NCERT Exemplar Class 10 Maths Solutions Chapter 7 Coordinate Geometry

Have you ever wondered in the world map, how we find India’s location? Or have you ever wondered why, while playing games like PUBG or BGMI, everyone has to set some coordinates before jumping down from the plane? In both situations, coordinate geometry, which is an integral part of Mathematics, is the solution. In a coordinate graph, there are two number lines, which are called axes: the x-axis is the horizontal number line, and the y-axis is the vertical number line.
Before planning to build the foundation of basic concepts about coordinate geometry, students always need to stay updated with the latest CBSE syllabus for class 10, as it changes frequently. After going through the NCERT textbooks, students should practice NCERT exemplar solutions to have a better grasp of this topic. There are step-by-step solutions along with the necessary formulae to make the learning easier for students.

Question:1

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

Solution:

We know that on the x-axis, y = 0

$\text{Hence, we have to find the distance of }P(2,3)\text{ from }Q(2,0).$

$(x_1,y_1)=(2,3),(x_2,y_2)=(2,0)$

$\text{Let the distance between them be }PQ.$

$\text{Using the distance formula:}$

$PQ=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$PQ=\sqrt{(2-2{)}^2+(0-3{)}^2}$

$PQ=\sqrt{0+9}=\sqrt{9}=3$

$PQ=3$

Question:2

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

Solution:

$\text{The given points are }A(0,6),B(0,-2)$

$(x_1,y_1)=(0,6),(x_2,y_2)=(0,-2)$

$\text{Let the distance between them be }AB$

$AB=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$

$\Rightarrow AB=\sqrt{(0-0{)}^2+(-2-6{)}^2}$

$\Rightarrow AB=\sqrt{(-8{)}^2}$

$\Rightarrow AB=\sqrt{64}$

$\Rightarrow AB=8$

Question:3

The distance of the point P (–6, 8) from the origin is
(A) 8
(B)$2\sqrt{7}$
(C) 10
(D) 6

Solution:

$$\begin{gathered} \text{The given point is P(-6, 8) origin (0,0)} \\ (x,y)=(-6,8) \end{gathered}$$

$\text{ Let the distance from origin be OP.}$
$OP=\sqrt{x^2+y^2}$

$\Rightarrow OP=\sqrt{{(-6)}^2+{(8)}^2}$

$$\begin{gathered} =\sqrt{36+64}=\sqrt{100}=10 \\ \Rightarrow OP=10 \end{gathered}$$

Question:4

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

Solution:

$\text{The given points are A (0, 5), the B (-5, 0)}$

$(x_1,y_1)=(0,5)(x_2,y_2)=(-5,0)$

$\text{Let the distance between the points is AB.}$

$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

$\Rightarrow AB=\sqrt{{(-5-0)}^2+{(0-5)}^2}$

$\Rightarrow AB=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$

$\Rightarrow AB=5\sqrt{2}$

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

Solution:

Given: Vertices of rectangle A(0, 3), O(0, 0), B(5, 0)
To find the length of the diagonal, we have to find the distance between A and B.
$\text{Let the length of diagonal is AB},x_1=0,x_2=5,y_1=3,y_2=0$

Hence, the length of the 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) $7+\sqrt{5}$

Solution:

Given: Vertices of the triangle are A (0, 4), B(0, 0) and C (3, 0)
$\mathrm{AB}=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}$

$({\mathrm{x}}_1,{\mathrm{y}}_1)=(0,4)$

$({\mathrm{x}}_2,{\mathrm{y}}_2)=(0,0)$

$\mathrm{AB}=\sqrt{(0-0{)}^2+(0-4{)}^2}=\sqrt{(4{)}^2}=4$

$({\mathrm{x}}_1,{\mathrm{y}}_1)=(0,0)$

$({\mathrm{x}}_2,{\mathrm{y}}_2)=(3,0)$

$\mathrm{BC}=\sqrt{(3-0{)}^2+(0-0{)}^2}=\sqrt{(3{)}^2}=3$

$({\mathrm{x}}_1,{\mathrm{y}}_1)=(3,0)$

$({\mathrm{x}}_2,{\mathrm{y}}_2)=(0,4)$

$\mathrm{CA}=\sqrt{(0-3{)}^2+(4-0{)}^2}=\sqrt{9+16}$
$\mathrm{CA}=\sqrt{25}=5$
Perimeter of $\mathrm{\Delta }$ABC = AB + BC + AC = 4 + 3 + 5 = 12
Hence perimeter of $\mathrm{\Delta }$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

Solution:

The given vertices are $A(3,0),B(7,0)$ and $C(8,4)$

Area of $\mathrm{\Delta }ABC$
$=\frac{1}{2}[x_1(y_2–y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$

$=\frac{1}{2}[3(0-4)+7(4-0)+8(0-0)]$

$=\frac{1}{2}[-12+28+0]$

$=\frac{1}{2}[16]=8$
Area of $\mathrm{\Delta }ABC=8$

Question:8

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

Solution:

The given points are $(-4,0),(4,0),(0,3)$
$\begin{array}{rl}& \mathrm{AB}=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}\\ & ({\mathrm{x}}_1,{\mathrm{y}}_1)=(-4,0)({\mathrm{x}}_2,{\mathrm{y}}_2)=(4,0)\\ & \mathrm{AB}=\sqrt{(4-(-4{)}^2+(0-0{)}^2}\\ & \mathrm{AB}=\sqrt{(8{)}^2}\\ & \Rightarrow \mathrm{AB}=8\\ & ({\mathrm{x}}_1,{\mathrm{y}}_1)=(4,0)({\mathrm{x}}_2,{\mathrm{y}}_2)=(0,3)\\ & \mathrm{BC}=\sqrt{(0-4{)}^2+(3-0{)}^2}\\ & \mathrm{BC}=\sqrt{16+9}=\sqrt{25}=5\\ & ({\mathrm{x}}_1,{\mathrm{y}}_1)=(0,3)({\mathrm{x}}_2,{\mathrm{y}}_2)=(-4,0)\\ & \mathrm{AC}=\sqrt{(0+4{)}^2+(3-0{)}^2}\\ & \mathrm{AC}=\sqrt{16+9}=\sqrt{25}=5\end{array}$

$\text{ In this triangle, }\mathrm{BC}=\mathrm{AC}$

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

Solution:

Here the points are A(7, -6), B(3, 4,) and the ratio is 1: 2.
$(x_1,y_1)=(7,-6)(x_2,y_2)=(3,4)$
Let point which divides the line is (x, y)
$(\mathrm{x},\mathrm{y})$

$=(\frac{{\mathrm{m}}_1{\mathrm{x}}_2+{\mathrm{m}}_2{\mathrm{x}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2},\frac{{\mathrm{m}}_1{\mathrm{y}}_2+{\mathrm{m}}_2{\mathrm{y}}_1}{{\mathrm{m}}_1+{\mathrm{m}}_2})$

$(\mathrm{x},\mathrm{y})=(\frac{1\times 3+2\times 7}{1+2},\frac{14+2\times -6}{1+2})$

$(\mathrm{x},\mathrm{y})=(\frac{3+14}{3},\frac{4-12}{3})$

$(\mathrm{x},\mathrm{y})=(\frac{17}{3},\frac{-8}{3})$
Here, x is positive, and e and y are negative.
Hence, the point lies in the 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)

Solution:

The point which lies on the perpendicular bisector is the coordinate of the mid-point on the line joining the points A(–2, –5), B(2, 5).
Let this point is (x, y)
$({\mathrm{x}}_1,{\mathrm{y}}_1)=(-2,-5)({\mathrm{x}}_2,{\mathrm{y}}_2)=(2,5)$

$(\mathrm{x},\mathrm{y})=(\frac{{\mathrm{x}}_1+{\mathrm{x}}_2}{2},\frac{{\mathrm{y}}_1+{\mathrm{y}}_2}{2})$

$$\begin{gathered} (\mathrm{x},\mathrm{y})=(\frac{-2+2}{2},\frac{-5+5}{2}) \\ (\mathrm{x},\mathrm{y})=(0,0) \end{gathered}$$
Hence, the required point is (0, 0)

Question:11

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

Solution:

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
$(\frac{-2+8}{2},\frac{3+3}{2})$

$$\begin{gathered} =(\frac{6+x}{2},\frac{7+y}{2}) \\ (\frac{6}{2},\frac{6}{2}) \end{gathered}$$

$=(\frac{6+x}{2},\frac{7+y}{2})$

$(3,3)=(\frac{6+x}{2},\frac{7+y}{2})$

$\frac{6+x}{2}=3\frac{7+y}{2}=3$

$$\begin{gathered} 6+x=67+y=6 \\ x=0y=-1 \end{gathered}$$
$\text{ Hence, }D=(0,-1)$
Option B is correct.

Question:12

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

(B) $\mathrm{AP}=\frac{1}{3}\mathrm{PB}$

(C) $\mathrm{PB}=\frac{1}{3}\mathrm{AB}$

(D) $\mathrm{AP}=\frac{1}{2}\mathrm{AB}$

Solution:

Let P divide AB in the ratio k: 1
$({\mathrm{x}}_1,{\mathrm{y}}_1)=(4,2)$
$({\mathrm{x}}_2,{\mathrm{y}}_2)=(8,4)$
$P=(\frac{k(8)+1(4)}{k+1},\frac{k(4)+1(2)}{k+1})$
By using section formula
Here,
$\mathrm{P}=(2,1)$
$(2,1)=(\frac{8k+4}{k+1},\frac{4k+12}{k+1})$
Compare x co-ordinate
$\frac{8k+4}{k+1}=2$

$8k+4=2k+2$
$6k=-2$

$k=\frac{-2}{6}=\frac{-1}{3}$

$\text{Here, k is negative. Hence, P divides AB in the ratio 1: 3 externally.}$

$\frac{\mathrm{AP}}{\mathrm{PB}}=\frac{1}{3}$

$\mathrm{AP}=\frac{1}{3}\mathrm{PB}$

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

Solution:

$$\begin{gathered} \mathrm{Q}(-6,5)\mathrm{R}(-2,3) \\ ({\mathrm{x}}_1,{\mathrm{y}}_1)({\mathrm{x}}_2,{\mathrm{y}}_2) \end{gathered}$$
By using the mid-point formula
$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$P=(\frac{-6-2}{2},\frac{5+3}{2})$
$$\begin{gathered} (\frac{a}{3},4)=(\frac{-8}{2},\frac{8}{2}) \\ (\frac{a}{3},4)=(-4,4) \end{gathered}$$
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) and B (4, 6) cuts the y-axis at
(A) (0, 13)
(B) (0, –13)
(C) (0, 12)
(D) (13, 0)

Solution:
$\text{At y-axis},x=0$
$\therefore$ The point P is $(0,y).$
$A(1,5)$ and $B(4,6)$
$\therefore$ AP = BP
Squaring both sides, we get,
$A{P}^2=B{P}^2$
$\Rightarrow (x_1-0{)}^2+(y_1+y{)}^2=(x_2-0{)}^2+(x_2-y{)}^2$
$(\text{Because distance formula}=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2})$
$\Rightarrow (1-0{)}^2+(5-y{)}^2=(4-0{)}^2+(6-y{)}^2$
$\Rightarrow (1{)}^2+(5{)}^2+(y{)}^2-2\times 5\times y=(4{)}^2+(6{)}^2+(y{)}^2-2\times 6\times y$
Using $\{(a-b{)}^2=a^2+b^2-2ab\}$
$\Rightarrow 1+25+y-10y=16+36+y-12y$
$\Rightarrow 26-10y+12y=52$
$\Rightarrow 2y=26$
$\Rightarrow y=13$
Hence, point P is (0, 13).
Therefore, option A is correct.

Question:15

The coordinates of the point which is equidistant from the three vertices of the ΔAOB, as shown in the figure, is
$(A)(x,y)$
$(B)(y,x)$
$(C)\frac{x}{2},\frac{y}{2}$
$(D)\frac{y}{2},\frac{x}{2}$

Solution:
In the given figure, it is clear that DAOB is a right-angle triangle.
In a right-angle triangle, the mid-point of the hypotenuse is equidistant from the three vertices. Thus, coordinates must be mid-point of AB.
$A(0,2y),B(2x,0)$
$(x_1,y_1)=(0,2y),(x_2,y_2)=(2x,0)$
Now find the mid-point of AB using the mid-point formula.
$(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
$(\frac{0+2x}{2},\frac{2y+0}{2})=(x,y)$
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 :
$$\begin{gathered} (A)\frac{-3}{4},1 \\ (B)2,\frac{7}{3} \\ (C)5,\frac{-1}{2} \\ (D)(-6,\frac{5}{2}) \end{gathered}$$

Solution:

(A) Distance of the point (-3/4, 1) from (0,0) is
$=\sqrt{{(\frac{-3}{4}-0)}^2+(1-0{)}^2}$
$=\sqrt{\frac{9}{16}+1}$
$=\sqrt{\frac{25}{16}}=\frac{5}{4}=1.25\mathrm{units}$
The distance is 1.25<6.5. so the point (-3/4, 1) is lie

( B) Distance point (2, 7/3) from (0,0) is
$=\sqrt{{(\frac{7}{3}-0)}^2+(2-0{)}^2}$
$=\sqrt{\frac{49}{9}+4}$
$=\sqrt{\frac{85}{9}}=\frac{9.2195}{3}=3.0731<6.25$
So the point is lie

(C) Distance point (5,-1/2) from (0,0) is
$=\sqrt{{(-\frac{1}{2}-0)}^2+(5-0{)}^2}$
$=\sqrt{\frac{1}{4}+25}$
$=\sqrt{\frac{101}{4}}=\frac{10.0498}{2}=5.0249<6.25$
So the point lies in the interior of the circle.

( D)
The circle passes through (13/2, 0) and has a centre (0,0)
$({\mathrm{x}}_1,{\mathrm{y}}_1)=(13/2,0)({\mathrm{x}}_2,{\mathrm{y}}_2)=(0,0)$
$\text{Apply distance formula}=\sqrt{{({\mathrm{x}}_2-{\mathrm{x}}_1)}^2+{({\mathrm{y}}_2-{\mathrm{y}}_1)}^2}$
$\text{Radius}=\sqrt{(6.5-0{)}^2+(0-0{)}^2}$
$=\sqrt{(6.5{)}^2}$
$=6.5$
$=5^2+{0.5}^2-{6.5}^2(\text{Negative)}$
$(-6,\frac{5}{2})=(-6{)}^2+(2.5{)}^2-6.5$
$=6^2+{2.5}^2-6.5\text{(positive)}$
Hence, point D is the correct answer.

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)

Solution:

Given Points:
Point $P=(0,y)$ is on the $y$-axis, so $x=0$.
Point $Q=(x,0)$ is on the x-axis, so $y=0$.
Midpoint Formula:
The midpoint $M$ of two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by:
$M=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Here, the midpoint is given as $(2,-5)$, and we know $P=(0,y)$ and $Q=(x,0)$.
Substituting the given points into the midpoint formula, we get,
$(\frac{0+x}{2},\frac{y+0}{2})=(2,-5)$
This simplifies to:
$\frac{x}{2}=2\text{ and }\frac{y}{2}=-5$
So, $x=4$ and $y=-10$.
Thus, the coordinates of the points are:
- $P=(0,-10)$
- $Q=(4,0)$
So, the points are $P(0,-10)$ and $Q(4,0)$.
Hence, the correct answer is option D.

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

Solution:
Vertices are (a, b + c), (b, c + a), (c, a + b)
Here $A(x_1,y_1)=(a,b+c)$
$$\begin{gathered} B(x_2,y_2)=(b,c+a) \\ C(x_3,y_3)=(c,a+b) \end{gathered}$$
We know that
Area of triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[a(c+a-a-b)+b(a+b-b-c)+c(b+c-c-a)]$
$=\frac{1}{2}[a(c-b)+b(a-c)+c(b-a)]$
$=\frac{1}{2}[ac-ab+ab-bc+bc-ac]$
$=0$
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

Solution:

Points are A(4, p) and B(1, 0)
Distance
$(AB)=5$ [Given]
Here
${x}_1=4x_2=1$
$y_1=p{y}_2=0$
Using distance formula,
$AB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
Squaring both sides
$(1-4{)}^2+(0-p{)}^2=(5{)}^2$
$\Rightarrow (-3{)}^2+(p{)}^2=25$
$\Rightarrow 9+p^2=25$
$\Rightarrow p^2=16$
$\Rightarrow p=\pm 4$
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

Solution:

If points A(1, 2), O(0, 0), and C (a, b) are collinear, then the area of the triangle formed by these points must be zero.
$A(1,2)=A(x_1,y_1)$
$O(0,0)=O(x_2,y_2)$
$C(a,b)=C(x_3,y_3)$
Area of triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$\Rightarrow \frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$
$\Rightarrow [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$
$\Rightarrow [1(0-b)+0(b-2)+a(2-0)]=0$
$\Rightarrow [-b+0+2a]=0$
$\Rightarrow 2a=b$
Hence, option C is correct.

Question:1

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

Answer [False]
Solution:
$DistanceFormula=\sqrt{{\left(x_2-x_1\right)}^2+{(y_2-y_1)}^2}$
In $△$ABC,
$AB=\sqrt{{(2+2)}^2+{(0-0)}^2}=\sqrt{{\left(4\right)}^2}=4$
$BC=\sqrt{{(0-2)}^2+{(2-0)}^2}=\sqrt{(4+4)}=\sqrt{8}=2\sqrt{2}$
$AC=\sqrt{{(0+2)}^2+{(2-0)}^2}=\sqrt{(4+4)}=2\sqrt{2}$
In $△$DEF,
$DE=\sqrt{{(4+4)}^2+{(0-0)}^2}=\sqrt{{\left(8\right)}^2}=8$
$EF=\sqrt{{(0-4)}^2+{(4-0)}^2}=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$
$DF=\sqrt{{(0+4)}^2+{(4-0)}^2}=\sqrt{16+16}=4\sqrt{2}$
$Here,\frac{AB}{DE}=\frac{BC}{EF}=\frac{AC}{DF}$
$\frac{4}{8}=\frac{2\sqrt{2}}{4\sqrt{2}}=\frac{1}{2}$,
Hence, $△$ABC is similar to $△$DEF i.e $△ABC\sim △DEF$

Question:2

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

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) lies on line AB if the area of triangle ABP =0
$\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_2+)+x_3\left(y_1-y_2\right)]=0$
$\frac{1}{2}[-4(-6-2)-4(2-6)-4(6+6)]=0$
$\frac{1}{2}[-4(-8)-4(-4)-4\left(12\right)]=0$
$[+32+16-48]=0$
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. [False]
Solution:
The given points area (0, 5), (0, –9) and (3, 6)
If the point area is collinear, then the area of a triangle is 0.
x₁ =0, x₂ =0, x₃ =3
y₁=5, y₂=9, y₃=6
We know that the Area of a triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[0(-9-6)+0(6-5)+3(5+9)]$
$=\frac{1}{2}[0+0+42]$
$=\frac{42}{2}=21$
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 the y–axis and the perpendicular bisector of the line segment joining the points A (–1, 1) and B (3, 3).

Answer: False
Solution:
If the point P is a perpendicular bisector of the line joining the points A(–1, 1) and B(3, 3), then it must be the midpoint of AB.
Mid-point of $AB=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
(x₁, y₁) = (-1, 1) (x₂, y₂) = (3, 3)
$=(\frac{-1+3}{2},\frac{1+3}{2})=(\frac{2}{2},\frac{4}{2})=(1,2)$
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. [True]
Solution:
If they are not the vertices of a triangle, then
Area of $\mathrm{△}$ABC = 0
x₁ =3 x₂ =12 x₃ =0
y₁=1 y₂=-2 y₃=2
Let us find the area of $\mathrm{△}$ABC
Area of a $\mathrm{△}$ABC
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[3(-2-2)+12(2-1)+0(1+2)]$
$=\frac{1}{2}[3(-4)+12\left(1\right)+0]$
$=\frac{1}{2}[-12+12]$
$=\frac{0}{2}$
= 0
Area of $\mathrm{△}$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. [False]
Solution:
The given points area A(4, 3), B(6, 4), C(5, –6), D(–3, 5)
$DistancebetweenAB=\sqrt{{(6-4)}^2+{(4-3)}^2}$
$AB=\sqrt{{\left(2\right)}^2+{\left(1\right)}^2}=\sqrt{5}$
$DistancebetweenBC=\sqrt{{(5-6)}^2+{(-6-4)}^2}$
$BC=\sqrt{1+100}=\sqrt{101}$
$DistancebetweenCD=\sqrt{{(-3-5)}^2+{(5+6)}^2}$
$CD=\sqrt{64+121}=\sqrt{185}$
$DistancebetweenDA=\sqrt{{(4+3)}^2+{(3-5)}^2}$
$DA=\sqrt{49+4}=\sqrt{53}$
Here, opposite sides are not equal, i.e.
$AB\ne CD,BC\ne DA$
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. Point Q (6, 8) lies outside the circle.

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
$OP=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$OP=\sqrt{{(5-0)}^2+{(0-0)}^2}$
$OP=\sqrt{{\left(5\right)}^2}=5$
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 greater than the radius of the circle.
(x₁, y₁) = (0, 0) (x₂, y₂) = (6, 8)
$OQ=\sqrt{{(6-0)}^2+{(8-0)}^2}$
$OQ=\sqrt{36+64}$
$=\sqrt{100}=10$
Here, point OQ is greater than the radius of the circle.
Hence, point Q(6, 8) lies outside the circle

Question:8

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

Answer. [False]
Solution:
If point A(2, 7) is a bisector, then it must be the mid-point of the line joining the points P(6, 5) and Q(0, –4)
Mid-point of $PQ=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
(x₁, y₁) = (6, 5) (x₂, y₂) = (0, -4)
$=(\frac{6+0}{2},\frac{5-4}{2})=(3,\frac{1}{2})$
Hence, A does not lie on the 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. [True]
Solution:
Let the two point of trisection are C, D $PointC=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
(x₁, y₁) = (7, -2) (x₂, y₂) = (1, -5)
m₁ = 1, m₂ = 2 (Because C divide AB in ratio 1:2 )
$C=(\frac{1\times 1+2\times 7}{1+2},\frac{1\times -5+2\times -2}{1+2})$
$C=(\frac{1+14}{3},\frac{-5-4}{3})$
$C=(5,-3)$
$PointD=(\frac{n_1{x}_2+n_2{x}_1}{n_1+n_2},\frac{n_1{y}_2+n_2{y}_1}{n_1+n_2})$
n₁ = 2, n₂ = 1 (Because D divide AB in ratio 2:1)
$D=(\frac{2\times 1+1\times 7}{2+1},\frac{2\times -5+1\times -2}{2+1})$
$D=(\frac{2+7}{3},\frac{-10-2}{3})$
$D=(3,-4)$
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. [True]
Solution:
If the points A (–6, 10), B(–4, 6) and C(3, –8) are collinear then area of $\mathrm{△}$ABC = 0
$\begin{array}{rl}& \text{ Area of a }\mathrm{△}ABC\\ & =\frac{1}{2}[-6(6+8)+(-4)(-8-10)+3(10-6)]\end{array}$
$=\frac{1}{2}[-84+72+12]$
Area of $\mathrm{△}$ABC = 0
Hence, A, B and C are collinear.
$DistancebetweenAB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$AB=\sqrt{{(-4+6)}^2+{(6-10)}^2}$
$AB=\sqrt{4+16}=\sqrt{20}=2\sqrt{5}$
$DistancebetweenAC=\sqrt{{(3+6)}^2+{(-8-10)}^2}$
$=\sqrt{81+324}$
$=\sqrt{405}$
$AC=9\sqrt{5}$
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. [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.
$DistancebetweenPC=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
(x₁, y₁) = (-2, 4) (x₂, y₂) = (3, 5)
$PC=\sqrt{{(3+2)}^2+{(5-4)}^2}$
$PC=\sqrt{25+1}=\sqrt{26}$
$PC\ne radius\left(6\right)$
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. [True]
Solution:
The given points are A(–1, –2), B(4, 3), C(2, 5) and D(–3, 0)
$LengthofAB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$AB=\sqrt{{(4+1)}^2+{(3+2)}^2}$
$AB=\sqrt{50}=5\sqrt{2}$
$LengthofBC=\sqrt{{(2-4)}^2+{(5-3)}^2}$
$BC=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
$LengthofCD=\sqrt{{(-3-2)}^2+{(0-5)}^2}$
$CD=\sqrt{25+25}=\sqrt{50}=5\sqrt{2}$
$LengthofDA=\sqrt{{(-1+3)}^2+{(-2-0)}^2}$
$DA=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}$
AB = CD , BC = DA
$LengthofAC=\sqrt{{(2+1)}^2+{(5+2)}^2}$
$AC=\sqrt{9+49}=\sqrt{58}$
$LengthofBD=\sqrt{{(-3-4)}^2+{(0-3)}^2}$
$BD=\sqrt{49+9}=\sqrt{58}$
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).

Solution:
Given vertices are A(–5, 6), B(–4, –2), C(7, 5)
$Distanceformula=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$DistanceofAB=\sqrt{{(-4-(-5))}^2+{(-2-6)}^2}$
$=\sqrt{{\left(1\right)}^2+{(-8)}^2}$
$=\sqrt{1+64}$
$=\sqrt{65}$
$DistanceofBC=\sqrt{{(7-(-4))}^2+{(5-(-2))}^2}$
$=\sqrt{{\left(11\right)}^2+{\left(7\right)}^2}$
$=\sqrt{121+49}$
$=\sqrt{170}$
$DistanceofAC=\sqrt{{(7-(-5))}^2+{(5-6)}^2}$
$=\sqrt{{\left(12\right)}^2+{(-1)}^2}$
$=\sqrt{144+1}$
$=\sqrt{145}$
$AB\ne BC\ne AC$
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

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

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?

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

Solution:

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

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,
$P{A}^2=P{B}^2$
${(-5-x)}^2+{(4-y)}^2={(-1-x)}^2+{(6-y)}^2$
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 $AB=(\frac{-5-1}{2},\frac{4+6}{2})$
= (-3,5)
$\because Mid-point=\{\left(\frac{x_1+x_2}{2}\right),\left(\frac{y_1+y_2}{2}\right)\}$
Put point (–3, 5) in eqn. (1)
2(–3) + 5 + 1 = 0
– 6 + 6 = 0
0 = 0
The midpoint of AB satisfies 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.

Solution:
$\text{Use distance formula}=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
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
$\therefore$ $\mathrm{△}$ABQ is an isosceles triangle.

Question:7

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

Solution:
Area of the triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
Given points are A(5, 1), B(–2, –3), C(8, 2m)
If points are collinear, then the area of triangle = 0
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$
$\Rightarrow [x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$
$\Rightarrow$ 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 point A (2, – 4) is equidistant from P (3, 8) and Q (–10, y), find the value of y. Also, find the distance PQ.

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
$PQ=\sqrt{{(-10-3)}^2+{(-5-8)}^2}=\sqrt{169+169}=13\sqrt{2}units$
Case-I when y = –3
$PQ=\sqrt{{(-10-3)}^2+{(-3-8)}^2}=\sqrt{169+121}=\sqrt{290}units$

Question:9

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

Solution:
Area of the triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
Given vertices are (–8, 4), (–6, 6), (–3, 9)
x₁ =-8 x₂ =-6 x₃ = -3
y₁=4 y₂= 6 y₃= 9
We know that the area of triangle is
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[(-8)(6-9)+(-6)(9-4)+(-3)(4-6)]$
$=\frac{1}{2}[24+30+6]$
$=\frac{1}{2}\left[60\right]$
$=30sq.units$

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.

Solution:
$(X,Y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
Let the point on the 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 the section formula, we have
$p(x,0)=[\frac{-k-4}{k+1},\frac{7k-6}{k+1}]$
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-ordinates of $P(\frac{-34}{13},0)$

Question:11

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

Solution:
$P(X,Y)=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
Let the point $P(\frac{3}{4},\frac{5}{12})$ divides the line segment joining the points $A(\frac{1}{2},\frac{3}{2})$ and B(2, –5) in the ration k : 1.
(x₁, y₁) = (1/2, 3/2) (x₂, y₂) = (2, -5)
m₁ = k, m₂ = 1
Using the section formula, we have
$(\frac{3}{4},\frac{5}{12})=[\frac{2k+1/2}{k+1},\frac{-5k+3/2}{k+1}]$
$(\frac{3}{4},\frac{5}{12})=[\frac{4k+1}{2k+1},\frac{-10k+3}{2k+2}]$
$\frac{4k+1}{2k+1}=\frac{3}{4}$ $\frac{-10k+3}{2k+2}=\frac{5}{12}$
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 the line segment joining A(3a + 1, –3) and B(8a, 5)in the ratio 3: 1, find the values of a and b.

Solution:
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 the section formula, we have
$(9a-2,-b)=[\frac{3\left(8a\right)+1(3a+1)}{3+1},\frac{3\left(5\right)+1(-3)}{3+1}]$
$(9a-2,-b)=[\frac{24a+3a+1}{4},\frac{15-3}{4}]$
$(9a-2,-b)=[\frac{27a+1}{4},\frac{12}{4}]$
Equating the 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.

Solution:
Mid-point formula, $x=\frac{x_1+x_2}{2},y=\frac{y_1+y_2}{2}$
Point P (a, b) divides A(10, –6) and B(k, 4) into 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)
$AB=\sqrt{{(22-10)}^2+{(4+6)}^2}$
$=\sqrt{{\left(12\right)}^2+{\left(10\right)}^2}$
$=\sqrt{144+100}$
$=\sqrt{244}$
$=2\sqrt{61}$

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 a diameter $10\sqrt{2}$ units.

Solution:
Distance formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
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)
$=\sqrt{{(11-2a)}^2+{(-9-a+7)}^2}=\left(5\sqrt{2}\right)$
Squaring both sides
${(11-2a)}^2+{(-2-a)}^2={\left(5\sqrt{2}\right)}^2$
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.

Solution:
Here, points A(3, 2) and B(5, 1) are 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
$P(x,y)=[\frac{1\times 5+2\times 3}{1+2},\frac{1\times 1+2\times 2}{1+2}]$
$=[\frac{5+6}{3},\frac{1+4}{3}]$
$=[\frac{11}{3},\frac{5}{3}]$
Line is 3x – 18y + k = 0 ......(1)
$PutpointP(\frac{11}{3},\frac{5}{3})in\left(1\right)$
$Putx=\frac{11}{3},y=\frac{5}{3}$
$3\left(\frac{11}{3}\right)-18\left(\frac{5}{3}\right)+k=0$
11 – 30 + k = 0
–19 + k = 0
k = 19

Question:16

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

Solution:
D is the 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 the mid-point of BC; using the 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 the mid-point of AB; using the 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)
Using (5) and (7), we get
x₁ + x₃ = 7
x₁– x₃ = –15
2x₁ = –8
x₁ = –4
Putting 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 equations (2) and (4), we get
y₁ + y₂ = 5
y₃ + y₃ = 6
- - –
y₁ - y₃ = –1 ….(8)
Adding equations (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 the triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[(-4)(2-4)+\left(3\right)(4-3)+11(3-2)]$
$=\frac{1}{2}[(-4)(-2)+3\left(1\right)+11\left(1\right)]$
$=\frac{1}{2}[8+3+11]$
$=\frac{1}{2}\left[22\right]$
= 11 sq. units

Question:17

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 $\mathrm{△}$ABC.

Solution:
Distance formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$\mathrm{△}$ABC is a right 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 points B and C coincide.
$\therefore$ a = 2
Area of the triangle
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[2(5-5)+2(5-9)+5(9-5)]$
$=\frac{1}{2}[0-8+20]$
$=\frac{1}{2}\left[16\right]$
= 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 $PR=\frac{3}{5}PQ$

Solution:
Section formula $=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
According to the 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 the section formula, we have
(x₁, y₁) = (-1, 3) (x₂, y₂) = (2, 5)
m₁ = 3, m₂ = 2
$R(x,y)=(\frac{3\left(2\right)+2(-1)}{3+2},\frac{3\left(5\right)+2\left(3\right)}{3+2})$
$R(x,y)=(\frac{6-2}{5},\frac{15+6}{5})$
$R(x,y)=(\frac{4}{5},\frac{21}{5})$
Here co- ordinates of R is $(\frac{4}{5},\frac{21}{5})$

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.

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
$\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$
$[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]=0$
[(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, the values of k are 2, ½.

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.

Solution:
Section formula $=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
Let point p(x, y) divide 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 the section formula, we have
$P(x,y)=[\frac{2k+8}{k+1},\frac{k-9}{k+1}]$
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)
$(x,y)=[\frac{2\left(8\right)}{8+1},\frac{8-9}{8+1}]$
$=(\frac{16+8}{9},\frac{-1}{9})=(\frac{24}{9},\frac{-1}{9})=(\frac{8}{3},\frac{-1}{9})$
Required point is $P(\frac{8}{3},\frac{-1}{9})$

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.

Solution:
In equilateral triangle AB = BC = AC
$AC=\sqrt{{(x+4)}^2+{(y-3)}^2}$
$AC=\sqrt{x^2+16+8x+y^2+9-6y}$
$(\because {(a+b)}^2=a^2+b^2+2ab{(a-b)}^2=a^2+b^2-2ab)$
$BC=\sqrt{{(x-4)}^2+{(y-3)}^2}$
$BC=\sqrt{x^2+16-8x+y^2+9-6y}$ $(\because {(a-b)}^2=a^2+b^2-2ab)$
AC = BC
$\sqrt{x^2+16+8x+y^2+9-6y}=\sqrt{x^2+16-8x+y^2+9-6y}$
Squaring both sides,
8x + 8x = 0
16x = 0
x = 0
C = (0, y)
Length of AB $=\sqrt{{(4+4)}^2+{(3-3)}^2}$
$AB=\sqrt{{\left(8\right)}^2}=8$
AC = AB
$\sqrt{x^2+16+8x+y^2+9-6y}=8$
Put x = 0, squaring both sides.
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 a 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 $\mathrm{△}$ADE.

Solution:
Distance formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
The given points A(6, 1), B(8, 2) and C(9, 4) let D(x, y)
The diagonals of a parallelogram bisect each other.
Here mid-point of AC = the mid-point of BD
$(\frac{6+9}{2},\frac{1+4}{2})=(\frac{8+x}{2},\frac{2+y}{2})$
$(\frac{15}{2},\frac{5}{2})=(\frac{8+x}{2},\frac{2+y}{2})$
$\frac{15}{2}=\frac{8+x}{2}$ $\frac{2+y}{2}=\frac{5}{2}$
x = 7 y = 3
D (7, 3)
E is the mid-point of CD.
Let E(x₀, y₀)
$(x_0,y_0)=(\frac{7+9}{2},\frac{3+4}{2})$
$(x_0,y_0)=(\frac{16}{2},\frac{2}{7})$
$E=(8,\frac{7}{2})$
Area of $\mathrm{△}$ ADE
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[6(\frac{7}{2}-3)+8(3-1)+7(1-\frac{7}{2})]$
$=\frac{1}{2}[6\times \frac{1}{2}+8\times 2+7\times \frac{-5}{2}]$
$=\frac{1}{2}[3+16-\frac{35}{2}]$
$=\frac{1}{2}\left[\frac{6+32-35}{2}\right]$
Area of $\mathrm{△}$ ADE $=\frac{1}{2}\times \frac{3}{2}=\frac{3}{4}$ sq units

Question:3

(i) The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of $\mathrm{△}$ABC.The median from A meets BC at D. Find the coordinates of point D.
(ii) The points A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of $\mathrm{△}$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 $\mathrm{△}$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 $\mathrm{△}$ABC.What are the coordinates of the centroid of the triangle ABC?

(i) Solution:
D is the midpoint of BC.
Mid-point formula $=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
Coordinates of $D(x,y)=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$(By midpoint formula)

(ii) Solution:
Section Formula $=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
$D=(\frac{x_2+x_3}{2},\frac{y_2+y_3}{2})$ (By Midpoint formula)
$P=(\frac{m_1{x}_2+m_2{x}_1}{m_1+m_2},\frac{m_1{y}_2+m_2{y}_1}{m_1+m_2})$
$P=(\frac{2\times \frac{(x_2+x_3)}{2}+1\times x_1}{2+1},\frac{2\times \frac{(y_2+y_3)}{2}+1\times x_1}{2+1})$
$P=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

(iii) Solution:
E is mid-point of AC
$E=(\frac{x_1+x_3}{2},\frac{y_1+y_3}{2})$
Q divides BF at 2: 1
$Q=(\frac{2\times \frac{(x_1+x_3)}{2}+1\times x_2}{2+1},\frac{2\times \frac{(y_1+y_3)}{2}+1\times y_2}{2+1})$
$Q=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$
R divides CF at 2: 1
$R=(\frac{2\times \frac{(x_1+x_2)}{2}+1\times x_3}{2+1},\frac{2\times \frac{(y_1+y_2)}{2}+1\times y_3}{2+1})$
$R=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

(iv) Solution:
Co-ordinate of centroid $=(\frac{\text{Sum of all coordinates of all vertices}}{3},\frac{\text{Sum of all coordinates of all vertices}}{3})$
Centroid: The centroid is the centre point of the triangle, which is the intersection of the medians of a triangle.
In $\mathrm{△}ABC$, coordinates of centroid $=(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

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 the height of the parallelogram taking AB as a base.

Solution:
We know that diagonals bisect each other.
Hence, the mid-point of AC = the mid-point of BD
$(\frac{1+a}{2},\frac{-2+2}{2})=(\frac{2-4}{2},\frac{3-3}{2})$
$(\frac{1+a}{2},0)=(-1,0)$
$\frac{1+a}{2}=-1$
1 + a = –2
a = –3
C(–3, 2)
Area of $\mathrm{△}$ ABC
$=\frac{1}{2}[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]$
$=\frac{1}{2}[1(3-2)+2(2+2)+(-3)(-2-3)]$
$=\frac{1}{2}[1+2\left(4\right)+15]$
$=\frac{1}{2}\left[24\right]=12sq\cdot units$
Area of parallelogram = 2 × Area of $\mathrm{△}$ABC
Area of parallelogram = 2 × 12 = 24sq.units
Length of AB
$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$=\sqrt{{(2-1)}^2+{(3+2)}^2}$
$AB=\sqrt{1+25}=\sqrt{26}units$
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 the 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?

Solution:
Points are A(3, 5), B(7, 9), C(11, 5), D(7, 1)
$LengthofAB=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$AB=\sqrt{{(7-3)}^2+{(9-5)}^2}$
$AB=\sqrt{16+16}=\sqrt{32}=4\sqrt{2}$
Length of BC
$=\sqrt{{(11-7)}^2+{(5-9)}^2}$
$BC=\sqrt{16+16}$
$=\sqrt{32}$
$=4\sqrt{2}$
Length of CD
$=\sqrt{{(7-11)}^2+{(1-5)}^2}$
$CD=\sqrt{16+16}$
$=\sqrt{32}$
$=4\sqrt{2}$
Length of AD
$=\sqrt{{(3-7)}^2+{(5-1)}^2}$
$AD=\sqrt{16+16}$
$=\sqrt{32}$
$=4\sqrt{2}$
Length of AC
$=\sqrt{{(11-3)}^2+{(5-5)}^2}$
$AC=\sqrt{\left(8\right)}$
= 8
Length of BD
$=\sqrt{{(7-7)}^2+{(1-9)}^2}$
$BD=\sqrt{64}$
= 8
AB = BC = AD, AC = BD
Hence, ABCD is a square.
The diagonals cut each other at mid-point, which is the equidistance from all four corners of the square.
$Mid-pointofAC=(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2})$
(x1, y1) =(3, 5) (x2, y2) = (11, 5)
$AC=(\frac{3+11}{2},\frac{5+5}{2})$
$AC=(7,5)$
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.

Solution:

The given points are (2, 4), (5, 8), (13, 14), (13, 26)
Distance between house and bank
$=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$
$=\sqrt{{(5-2)}^2+{(8-4)}^2}=\sqrt{9+16}=5$
Distance between bank and school
$=\sqrt{{(13-5)}^2+{(14-8)}^2}=\sqrt{64+36}=\sqrt{100}=10$
Distance between school and office
$=\sqrt{{(13-13)}^2+{(26-14)}^2}=\sqrt{{\left(12\right)}^2}=12$
Distance between office and house
$=\sqrt{{(13-2)}^2+{(26-14)}^2}$
$=\sqrt{121+484}$
$=\sqrt{605}$
$=24\cdot 59$
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 the distance between two points if their coordinates are given (Most important formula of coordinate geometry).
• How to find out the area of a triangle if the coordinates of the three vertices are known.
• Class 10 Maths NCERT exemplar chapter 7 solutions discuss the section formula to find out the coordinate of one point dividing the line segment joining two points in the given ratio.

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Importance of solving NCERT Exemplar Class 10 Maths Solutions Chapter 7

These Class 10 Maths NCERT 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 the 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.

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