10th Maths Chapter 5 Coordinate Geometry

10th Maths Chapter 5 Coordinate Geometry solutions, with the chapter's formulas and every exercise worked step by step, as per the 2026-27 Samacheer Kalvi textbook.

This chapter uses coordinates to study points and straight lines in a plane. You learn to find the area of a triangle and a quadrilateral from the vertices, test whether three points lie on one line, and calculate the slope of a line. It also covers the different forms of the equation of a line, and the parallel and perpendicular conditions. The exercises ask you to form equations, find areas and missing coordinates, and locate points of intersection.

Exercises in this chapter

Key formulas

Area of a triangle with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃)
$\text{Area}=\frac{1}{2}\left\{(x_1y_2+x_2y_3+x_3y_1)-(x_2y_1+x_3y_2+x_1y_3)\right\}\ \text{sq.units}$
Collinearity condition (area)
$\text{Area of }\triangle ABC=0\quad\text{or}\quad x_1y_2+x_2y_3+x_3y_1=x_2y_1+x_3y_2+x_1y_3$
Collinearity condition (slope)
$\text{slope of }AB=\text{slope of }BC=\text{slope of }AC$
Area of a quadrilateral with vertices (x₁,y₁), (x₂,y₂), (x₃,y₃), (x₄,y₄)
$\text{Area}=\frac{1}{2}\left\{(x_1y_2+x_2y_3+x_3y_4+x_4y_1)-(x_2y_1+x_3y_2+x_4y_3+x_1y_4)\right\}\ \text{sq.units}$
Slope of a line making angle θ with the positive direction of the X axis
$m=\tan\theta$
Slope of a line joining two distinct points A(x₁,y₁) and B(x₂,y₂)
$m=\frac{y_2-y_1}{x_2-x_1}$
Slope of the line ax + by + c = 0
$m=\frac{-a}{b}$
General form of a straight line
$ax+by+c=0$
Point-slope form
$y-y_1=m(x-x_1)$
Slope-intercept form
$y=mx+c$
Two point form
$\frac{y-y_1}{y_2-y_1}=\frac{x-x_1}{x_2-x_1}$
Intercept form
$\frac{x}{a}+\frac{y}{b}=1$
Line parallel to the Y axis
$x=c$
Line parallel to the X axis
$y=b$
Condition for two straight lines to be parallel
$m_1=m_2$
Condition for two straight lines to be perpendicular
$m_1\times m_2=-1$

Chapter 5 one-mark practice

The area of triangle formed by the points $(-5,0)$ , $(0,-5)$ and $(5,0)$ is
1. (1) $0$ sq.units
2. (2) $25$ sq.units
3. (3) $5$ sq.units
4. (4) none of these
Show answer
Answer: (2) $25$ sq.units
1. The points $(-5,0)$ and $(5,0)$ lie on the X axis, so the base $=10$ units.
2. The third point $(0,-5)$ is at a distance $5$ from the X axis, so height $=5$ units.
3. Area $=\dfrac{1}{2}\times 10\times 5=25$ sq.units.
Question 1 of 10
A man walks near a wall, such that the distance between him and the wall is 10 units. Consider the wall to be the $Y$ axis. The path travelled by the man is
1. (1) $x = 10$
2. (2) $y = 10$
3. (3) $x = 0$
4. (4) $y = 0$
Show answer
Answer: (1) $x = 10$
1. The wall is the Y axis, and his distance from it is always $10$ units.
2. So his $x$ coordinate is always $10$ .
3. Path is the line $x=10$ .
Question 2 of 10
The straight line given by the equation $x = 11$ is
1. (1) parallel to $X$ axis
2. (2) parallel to $Y$ axis
3. (3) passing through the origin
4. (4) passing through the point $(0,11)$
Show answer
Answer: (2) parallel to $Y$ axis
1. $x=11$ means every point on it has $x$ coordinate $11$ , whatever be $y$ .
2. Such a line is vertical, hence parallel to the Y axis.
Question 3 of 10
If $(5,7)$ , $(3,p)$ and $(6,6)$ are collinear, then the value of $p$ is
1. (1) $3$
2. (2) $6$
3. (3) $9$
4. (4) $12$
Show answer
Answer: (3) $9$
1. Slope of the line joining $(5,7)$ and $(6,6)$ is $\dfrac{6-7}{6-5}=-1$ .
2. For collinearity, slope of $(5,7)$ and $(3,p)$ is also $-1$ .
3. $\dfrac{p-7}{3-5}=-1\Rightarrow p-7=2$
4. $p=9$
Question 4 of 10
The point of intersection of $3x - y = 4$ and $x + y = 8$ is
1. (1) $(5,3)$
2. (2) $(2,4)$
3. (3) $(3,5)$
4. (4) $(4,4)$
Show answer
Answer: (3) $(3,5)$
1. $3x-y=4$ and $x+y=8$ .
2. Adding: $4x=12\Rightarrow x=3$ .
3. Then $3+y=8\Rightarrow y=5$ .
4. Point of intersection is $(3,5)$ .
Question 5 of 10
The slope of the line joining $(12,3)$ , $(4,a)$ is $\dfrac{1}{8}$ . The value of ' $a$ ' is
1. (1) $1$
2. (2) $4$
3. (3) $-5$
4. (4) $2$
Show answer
Answer: (4) $2$
1. Slope $=\dfrac{a-3}{4-12}=\dfrac{a-3}{-8}$ .
2. $\dfrac{a-3}{-8}=\dfrac{1}{8}\Rightarrow a-3=-1$
3. $a=2$
Question 6 of 10
The slope of the line which is perpendicular to a line joining the points $(0,0)$ and $(-8,8)$ is
1. (1) $-1$
2. (2) $1$
3. (3) $\dfrac{1}{3}$
4. (4) $-8$
Show answer
Answer: (2) $1$
1. Slope of the line joining $(0,0)$ and $(-8,8)$ is $\dfrac{8-0}{-8-0}=-1$ .
2. Product of slopes of perpendicular lines $=-1$ .
3. Required slope $=\dfrac{-1}{-1}=1$ .
Question 7 of 10
If slope of the line $PQ$ is $\dfrac{1}{\sqrt{3}}$ then slope of the perpendicular bisector of $PQ$ is
1. (1) $\sqrt{3}$
2. (2) $-\sqrt{3}$
3. (3) $\dfrac{1}{\sqrt{3}}$
4. (4) $0$
Show answer
Answer: (2) $-\sqrt{3}$
1. The perpendicular bisector of $PQ$ is perpendicular to $PQ$ .
2. Slope of $PQ=\dfrac{1}{\sqrt{3}}$ .
3. Required slope $=-\dfrac{1}{(1/\sqrt{3})}=-\sqrt{3}$ .
Question 8 of 10
If $A$ is a point on the $Y$ axis whose ordinate is 8 and $B$ is a point on the $X$ axis whose abscissae is 5 then the equation of the line $AB$ is
1. (1) $8x + 5y = 40$
2. (2) $8x - 5y = 40$
3. (3) $x = 8$
4. (4) $y = 5$
Show answer
Answer: (1) $8x + 5y = 40$
1. $A$ is on the Y axis with ordinate $8$ , so $A(0,8)$ ; $B$ is on the X axis with abscissa $5$ , so $B(5,0)$ .
2. Intercept form: $\dfrac{x}{5}+\dfrac{y}{8}=1$ .
3. Multiplying by $40$ : $8x+5y=40$ .
Question 9 of 10
The equation of a line passing through the origin and perpendicular to the line $7x - 3y + 4 = 0$ is
1. (1) $7x - 3y + 4 = 0$
2. (2) $3x - 7y + 4 = 0$
3. (3) $3x + 7y = 0$
4. (4) $7x - 3y = 0$
Show answer
Answer: (3) $3x + 7y = 0$
1. Slope of $7x-3y+4=0$ is $\dfrac{-7}{-3}=\dfrac{7}{3}$ .
2. Slope of the perpendicular line $=-\dfrac{3}{7}$ .
3. Through the origin: $y=-\dfrac{3}{7}x\Rightarrow 3x+7y=0$ .
Question 10 of 10

Spotted a mistake? Email us at [email protected] and we will fix it.