – 16 = 0
k = 8
Hence, p divides the line in ration 8 : 1.
Put k = 5 in eqn. (1)
$\left ( x,y \right )= \left [ \frac{2\left ( 8 \right )}{8+1},\frac{8-9}{8+1} \right ]$
$= \left ( \frac{16+8}{9},\frac{-1}{9} \right )= \left ( \frac{24}{9} ,\frac{-1}{9}\right )= \left ( \frac{8}{3},\frac{-1}{9} \right )$
Required point is $P\left ( \frac{8}{3} ,\frac{-1}{9}\right )$

Question:1

If (– 4, 3) and (4, 3) are two vertices of an equilateral triangle, find the coordinates of the third vertex, given that the origin lies in the interior of the triangle.

Answer:

Solution

In equilateral triangle AB = BC = AC
$AC= \sqrt{\left ( x+4 \right )^{2}+\left ( y-3 \right )^{2}}$
$AC= \sqrt{x^{2}+16+8x+y^{2}+9-6y}$
$\left ( because \left ( a+b \right )^{2} = a^{2}+b^{2}+2ab\; \;\left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )$
$BC= \sqrt{\left ( x-4 \right )^{2}+\left ( y-3 \right )^{2}}$
$BC= \sqrt{x^{2}+16-8x+y^{2}+9-6y}$ $\left ( because \left ( a-b \right )^{2} = a^{2}+b^{2}-2ab \right )$
AC = BC
$\sqrt{x^{2}+16+8x+y^{2}+9-6y}= \sqrt{x^{2}+16-8x+y^{2}+9-6y}$
Squaring both side
8x + 8x = 0
16x = 0
x = 0
C = (0, y)
$Length\, of\, AB= \sqrt{\left ( 4+4 \right )^{2}+\left ( 3-3 \right )^{2}}$
$AB= \sqrt{\left ( 8 \right )^{2}}= 8$
AC = AB
$\sqrt{x^{2}+16+8x+y^{2}+9-6y}= 8$
Put x = 0, squaring both side
0 + 16 + 0 + y² + 9 – 6y = 64
y² – 6y + 25 – 64 = 0
y² – 6y – 39 = 0
$y= \frac{6\pm \sqrt{36+156}}{2}$
$y=\frac{6-8\sqrt{3}}{2}$ (For origin in the interior we take the only term with negative sign)
$y=3-4\sqrt{3}$

Question:2

A(6, 1), B(8, 2) and C(9, 4) are three vertices of a parallelogram ABCD. If E is the midpoint of DC, find the area of $\triangle$ ADE.

Answer:

Solution.
$Distance \, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
The given points A(6, 1), B(8, 2) and C(9, 4) let D(x, y)
59139
As the diagonal of a parallelogram bisect each other.
Here mid-point of AC = mid-point of BD
$\left ( \frac{6+9}{2},\frac{1+4}{2} \right )= \left ( \frac{8+x}{2},\frac{2+y}{2} \right )$
$\left ( \frac{15}{2},\frac{5}{2} \right )= \left ( \frac{8+x}{2},\frac{2+y}{2} \right )$
$\frac{15}{2}= \frac{8+x}{2}$ $\frac{2+y}{2}= \frac{5}{2}$
x = 7 y = 3
D (7, 3)
E is the mid-point of CD
Let E(x₀, y₀)
$\left ( x_{0},y_{0} \right )= \left ( \frac{7+9}{2}, \frac{3+4}{2} \right )$
$\left ( x_{0},y_{0} \right )= \left ( \frac{16}{2},\frac{2}{7} \right )$
$E= \left ( 8,\frac{7}{2} \right )$
$Area \, of \, \triangle ADE = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 6\left ( \frac{7}{2}-3 \right ) +8\left ( 3-1 \right )+7\left ( 1-\frac{7}{2} \right )\right ]$
$= \frac{1}{2}\left [ 6\times \frac{1}{2}+8\times 2+7\times \frac{-5}{2}\right ]$
$= \frac{1}{2}\left [ 3+16-\frac{35}{2}\right ]$
$= \frac{1}{2}\left [ \frac{6+32-35}{2}\right ]$
$Area\, of\, \triangle ADE= \frac{1}{2}\times \frac{3}{2}= \frac{3}{4}\, sq\cdot units$

Question:3

(i) The points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of $\triangle$ ABC.The median from A meets BC at D. Find the coordinates of the point D.
(ii) The points A(x₁, y₁), B(x₂, y₂) and C(x₃, y₃) are the vertices of $\triangle$ ABC.Find the coordinates of the point P on AD such that AP : PD = 2 : 1
(iii) The points A(x₁, y₁)B(x₂, y₂)and C(x₃, y₃)are the vertices of $\triangle$ ABC.Find the coordinates of points Q and R on medians BE and CF, respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1
(iv) The points A(x₁, y₁),B(x₂, y₂) and C(x₃, y₃) are the vertices of $\triangle$ ABC.What are the coordinates of the centroid of the triangle ABC?

Answer:

(i) Solution
591401
D is the mid-point of BC.
$Mid-point \,formula = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$
Coordinates of $D\left ( x,y \right )= \left ( \frac{x_{2}+x_{3}}{2} ,\frac{y_{2}+y_{3}}{2}\right )$ (By midpoint formula)
(ii) Solution
591402
$Section \, formula= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$D= \left ( \frac{x_{2}+x_{3}}{2},\frac{y_{2}+y_{3}}{2} \right )$ (By Midpoint formula)
$P= \left ( \frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}},\frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$P= \left ( \frac{2\times \frac{\left ( x_{2}+x_{3} \right )}{2}+1\times x_{1}}{2+1}, \frac{2\times \frac{\left ( y_{2}+y_{3} \right )}{2}+1\times x_{1}}{2+1} \right )$
$P= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
(iii) Solution

E is mid-point of AC
$E = \left ( \frac{x_{1}+x_{3}}{2} ,\frac{y_{1}+y_{3}}{2}\right )$
Q divides BF at 2 : 1
$Q= \left ( \frac{2\times \frac{\left ( x_{1}+x_{3} \right )}{2}+1\times x_{2}}{2+1}, \frac{2\times \frac{\left ( y_{1}+y_{3} \right )}{2}+1\times y_{2}}{2+1} \right )$
$Q= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
R divides CF at 2 : 1
$R= \left ( \frac{2\times \frac{\left ( x_{1}+x_{2} \right )}{2}+1\times x_{3}}{2+1}, \frac{2\times \frac{\left ( y_{1}+y_{2} \right )}{2}+1\times y_{3}}{2+1} \right )$
$R= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$
(iv) solution

$co-ordinate\, of\,centroid= \left ( \frac{Sum\,of\,all\,coordinates \, of\,all\,vertices\,}{3}, \frac{Sum\,of\,all\,coordinates \, of\,all\,vertices\,}{3}\right )$
Centroid: The centroid is the centre point of the triangle which is the intersection of the medians of a triangle.
$\triangle ABC\, coordinates\, of\, centroid= \left ( \frac{x_{1}+x_{2}+x_{3}}{3},\frac{y_{1}+y_{2}+y_{3}}{3} \right )$

Question:4

If the points A (1, –2), B (2, 3) C (a, 2) and D (– 4, –3) form a parallelogram, find the value of a and height of the parallelogram taking AB as base.

Answer:

Solution

We know that diagonals bisect each other
Hence, mid-point of AC = mid-point of BD
$\left ( \frac{1+a}{2},\frac{-2+2}{2} \right )= \left ( \frac{2-4}{2},\frac{3-3}{2} \right )$
$\left ( \frac{1+a}{2},0 \right )= \left ( -1,0 \right )$
$\frac{1+a}{2}= -1$
1 + a = –2
a = –3
C(–3, 2)
$Area \, of \, \triangle ABC = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 1\left ( 3-2 \right ) +2\left ( 2+2 \right )+\left ( -3 \right )\left ( -2-3 \right )\right ]$
$= \frac{1}{2}\left [ 1+2\left ( 4 \right )+15\right ]$
$= \frac{1}{2}\left [ 24 \right ]= 12sq\cdot units$
Area of parallelogram = 2 × Area of $\triangle$ ABC
Area of parallelogram = 2 × 12 = 24sq.units
$Length\, of\, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$= \sqrt{\left ( 2-1 \right )^{2}+\left ( 3+2 \right )^{2}}$
$AB=\sqrt{1+25}= \sqrt{26}units$
Area of parallelogram = Base × height
$\frac{24}{Base}= Height$
$Height= \frac{24}{AB}$
$Height= \frac{24}{\sqrt{26}}\times\frac{\sqrt{26}}{\sqrt{26}}$
$\frac{24\sqrt{26}}{13}units$

Question:5

Students of a school are standing in rows and columns in their playground for a drill practice. A, B, C and D are the positions of four students as shown in figure. Is it possible to place Jaspal in the drill in such a way that he is equidistant from each of the four students A, B, C and D? If so, what should be his position?

test

Answer:

Solution. Points are A(3, 5), B(7, 9), C(11, 5), D(7, 1)
$Length\, of\, AB= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$AB= \sqrt{\left ( 7-3 \right )^{2}+\left ( 9-5 \right )^{2}}$
$AB= \sqrt{16+16}= \sqrt{32}= 4\sqrt{2}$
$Length\, of\, BC= \sqrt{\left ( 11-7 \right )^{2}+\left ( 5-9 \right )^{2}}$
$BC= \sqrt{16+16}$
$= \sqrt{32}$
$=4 \sqrt{2}$
$Length\, of\, CD= \sqrt{\left ( 7-11 \right )^{2}+\left ( 1-5 \right )^{2}}$
$CD= \sqrt{16+16}$
$= \sqrt{32}$
$=4 \sqrt{2}$
$Length\, of\, AD= \sqrt{\left ( 3-7 \right )^{2}+\left ( 5-1 \right )^{2}}$
$AD= \sqrt{16+16}$
$= \sqrt{32}$
$=4 \sqrt{2}$
$Length\, of\, AC= \sqrt{\left ( 11-3 \right )^{2}+\left ( 5-5 \right )^{2}}$
$AC= \sqrt{\left ( 8 \right )}$
= 8
$Length\, of\, BD= \sqrt{\left ( 7-7 \right )^{2}+\left ( 1-9 \right )^{2}}$
$BD= \sqrt{64}$
= 8
AB = BC = AD, AC = BD
Hence, ABCD is square
The diagonals cut each other at mid-point, which is the equidistance from all four corners of square.
$Mid-point \, of \, AC = \left ( \frac{x_{1}+x_{2}}{2} ,\frac{y_{1}+y_{2}}{2}\right )$
(x1, y1) =(3, 5) (x2, y2) = (11, 5)
$AC= \left ( \frac{3+11}{2},\frac{5+5}{2} \right )$
$AC= \left ( 7,5 \right )$
This should be the position of Jaspal.

Question:6

Ayush starts walking from his house to the office. Instead of going to the office directly, he goes to a bank first, from there to his daughter’s school and then reaches the office. What is the extra distance travelled by Ayush in reaching his office? (Assume that all distances covered are in straight lines). If the house is situated at (2,4), bank at (5, 8), school at (13, 14) and office at(13, 26) and coordinates are in km.

Answer:

The given point are (2, 4), (5, 8), (13, 14), (13, 26)
Distance between house and bank
$= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$= \sqrt{\left ( 5-2 \right )^{2}+\left ( 8-4 \right )^{2}}= \sqrt{9+16}= 5$
Distance between bank and school
$= \sqrt{\left ( 13-5 \right )^{2}+\left ( 14-8 \right )^{2}}= \sqrt{64+36}= \sqrt{100}= 10$
Distance between school and office
$= \sqrt{\left ( 13-13 \right )^{2}+\left ( 26-14 \right )^{2}}= \sqrt{\left ( 12 \right )^{2}}= 12$
Distance between office and house
$= \sqrt{\left ( 13-2 \right )^{2}+\left ( 26-14 \right )^{2}}$
$= \sqrt{121+484}$
$= \sqrt{605}$
$= 24\cdot 59$
Total distance covered from house to bank, bank to school, school to office = 5 + 10 + 12 = 27
Extra distance covered = 27 – 24.59 = 2.41 km.

NCERT Exemplar Solutions Class 10 Maths Chapter 7 Important Topics:

How to find out distance between two points if their coordinates are given (Most important formula of coordinate geometry).
How to find out area of a triangle if the coordinates of the three vertices are known.
Class 10 Maths NCERT exemplar chapter 7 solutions discusses the section formula to find out the coordinate of one point dividing the line segment joining two points in the given ratio.

NCERT Class 10 Exemplar Solutions for Other Subjects:

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Chapter 1 Real Numbers

Chapter 2 Polynomials

Chapter 3 Pair of Linear Equations in Two Variables

Chapter 4 Quadratic Equations

Chapter 5 Arithmetic Progressions

Chapter 6 Triangles

Chapter 8 Introduction to Trigonometry & Its Equations

Chapter 9 Circles

Chapter 10 Constructions

Chapter 11 Areas related to Circles

Chapter 12 Surface Areas and Volumes

Chapter 13 Statistics and Probability

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 7:

These Class 10 Maths NCERT exemplar chapter 7 solutions provide an extended knowledge of coordinate geometry. In Class 9 students have studied about abscissa and ordinate. Students have also learnt about coordinate to represent any point. In this chapter of Class 10, student will learn to find out distance between two points if their coordinates are given which is the most useful formula of coordinate geometry. These solutions can be used as the reference material for better study of Coordinate Geometry based practice problems.

The NCERT exemplar Class 10 Maths chapter 7 solutions Coordinate Geometry will be adequate to solve reference books like RD Sharma Class 10 Maths, NCERT Class 10 Maths, RS Aggarwal Class 10 Maths etc.

NCERT exemplar Class 10 Maths solutions chapter 7 pdf download will be made available to help in resolving the issues encountered while solving the NCERT exemplar Class 10 Maths chapter 7.

Check Chapter-Wise Solutions of Book Questions

Chapter No.	Chapter Name
Chapter 1	Real Numbers
Chapter 2	Polynomials
Chapter 3	Pair of Linear Equations in Two Variables
Chapter 4	Quadratic Equations
Chapter 5	Arithmetic Progressions
Chapter 6	Triangles
Chapter 7	Coordinate Geometry
Chapter 8	Introduction to Trigonometry
Chapter 9	Some Applications of Trigonometry
Chapter 10	Circles
Chapter 11	Constructions
Chapter 12	Areas Related to Circles
Chapter 13	Surface Areas and Volumes
Chapter 14	Statistics
Chapter 15	Probability

Must, Read NCERT Solution Subject Wise

Check NCERT Notes Subject Wise

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. Which theorem is used to derive distance formula?

The distance formula enables us to find out the distance between two points of coordinates. This formula is derived by the Pythagoras theorem.

2. What is external division of a line segment?

If the point which divides the line segment lies outside the line segment and the ratio of its distance from the given two points of line segment are in the given ratio.

3. Can we find the area of any polygon if the coordinate of all vertices is given?

Yes, we can find out the area of any polygon. To calculate the area of any polygon we have to divide this area into different triangles and we can use the formula of the area of the triangle.

4. Is the chapter Coordinate Geometry important for Board examinations?

The chapter Coordinate Geometry holds around 6-8% weightage of the whole paper.

5. Will these problems of NCERT exemplar require any additional knowledge of Coordinate Geometry?

NCERT exemplar Class 10 Maths solutions chapter 7 help students with a multidimensional approach. It will help them to understand the concept of Coordinate Geometry quite easily.

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Questions related to CBSE Class 10th

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sadhu ashram ramgart road aligarh

If you're looking for directions or steps to reach Sadhu Ashram on Ramgart Road in Aligarh, here’s how you can get there:

Steps to Reach Sadhu Ashram, Ramgart Road, Aligarh:

Starting Point:
- Determine your starting point in Aligarh or the nearby area.
Use Google Maps:
- Open Google Maps on your phone or computer.
- Enter "Sadhu Ashram, Ramgart Road, Aligarh" as your destination.
- Follow the navigation instructions provided by Google Maps.
By Local Transport:
- Auto-rickshaw/Taxi: Hire an auto-rickshaw or taxi and inform the driver about your destination. Most local drivers should be familiar with Sadhu Ashram.
- Bus: Check if there are any local buses that operate on Ramgart Road. Ask the bus conductor if the bus stops near Sadhu Ashram.
Landmarks to Look For:
- As you approach Ramgart Road, look for local landmarks that might help you confirm you’re on the right path, such as known shops, temples, or schools nearby.
Ask for Directions:
- If you're unsure, ask locals for directions to Sadhu Ashram on Ramgart Road. It's a known location in the area.
Final Destination:
- Once you reach Ramgart Road, Sadhu Ashram should be easy to spot. Look for any signage or ask nearby people to guide you to the exact location.

If you need detailed directions from a specific location or more information about Sadhu Ashram, feel free to ask

Read Complete Answer

sai raja 01 September,2024

show the previous year exams ?

Hello Aspirant, Hope your doing great, your question was incomplete and regarding what exam your asking.

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SWETCHCHA MISKA 10 July,2024

as an icse student in class 10 is above 80 percent good enough to get into commerce in 11th cbse

Yes, scoring above 80% in ICSE Class 10 exams typically meets the requirements to get into the Commerce stream in Class 11th under the CBSE board . Admission criteria can vary between schools, so it is advisable to check the specific requirements of the intended CBSE school. Generally, a good academic record with a score above 80% in ICSE 10th result is considered strong for such transitions.

Read Complete Answer

Manasvini Gupta 23 July,2024

i had cleared 10th from cbse in 2022 2 years ago now i wish to apply for 12th as private candidate in 2025. can i do so?

hello Zaid,

Yes, you can apply for 12th grade as a private candidate .You will need to follow the registration process and fulfill the eligibility criteria set by CBSE for private candidates.If you haven't given the 11th grade exam ,you would be able to appear for the 12th exam directly without having passed 11th grade. you will need to give certain tests in the school you are getting addmission to prove your eligibilty.

best of luck!

Read Complete Answer

Ashvadha 03 July,2024

i had cleared 10th in 2022 from cbse can i apply for 12th as private candidate this year?

According to cbse norms candidates who have completed class 10th, class 11th, have a gap year or have failed class 12th can appear for admission in 12th class.for admission in cbse board you need to clear your 11th class first and you must have studied from CBSE board or any other recognized and equivalent board/school.

You are not eligible for cbse board but you can still do 12th from nios which allow candidates to take admission in 12th class as a private student without completing 11th.

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sadafreen356 03 July,2024

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Central Board of Secondary Education Class 10th Examination

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Option 1) $0.34\; J$	Option 2) $0.16\; J$
Option 3) $1.00\; J$	Option 4) $0.67\; J$

Option 1) $2,000 \; J - 5,000\; J$	Option 2) $200 \, \, J - 500 \, \, J$
Option 3) $2\times 10^{5}J-3\times 10^{5}J$	Option 4) $20,000 \, \, J - 50,000 \, \, J$

Option 1) $K/2\,$	Option 2) $\; K\;$
Option 3) $zero\;$	Option 4) $K/4$

Option 1) $11.2\, L\, H_{2(g)}$ at STP is produced for every mole $HCL_{(aq)}$ consumed	Option 2) $6L\, HCl_{(aq)}$ is consumed for ever $3L\, H_{2(g)}$ produced
Option 3) $33.6 L\, H_{2(g)}$ is produced regardless of temperature and pressure for every mole Al that reacts	Option 4) $67.2\, L\, H_{2(g)}$ at STP is produced for every mole Al that reacts .

Option 1) 0.02	Option 2) 3.125 × 10^-2
Option 3) 1.25 × 10^-2	Option 4) 2.5 × 10^-2

Option 1) decrease twice	Option 2) increase two fold
Option 3) remain unchanged	Option 4) be a function of the molecular mass of the substance.

Option 1) Molality	Option 2) Weight fraction of solute
Option 3) Fraction of solute present in water	Option 4) Mole fraction.

Option 1) twice that in 60 g carbon	Option 2) 6.023 × 10²²
Option 3) half that in 8 g He	Option 4) 558.5 × 6.023 × 10²³

Option 1) less than 3	Option 2) more than 3 but less than 6
Option 3) more than 6 but less than 9	Option 4) more than 9

NCERT Exemplar Class 10 Maths Solutions Chapter 7 Coordinate Geometry

NCERT Exemplar Solutions Class 10 Maths Chapter 7 Important Topics:

NCERT Class 10 Exemplar Solutions for Other Subjects:

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NCERT Class 10 Maths Exemplar Solutions for Other Chapters:

Features of NCERT Exemplar Class 10 Maths Solutions Chapter 7:

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