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In the world of coordinate geometry, every point has a story, and every line has a direction. In real life, people use coordinate geometry in navigation, GPS tracking, computer graphics, game development, and for designing various structures. Coordinate Geometry is an integral part of mathematics. This chapter is the link between geometry and algebra. It deals with geometrical figures in coordinate systems and coordinates like the x-axis and y-axis to locate the exact position of the points in two-dimensional planes. This chapter also consists of an introductory phase into coordinate geometry, which gradually makes students aware of topics like distance formulas and section formulas. These NCERT solutions for Class 10 Maths will offer a systematic and structured approach to the exercise problems in the NCERT textbook to prepare well for your board exams by providing detailed solutions to all the exercise questions.
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Many toppers rely on NCERT Solutions since they are designed as per the latest syllabus. This chapter can be challenging for students who don't have their basic concepts clear. This article allows every student to understand all the necessary concepts and strengthen their knowledge of coordinate geometry. These NCERT solutions for Class 10 also provide a valuable resource for the students to enhance their performance in their board exams. Refer to this NCERT article for the up-to-date NCERT syllabus, notes, and PDF resources.
The NCERT Solutions for Class 10 Maths Chapter 7 have been prepared by Careers360 experts to make learning simpler and to help you score better in exams. You can also download the solutions in PDF format.
Class 10 Maths Chapter 7 Solutions Exercise: 7.1 Page number: 105-106 Total questions: 10 |
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Q1 (i): Find the distance between the following pairs of points : (2, 3), (4, 1)
Answer:
Given points: (2, 3), (4, 1)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(4-2)^2+(1-3)^2} = \sqrt{4+4} = 2\sqrt{2}$
Q1 (ii): Find the distance between the following pairs of points : (– 5, 7), (– 1, 3)
Answer:
Given points: (– 5, 7), (– 1, 3)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(-1+5)^2+(3-7)^2} = \sqrt{16+16} = 4\sqrt{2}$
Q1 (iii): Find the distance between the following pairs of points :(a, b), (– a, – b)
Answer:
Given points: (a, b), (– a, – b)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(-a-a)^2+(-b-b)^2} = \sqrt{4(a^2+b^2)} = 2\sqrt{a^2+b^2}$
Answer:
Given points: (0, 0) and (36, 15)
Distance between the points will be: $(x_{1},y_{1})$ and $(x_{2},y_{2})$
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$⇒D= \sqrt{(36-0)^2+(15-0)^2} =\sqrt{1296+225} = \sqrt{1521} = 39$
The distance between the two towns A and B is, thus, 39 km for the given town location
$(0,0)$ and $(36,15)$.
Q3: Determine if the points (1, 5), (2, 3), and (– 2, – 11) are collinear.
Answer:
Let the points (1, 5), (2, 3), and (– 2, – 11) represent the vertices A, B, and C of the given triangle, respectively.
$A = (1,5),\ B = (2,3),\ C = (-2,-11)$
Therefore,
$AB = \sqrt{(1-2)^2+(5-3)^2} = \sqrt{5}$
$BC = \sqrt{(2-(-2))^2+(3-(-11))^2} = \sqrt{4^2+14^2} = \sqrt{16+196} = \sqrt{212}$ $CA = \sqrt{(1-(-2))^2+(5-(-11))^2} = \sqrt{3^2+16^2} = \sqrt{9+256} = \sqrt{265}$ Since these are not satisfied.
$AB+BC \neq CA$
$BA+AC \neq BC$
$BC+CA \neq BA$
As these cases are not satisfied.
Hence, the points are not collinear.
Q4: Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Answer:
The distance between two points $A(x_{1},y_{1})$ and $B(x_{2},y_{2})$ is given by:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
So, we have the following points: (5, – 2), (6, 4) and (7, – 2) assuming it to be the vertices of triangles A, B, and C, respectively.
$AB = \sqrt{(5-6)^2+(-2-4)^2} = \sqrt{1+36} = \sqrt{37}$
$BC = \sqrt{(6-7)^2+(4+2)^2} = \sqrt{1+36} = \sqrt{37}$
$CA = \sqrt{(5-7)^2+(-2+2)^2} = \sqrt{4+0} = 2$
Therefore, AB = BC
Here two sides are equal in length.
Therefore, ABC is an isosceles triangle.
Answer:
The coordinates of the points:
$A(3,4),\ B(6,7),\ C(9,4),$ and $D(6,1)$ are the positions of 4 friends.
The distance between two points $A(x_{1},y_{1}),$ and $B(x_{2},y_{2})$ is given by:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
Hence,
$AB = \sqrt{(3-6)^2+(4-7)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
$BC = \sqrt{(6-9)^2+(7-4)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
$CD = \sqrt{(9-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
$AD= \sqrt{(3-6)^2+(4-1)^2} = \sqrt{9+9} = \sqrt{18} = 3\sqrt2$
And the lengths of diagonals:
$AC = \sqrt{(3-9)^2+(4-4)^2} =\sqrt{36+0} = 6$
$BD = \sqrt{(6-6)^2+(7-1)^2} =\sqrt{36+0} = 6$
So, here it can be seen that all sides of quadrilateral ABCD are of the same length and diagonals are also having the same length.
Therefore, quadrilateral ABCD is a square, and Champa is saying right.
Answer:
Let the given points $(-1,-2),\ (1,0),\ (-1,2),$ and $(-3,0)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$AB= \sqrt{(-1-1)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
$BC= \sqrt{(1+1)^2+(0-2)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
$CD= \sqrt{(-1+3)^2+(2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
$AD= \sqrt{(-1+3)^2+(-2-0)^2} =\sqrt{4+4} = \sqrt{8} = 2\sqrt2$
Finding the length of the diagonals:
$AC= \sqrt{(-1+1)^2+(-2-2)^2} =\sqrt{0+16} = 4$
$BD= \sqrt{(1+3)^2+(0-0)^2} =\sqrt{16+0} = 4$
It is clear that all sides are of the same lengths, and also, the diagonals have the same lengths.
Hence, the given quadrilateral is a square.
Q6 (ii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (–3, 5), (3, 1), (0, 3), (–1, – 4)
Answer:
Let the given points $(-3,5),\ (3,1),\ (0,3),$ and $(-1,-4)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$AB= \sqrt{(-3-3)^2+(5-1)^2} =\sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$
$BC= \sqrt{(3-0)^2+(1-3)^2} =\sqrt{9+4} = \sqrt{13}$
$CD= \sqrt{(0+1)^2+(3+4)^2} =\sqrt{1+49} = \sqrt{50} = 5\sqrt2$
$AD= \sqrt{(-3+1)^2+(5+4)^2} =\sqrt{4+81} = \sqrt{85}$
All the sides of the given quadrilateral have different lengths.
Therefore, it is only a general quadrilateral and not a specific one like a square, rectangle, etc.
Q6 (iii): Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (4, 5), (7, 6), (4, 3), (1, 2)
Answer:
Let the given points $(4,5),\ (7,6),\ (4,3),\ (1,2)$ represent the vertices A, B, C, and D of the given quadrilateral, respectively.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
$AB= \sqrt{(4-7)^2+(5-6)^2} =\sqrt{9+1} = \sqrt{10}$
$BC= \sqrt{(7-4)^2+(6-3)^2} =\sqrt{9+9} = \sqrt{18}$
$CD= \sqrt{(4-1)^2+(3-2)^2} =\sqrt{9+1} = \sqrt{10}$
$AD= \sqrt{(4-1)^2+(5-2)^2} =\sqrt{9+9} = \sqrt{18}$
And the diagonals:
$AC =\sqrt{(4-4)^2+(5-3)^2} = \sqrt{0+4} = 2$
$BD =\sqrt{(7-1)^2+(6-2)^2} = \sqrt{36+16} = \sqrt{52} = 2\sqrt{13}$
Here, we can observe that the opposite sides of this quadrilateral are of the same length.
However, the diagonals are of different lengths.
Therefore, the given points are the vertices of a parallelogram.
Q7: Find the point on the x-axis which is equidistant from (2, –5) and (–2, 9).
Answer:
Let the point which is equidistant from $A(2,-5)\ and \ B(-2,9)$ be $X(x,0)$ as it lies on X-axis.
Then, we have,
Distance AX $= \sqrt{(x-2)^2+(0+5)^2}$
and Distance BX $= \sqrt{(x+2)^2+(0+9)^2}$
According to the question, these distances are equal in length.
Hence, we have,
$\sqrt{(x-2)^2+(0+5)^2}$ $= \sqrt{(x+2)^2+(0+9)^2}$
Squaring both sides, we get,
$(x-2)^2+25 = (x+2)^2+81$
$⇒x^2-4x+4+25=x^2+4x+4+81$
$⇒4x+4x=-81+25$
$⇒8x=-56$
$\therefore x=-7$
Hence, the point is $X(-7,0)$.
Q8: Find the values of y for which the distance between the points P(2, – 3) and Q(10, y) is 10 units.
Answer:
Given the distance between the points $P(2,-3)$ and $Q(10,y)$ is 10 units.
The distance formula :
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
So, given $PQ = 10$ units
$PQ= \sqrt{(10-2)^2+(y-(-3))^2} = 10$
After squaring both sides,
$\Rightarrow (10-2)^2+(y-(-3))^2 = 100$
$\Rightarrow (y+3)^2 = 100 - 64$
$\Rightarrow y+3 = \pm 6$
$\Rightarrow y = 6 - 3$ or $y = -6-3$
Therefore, the values are $y = 3$ or $y=-9$.
Answer:
Given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$ .
Then, the distances $PQ = RQ$.
Distance $PQ = \sqrt{(5-0)^2+(-3-1)^2} = \sqrt{25+16} = \sqrt{41}$
Distance $RQ = \sqrt{(x-0)^2+(6-1)^2} = \sqrt{x^2+25}$
$\Rightarrow \sqrt{x^2+25} = \sqrt{41}$
Squaring both sides, we get,
$⇒x^2+25=41$
$\Rightarrow x^2 = 16$
$\Rightarrow x = \pm 4$
The points are: $R(4,6)\ or\ R(-4,6.)$
CASE I: when R is $(4,6)$
The distances QR and PR.
$QR = \sqrt{(0-4)^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$
$PR = \sqrt{(5-4)^2+(-3-6)^2} = \sqrt{1^2+(-9)^2} = \sqrt{1+81} = \sqrt{82}$
CASE II: when R is $(-4,6)$
The distances QR and PR.
$QR = \sqrt{(0-(-4))^2+(1-6)^2} = \sqrt{16+25} = \sqrt{41}$
$PR = \sqrt{(5-(-4))^2+(-3-6)^2} = \sqrt{9^2+(-9)^2} = \sqrt{81+81} = 9\sqrt{2}$
Answer:
Let the point $P(x,y )$ is equidistant from $A(3,6)$ and $B(-3,4)$ .
Then, the distances $AP =BP$
$AP = \sqrt{(x-3)^2+(y-6)^2}$ and $BP = \sqrt{(x-(-3))^2+(y-4)^2}$
$\Rightarrow \sqrt{(x-3)^2+(y-6)^2} = \sqrt{(x-(-3))^2+(y-4)^2}$
Squaring both sides, we obtain
$\Rightarrow (x-3)^2+(y-6)^2= (x+3)^2+(y-4)^2$
$\Rightarrow (2x)(-6)+(2y-10)(-2)= 0$ $\left [\because a^2-b^2 = (a+b)(a-b) \right ]$
$\Rightarrow -12x-4y+20 = 0$
$\Rightarrow 3x+y-5 = 0$
Thus, the relation is $3x+y-5 = 0$ between x and y.
Class 10 Maths Chapter 7 Solutions Exercise: 7.2 Page number: 111 Total questions: 10 |
Q1: Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Answer:
Let the coordinates of point $P(x,y)$ which divides the line segment joining the points $A(-1,7)$ and $B(4,-3)$ , internally, in the ratio $m_{1}:m_{2}$ then,
Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Substituting the values in the formula:
Here, $m_{1}:m_{2} = 2:3$
$\Rightarrow \left (\frac{2(4)+3(-1)}{2+3} , \frac{2(-3)+3(7)}{2+3} \right )$
$\Rightarrow \left (\frac{5}{5} , \frac{15}{5} \right )$
$⇒(1,3)$
Hence, the coordinate is $P \left (1, 3 \right )$.
Q2: Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Answer:
Let the trisection of the line segment $A(4,-1)$ and $B(-2,-3)$ have the points $P(x_{1},y_{1})$ and $Q(x_{2},y_{2})$
Then,
Section formula: $\left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
By observation point, P divides AB internally in the ratio $1:2$.
Hence, $m:n = 1:2$
Substituting the values in the equation, we get;
$\Rightarrow P\left (\frac{1(-2)+2(4)}{1+2} , \frac{1(-3)+2(-1)}{1+2} \right )$
$\Rightarrow P \left (\frac{-2+8}{3} , \frac{-3-2}{3} \right )$
$\Rightarrow P \left (2 , \frac{-5}{3} \right )$
And by observation point Q, divides AB internally in the ratio $2:1$
Hence, $m:n = 2:1$
Substituting the values in the equation above, we get
$\Rightarrow Q\left (\frac{2(-2)+1(4)}{2+1} , \frac{2(-3)+1(-1)}{2+1} \right )$
$\Rightarrow Q \left (\frac{-4+4}{3} , \frac{-6-1}{3} \right )$
$\Rightarrow Q\left (0 , \frac{-7}{3} \right )$
Hence, the points of trisections are $P \left (2 , \frac{-5}{3} \right )$ and $Q\left (0 , \frac{-7}{3} \right )$
Answer:
Niharika posted the green flag at the distance P, i.e.,
$\frac{1}{4}\times100\ m = 25\ m$ from the starting point of $2^{nd}$ line.
Therefore, the coordinates of this point $P$ are $(2,25).$
Similarly, Preet posted red flag at $\frac{1}{5}$ of the distance Q i.e.,
$\frac{1}{5}\times100\ m = 20\ m$ from the starting point of $8^{th}$ line.
Therefore, the coordinates of this point Q are $(8,20)$.
The distance $PQ$ is given by,
$PQ = \sqrt{(8-2)^2+(25-20)^2} = \sqrt{36+25} = \sqrt{61} m$
and the point at which Rashmi should post her Blue Flag is the mid-point of the line joining these points. Let this point be $R(x,y)$.
Then, by Section Formula,
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$⇒x = \frac{2+8}{2},\ y = \frac{25+20}{2}$
$⇒x = 5,\ y = 22.5$
Therefore, Rashmi should post her Blue Flag at 22.5 m on the 5th line.
Answer:
Let the ratio be $k:1$
Then, by section formula:
$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$
Given point $P(x,y) = (-1,6)$
$-1 = \frac{6k-3}{k+1}$
$\Rightarrow -k-1 = 6k-3$
$\Rightarrow k = \frac{2}{7}$
Hence, the point $P$ divides the line AB in the ratio $2:7$.
Answer:
Let the point on the x-axis be $P(x,0)$, and it divides it in the ratio $k:1$.
Then, we have
Section formula:
$P(x,y) = \left (\frac{kx_{2}+x_{1}}{k+1} , \frac{ky_{2}+y_{1}}{k+1} \right )$
$⇒ \frac{ky_{2}+y_{1}}{k+1} = 0$
$⇒k =-\frac{y_{1}}{y_{2}}$
Hence, the value of k will be: $k =-\frac{-5}{5}= 1$
Therefore, the x-axis divides the line in the ratio $1:1$, and the point will be,
Putting the value of $k=1$ in the section formula.
$P(x,0) = \left ( \frac{x_{2}+x_{1}}{2}, 0 \right )$
$P(x,0) = \left ( \frac{1-4}{2}, 0 \right ) = \left ( \frac{-3}{2}, 0 \right )$
Answer:
Let the given points $A(1,2),\ B(4,y),\ C(x,6),\ D(3,5)$
Since the diagonals of a parallelogram bisect each other. Intersection point O of diagonals AC and BD also divides these diagonals.
Therefore, O is the midpoint of AC and BD.
The coordinates of the point O when it is the mid-point of AC.
$\left ( \frac{1+x}{2}, \frac{2+6}{2} \right ) \Rightarrow \left ( \frac{x+1}{2}, 4 \right )$
The coordinates of the point O when it is the mid-point of BD.
$\left ( \frac{4+3}{2}, \frac{5+y}{2} \right ) \Rightarrow \left ( \frac{7}{2}, \frac{5+y}{2} \right )$
Since both coordinates are of the same point O.
Therefore,
$\frac{x+1}{2} =\frac{7}{2}$ and $4 = \frac{5+y}{2}$
Or,
$x = 6\ and\ y = 3$
Answer:
As the centre point $C(2,-3)$ will be the midpoint of the diameter AB.
Then, the coordinates of point A will be $A(x,y)$.
Given point $B(1,4)$ .
Therefore,
$(2,-3) = \left ( \frac{x+1}{2}, \frac{y+4}{2} \right )$
$\frac{x+1}{2} = 2\ and\ \frac{y+4}{2} = -3$
$\Rightarrow x = 3\ and\ y = -10$ .
Therefore, the coordinates of A are $(3,-10).$
Answer:
From the figure:
As $AP = \frac{3}{7}AB$
$\Rightarrow PB = \frac{4}{7}AB$ hence the ratio is 3:4,
Now, from the section formula, we can find the coordinates of Point P.
Section Formula:
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
$⇒P(x,y)= \left (\frac{3(2)+4(-2)}{3+4} , \frac{3(-4)+4(-2)}{3+4} \right )$
$⇒P(x,y)= \left (\frac{6-8}{7} , \frac{-12-8}{7} \right )$
$⇒P(x,y)= \left (\frac{-2}{7} , \frac{-20}{7} \right )$
Answer:
From the figure:
Points C, D, and E divide the line segment AB into four equal parts.
Now, from the section formula, we can find the coordinates of Points C, D, and E.
Section Formula:
$P(x,y)= \left (\frac{m_{1}x_{2}+m_{2}x_{1}}{m_{1}+m_{2}} , \frac{m_{1}y_{2}+m_{2}y_{1}}{m_{1}+m_{2}} \right )$
Here, point D divides the line segment AB into two equal parts; hence
$D(x_{2},y_{2})= \left (\frac{-2+2}{2} , \frac{2+8}{2} \right )$
$⇒D(x_{2},y_{2})= \left (0 , 5 \right )$
Now, point C divides the line segment AD into two equal parts hence
$C(x_{1},y_{1})= \left (\frac{-2+0}{2} , \frac{2+5}{2} \right )$
$⇒C(x_{2},y_{2})= \left (-1 , \frac{7}{2} \right )$
Also, point E divides the line segment DB into two equal parts hence
$E(x_{1},y_{1})= \left (\frac{2+0}{2} , \frac{8+5}{2} \right )$
$⇒E(x_{2},y_{2})= \left (1 , \frac{13}{2} \right )$
Answer:
From the figure:
Let the vertices of the rhombus be:
$A(3,0),\ B(4,5),\ C(-1,4),\ D(-2,-1)$
Area of the rhombus ABCD is given by:
$= \frac{1}{2}\times$ Product of length of diagonals
Hence, we have to find the lengths of the diagonals AC and BD of the rhombus.
The distance formula:
$D = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2}$
Length of the diagonal AC:
$AC = \sqrt{(3-(-1))^2+(0-4)^2} = \sqrt{16+16} = 4\sqrt{2}$
Length of the diagonal BD:
$BD = \sqrt{(4-(-2))^2+(5-(-1))^2} = \sqrt{36+36} = 6\sqrt{2}$
Thus, the area will be,
$= \frac{1}{2}\times (AC)\times(BD)$
$= \frac{1}{2}\times (4\sqrt{2})\times(6\sqrt{2}) = 24$ square units
Also, read,
Question:
Find the coordinates of the points where the graph $57x – 19y = 399$ cuts the coordinate axes.
Answer:
$57x-19y=399$
⇒ $\frac{57x}{399}-\frac{19y}{399} = 1$
⇒ $\frac{x}{7}-\frac{y}{21} = 1$ -------------(i)
Comparing it with the equation of a line:
$\frac{x}{a}+\frac{y}{b} = 1$
x-intercept = $a$ = 7
y-intercept = $b$ = –21
The line, $57x-19y=399$, cuts the x-axis at (7, 0) and the y-axis at (0, –21)
Hence, the correct answer is 'x-axis at (7, 0) and y-axis at (0, –21)'.
Topics you will learn in NCERT Class 10 Maths Chapter 7 Coordinate Geometry include:
For a line defined by two points A$(x_1,y_1)$ and B$(x_2,y_2)$, the distance between these points can be calculated using the formula:
Distance AB = $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$
When a point P divides a line AB, with A(x1, y1) and B(x2, y2) as endpoints, in a ratio of m:n, the coordinates of point P can be found using the formula:
Point P = $\{\frac{mx_2+nx_1}{m+n},\frac{my_2+ny_1}{m+n} \}$
The midpoint of a line AB, defined by A(x1, y1) and B(x2, y2), can be determined using the following formula:
Midpoint P = $\{ \frac{x_1+x_2}2,\frac{(y_1+y_2}2\}$
For students' preparation, Careers360 has gathered all Class 10 Maths NCERT solutions here for quick and convenient access.
Also, read,
Here is the latest NCERT syllabus, which is very useful for students before strategising their study plan. Also, links to some reference books which are important for further studies.
After solving all the exercises, if students want to practice more, then the exemplar solutions can be practised. For that, the following links can be used.
Frequently Asked Questions (FAQs)
Let the two points in a Cartesian plane be $(x_1,y_1)$ and $(x_2,y_2)$.
Then their midpoint $M(x,y)=M\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$
In class 10 Maths Chapter 7 following topics of coordinate geometry are covered.
Let the two points in a Cartesian plane be $(x_1,y_1)$ and $(x_2,y_2)$.
So, their distance = $\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$
If a point $P(x, y)$ divides the line segment joining $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$ in the ratio $m: n$, then the coordinates of $P$ are:
$x=\frac{m x_2+n x_1}{m+n}, \quad y=\frac{m y_2+n y_1}{m+n}$
The formula of the area of a triangle can be used to prove that three points $A(x_1,y_1), B(x_2,y_2)$ and $C(x_3,y_3)$ are collinear.
If the area of the triangle formed by these three points is zero(0), then the points are collinear.
Area of a triangle $=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right|$
On Question asked by student community
Hello,
Yes, you can give the CBSE board exam in 2027.
If your date of birth is 25.05.2013, then in 2027 you will be around 14 years old, which is the right age for Class 10 as per CBSE rules. So, there is no problem.
Hope it helps !
Hello! If you selected “None” while creating your APAAR ID and forgot to mention CBSE as your institution, it may cause issues later when linking your academic records or applying for exams and scholarships that require school details. It’s important that your APAAR ID correctly reflects your institution to avoid verification problems. You should log in to the portal and update your profile to select CBSE as your school. If the system doesn’t allow editing, contact your school’s administration or the APAAR support team immediately so they can correct it for you.
Hello Aspirant,
Here's how you can find it:
School ID Card: Your registration number is often printed on your school ID card.
Admit Card (Hall Ticket): If you've received your board exam admit card, the registration number will be prominently displayed on it. This is the most reliable place to find it for board exams.
School Records/Office: The easiest and most reliable way is to contact your school office or your class teacher. They have access to all your official records and can provide you with your registration number.
Previous Mark Sheets/Certificates: If you have any previous official documents from your school or board (like a Class 9 report card that might have a student ID or registration number that carries over), you can check those.
Your school is the best place to get this information.
Hello,
It appears you are asking if you can fill out a form after passing your 10th grade examination in the 2024-2025 academic session.
The answer depends on what form you are referring to. Some forms might be for courses or examinations where passing 10th grade is a prerequisite or an eligibility criteria, such as applying for further education or specific entrance exams. Other forms might be related to other purposes, like applying for a job, which may also have age and educational requirements.
For example, if you are looking to apply for JEE Main 2025 (a competitive exam in India), having passed class 12 or appearing for it in 2025 are mentioned as eligibility criteria.
Let me know if you need imformation about any exam eligibility criteria.
good wishes for your future!!
Hello Aspirant,
"Real papers" for CBSE board exams are the previous year's question papers . You can find these, along with sample papers and their marking schemes , on the official CBSE Academic website (cbseacademic.nic.in).
For notes , refer to NCERT textbooks as they are the primary source for CBSE exams. Many educational websites also provide chapter-wise revision notes and study material that align with the NCERT syllabus. Focus on practicing previous papers and understanding concepts thoroughly.
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