Class 10 Mathematics - NCERT Mathematics
NCERT Solutions for Class 10 Maths Chapter 7: Coordinate Geometry
Chapter 7 covers Coordinate Geometry, where students learn to find the distance between two points, the coordinates of a point that divides a line segment in a given ratio (section formula), and the area of a triangle using coordinates. These formulas are essential for solving real-world geometry problems on the Cartesian plane. This chapter is part of the Coordinate Geometry unit which carries 6 marks in the CBSE Class 10 board exam 2027.
Syllabus Update 2026-27
As per the CBSE 2026-27 syllabus, Chapter 7 Coordinate Geometry has 3 exercises: Exercise 7.1 (Distance Formula), Exercise 7.2 (Section Formula), and Exercise 7.3 (Area of a Triangle). The older Exercise 7.4 has been removed from the CBSE syllabus and will NOT be asked in the 2027 board exam. However, solutions for Ex 7.4 are provided below for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board) who may still follow the 4-exercise format.
Table of Contents
Exercise 7.1 - Distance Formula
5 questions
Exercise 7.2 - Section Formula
5 questions
Exercise 7.3 - Area of a Triangle
5 questions
Old Exercise 7.4 (For State Boards)
3 questions • This exercise has been removed from the CBSE 2026-
Previous Year Questions
16 questions (CBSE 2019-2026)
MCQ Practice
12 questions
7.1Exercise 7.1 - Distance Formula
Q1Find the distance between the following pairs of points: (i) (2, 3) and (4, 1) (ii) (-5, 7) and (-1, 3) (iii) (a, b) and (-a, -b)
Using the distance formula: d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
(i) (2, 3) and (4, 1):
d = sqrt((4 - 2)^2 + (1 - 3)^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2) units
(ii) (-5, 7) and (-1, 3):
d = sqrt((-1 - (-5))^2 + (3 - 7)^2) = sqrt((4)^2 + (-4)^2) = sqrt(16 + 16) = sqrt(32) = 4sqrt(2) units
(iii) (a, b) and (-a, -b):
d = sqrt((-a - a)^2 + (-b - b)^2) = sqrt((-2a)^2 + (-2b)^2) = sqrt(4a^2 + 4b^2) = 2sqrt(a^2 + b^2) units
Q3Determine if the points (1, 5), (2, 3) and (-2, -11) are collinear.
Let A = (1, 5), B = (2, 3), C = (-2, -11).
AB = sqrt((2 - 1)^2 + (3 - 5)^2) = sqrt(1 + 4) = sqrt(5)
BC = sqrt((-2 - 2)^2 + (-11 - 3)^2) = sqrt(16 + 196) = sqrt(212) = 2sqrt(53)
AC = sqrt((-2 - 1)^2 + (-11 - 5)^2) = sqrt(9 + 256) = sqrt(265)
For collinearity, AB + BC should equal AC (or similar combination).
sqrt(5) + 2sqrt(53) = sqrt(5) + 2sqrt(53)
Checking: (sqrt(5))^2 + ... Let us verify: AB^2 + BC^2 = 5 + 212 = 217, but AC^2 = 265.
Also AB + BC = sqrt(5) + sqrt(212) approximately 2.236 + 14.56 = 16.8, but AC = sqrt(265) approximately 16.28.
Since AB + BC is not equal to AC (and no other combination works), the points are NOT collinear.
Q4Check whether (5, -2), (6, 4) and (7, -2) are the vertices of an isosceles triangle.
Let A = (5, -2), B = (6, 4), C = (7, -2).
AB = sqrt((6 - 5)^2 + (4 - (-2))^2) = sqrt(1 + 36) = sqrt(37)
BC = sqrt((7 - 6)^2 + (-2 - 4)^2) = sqrt(1 + 36) = sqrt(37)
AC = sqrt((7 - 5)^2 + (-2 - (-2))^2) = sqrt(4 + 0) = 2
Since AB = BC = sqrt(37), two sides are equal.
Yes, the points form an isosceles triangle (with AB = BC).
Q6Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer: (-1, -2), (1, 0), (-1, 2), (-3, 0)
Let A = (-1, -2), B = (1, 0), C = (-1, 2), D = (-3, 0).
AB = sqrt((1-(-1))^2 + (0-(-2))^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2)
BC = sqrt((-1-1)^2 + (2-0)^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2)
CD = sqrt((-3-(-1))^2 + (0-2)^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2)
DA = sqrt((-1-(-3))^2 + (-2-0)^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2)
All four sides are equal: AB = BC = CD = DA = 2sqrt(2).
Diagonals:
AC = sqrt((-1-(-1))^2 + (2-(-2))^2) = sqrt(0 + 16) = 4
BD = sqrt((-3-1)^2 + (0-0)^2) = sqrt(16 + 0) = 4
Diagonals are also equal: AC = BD = 4.
Since all sides are equal AND both diagonals are equal, the quadrilateral is a square.
Q9Find the value of y for which the distance between the points P(2, -3) and Q(10, y) is 10 units.
Distance PQ = 10
sqrt((10 - 2)^2 + (y - (-3))^2) = 10
sqrt(64 + (y + 3)^2) = 10
Squaring both sides:
64 + (y + 3)^2 = 100
(y + 3)^2 = 36
y + 3 = +6 or y + 3 = -6
y = 3 or y = -9
7.2Exercise 7.2 - Section Formula
Q1Find the coordinates of the point which divides the join of (-1, 7) and (4, -3) in the ratio 2:3.
Using the section formula, a point dividing the line joining (x1, y1) and (x2, y2) in ratio m:n is:
P = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n))
Here (x1, y1) = (-1, 7), (x2, y2) = (4, -3), m:n = 2:3
x = (2(4) + 3(-1))/(2 + 3) = (8 - 3)/5 = 5/5 = 1
y = (2(-3) + 3(7))/(2 + 3) = (-6 + 21)/5 = 15/5 = 3
The required point is (1, 3).
Q2Find the coordinates of the points of trisection of the line segment joining (4, -1) and (-2, -3).
Points of trisection divide the segment into 3 equal parts.
Point P divides AB in ratio 1:2:
x = (1(-2) + 2(4))/(1 + 2) = (-2 + 8)/3 = 6/3 = 2
y = (1(-3) + 2(-1))/(1 + 2) = (-3 - 2)/3 = -5/3
P = (2, -5/3)
Point Q divides AB in ratio 2:1:
x = (2(-2) + 1(4))/(2 + 1) = (-4 + 4)/3 = 0/3 = 0
y = (2(-3) + 1(-1))/(2 + 1) = (-6 - 1)/3 = -7/3
Q = (0, -7/3)
The points of trisection are (2, -5/3) and (0, -7/3).
Q4Find the ratio in which the line segment joining the points (-3, 10) and (6, -8) is divided by (-1, 6).
Let the point (-1, 6) divide the line joining A(-3, 10) and B(6, -8) in the ratio k:1.
Using section formula for x-coordinate:
-1 = (k(6) + 1(-3))/(k + 1)
-1(k + 1) = 6k - 3
-k - 1 = 6k - 3
-1 + 3 = 6k + k
2 = 7k
k = 2/7
So the ratio is 2/7 : 1 = 2 : 7.
Verification using y-coordinate:
y = (2(-8) + 7(10))/(2 + 7) = (-16 + 70)/9 = 54/9 = 6 ✓
Q6If (1, 2), (4, y), (x, 6) and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
In a parallelogram, diagonals bisect each other. So the midpoint of diagonal AC equals the midpoint of diagonal BD.
A = (1, 2), B = (4, y), C = (x, 6), D = (3, 5)
Midpoint of AC = ((1 + x)/2, (2 + 6)/2) = ((1 + x)/2, 4)
Midpoint of BD = ((4 + 3)/2, (y + 5)/2) = (7/2, (y + 5)/2)
Equating x-coordinates:
(1 + x)/2 = 7/2
1 + x = 7
x = 6
Equating y-coordinates:
4 = (y + 5)/2
8 = y + 5
y = 3
Q8The three vertices of a triangle are A(2, 3), B(4, -1), and C(1, 2). Find the coordinates of the centroid of the triangle.
The centroid of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is:
G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)
A = (2, 3), B = (4, -1), C = (1, 2)
G = ((2 + 4 + 1)/3, (3 + (-1) + 2)/3)
G = (7/3, 4/3)
The centroid is (7/3, 4/3).
7.3Exercise 7.3 - Area of a Triangle
Q1Find the area of the triangle whose vertices are: (i) (2, 3), (-1, 0), (2, -4) (ii) (-5, -1), (3, -5), (5, 2)
Using area formula: Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
(i) (2, 3), (-1, 0), (2, -4):
Area = (1/2)|2(0 - (-4)) + (-1)((-4) - 3) + 2(3 - 0)|
= (1/2)|2(4) + (-1)(-7) + 2(3)|
= (1/2)|8 + 7 + 6|
= (1/2)(21)
= 10.5 sq. units
(ii) (-5, -1), (3, -5), (5, 2):
Area = (1/2)|(-5)(-5 - 2) + 3(2 - (-1)) + 5((-1) - (-5))|
= (1/2)|(-5)(-7) + 3(3) + 5(4)|
= (1/2)|35 + 9 + 20|
= (1/2)(64)
= 32 sq. units
Q2In each of the following find the value of k, for which the points are collinear: (i) (7, -2), (5, 1), (3, k) (ii) (8, 1), (k, -4), (2, -5)
For collinear points, the area of the triangle formed = 0.
(i) (7, -2), (5, 1), (3, k):
Area = (1/2)|7(1 - k) + 5(k - (-2)) + 3((-2) - 1)| = 0
|7 - 7k + 5k + 10 + (-9)| = 0
|7 - 7k + 5k + 10 - 9| = 0
|-2k + 8| = 0
-2k + 8 = 0
k = 4
(ii) (8, 1), (k, -4), (2, -5):
Area = (1/2)|8(-4 - (-5)) + k((-5) - 1) + 2(1 - (-4))| = 0
|8(1) + k(-6) + 2(5)| = 0
|8 - 6k + 10| = 0
|18 - 6k| = 0
18 - 6k = 0
k = 3
Q3Find the area of the triangle formed by joining the midpoints of the sides of the triangle whose vertices are (0, -1), (2, 1) and (0, 3). Find the ratio of the area of the triangle formed to the area of the given triangle.
Vertices: A(0, -1), B(2, 1), C(0, 3)
Midpoints:
D (midpoint of AB) = ((0+2)/2, (-1+1)/2) = (1, 0)
E (midpoint of BC) = ((2+0)/2, (1+3)/2) = (1, 2)
F (midpoint of AC) = ((0+0)/2, (-1+3)/2) = (0, 1)
Area of triangle ABC:
= (1/2)|0(1 - 3) + 2(3 - (-1)) + 0((-1) - 1)|
= (1/2)|0 + 8 + 0| = 4 sq. units
Area of triangle DEF:
= (1/2)|1(2 - 1) + 1(1 - 0) + 0(0 - 2)|
= (1/2)|1 + 1 + 0| = 1 sq. unit
Ratio = Area of DEF : Area of ABC = 1 : 4
(The triangle formed by joining midpoints is always 1/4 of the original triangle.)
Q4Find the area of the quadrilateral whose vertices, taken in order, are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Vertices: A(-4, -2), B(-3, -5), C(3, -2), D(2, 3)
Divide the quadrilateral into two triangles by diagonal AC.
Area of triangle ABC:
= (1/2)|(-4)(-5 - (-2)) + (-3)((-2) - (-2)) + 3((-2) - (-5))|
= (1/2)|(-4)(-3) + (-3)(0) + 3(3)|
= (1/2)|12 + 0 + 9|
= (1/2)(21) = 10.5 sq. units
Area of triangle ACD:
= (1/2)|(-4)((-2) - 3) + 3(3 - (-2)) + 2((-2) - (-2))|
= (1/2)|(-4)(-5) + 3(5) + 2(0)|
= (1/2)|20 + 15 + 0|
= (1/2)(35) = 17.5 sq. units
Area of quadrilateral ABCD = 10.5 + 17.5 = 28 sq. units
Q5You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for the triangle with vertices A(4, -6), B(3, -2) and C(5, 2).
Let D be the midpoint of BC.
D = ((3+5)/2, (-2+2)/2) = (4, 0)
Area of triangle ABD (vertices A(4,-6), B(3,-2), D(4,0)):
= (1/2)|4(-2 - 0) + 3(0 - (-6)) + 4((-6) - (-2))|
= (1/2)|4(-2) + 3(6) + 4(-4)|
= (1/2)|-8 + 18 - 16|
= (1/2)|-6| = 3 sq. units
Area of triangle ACD (vertices A(4,-6), C(5,2), D(4,0)):
= (1/2)|4(2 - 0) + 5(0 - (-6)) + 4((-6) - 2)|
= (1/2)|8 + 30 - 32|
= (1/2)|6| = 3 sq. units
Area of ABD = Area of ACD = 3 sq. units.
The median AD divides triangle ABC into two triangles of equal area. Verified.
old-7.4Old Exercise 7.4 (For State Boards)
This exercise has been removed from the CBSE 2026-27 syllabus. It is provided here for UP Board, Bihar Board, MP Board, and other state board students who still follow the older textbook.
Q1Determine the ratio in which the line 2x + y - 4 = 0 divides the line segment joining the points A(2, -2) and B(3, 7).
Let the line 2x + y - 4 = 0 divide AB in the ratio k:1.
Point of division = ((3k + 2)/(k + 1), (7k - 2)/(k + 1))
This point lies on the line 2x + y - 4 = 0:
2((3k + 2)/(k + 1)) + ((7k - 2)/(k + 1)) - 4 = 0
(6k + 4 + 7k - 2)/(k + 1) = 4
(13k + 2)/(k + 1) = 4
13k + 2 = 4k + 4
9k = 2
k = 2/9
The ratio is 2 : 9.
Q2Find a relation between x and y if the points (x, y), (1, 2) and (7, 0) are collinear.
For collinear points, area = 0:
(1/2)|x(2 - 0) + 1(0 - y) + 7(y - 2)| = 0
|2x - y + 7y - 14| = 0
|2x + 6y - 14| = 0
2x + 6y - 14 = 0
x + 3y - 7 = 0
x + 3y = 7
Q5Find the area of a rhombus if its vertices are (3, 0), (4, 5), (-1, 4) and (-2, -1) taken in order.
For a rhombus, Area = (1/2) x d1 x d2 where d1, d2 are the diagonals.
Vertices: A(3, 0), B(4, 5), C(-1, 4), D(-2, -1)
Diagonal AC = sqrt((-1-3)^2 + (4-0)^2) = sqrt(16 + 16) = sqrt(32) = 4sqrt(2)
Diagonal BD = sqrt((-2-4)^2 + (-1-5)^2) = sqrt(36 + 36) = sqrt(72) = 6sqrt(2)
Area = (1/2) x 4sqrt(2) x 6sqrt(2) = (1/2) x 48 = 24 sq. units
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Ask AI TutorPrevious Year Board Exam Questions
Find the ratio in which the point (-3, k) divides the line segment joining the points (-5, -4) and (-2, 3). Also find the value of k.
Let (-3, k) divide the join of A(-5, -4) and B(-2, 3) in the ratio m:1.
Using section formula for x-coordinate: -3 = (m(-2) + 1(-5))/(m + 1)
-3(m + 1) = -2m - 5
-3m - 3 = -2m - 5
-m = -2
m = 2
Ratio = 2:1.
Now k = (2(3) + 1(-4))/(2 + 1) = (6 - 4)/3 = 2/3.
Ratio is 2:1 and k = 2/3.
Find the distance between the points A(2, -1) and B(5, 3). Hence show that the points A(2, -1), B(5, 3) and C(11, 11) are collinear.
AB = sqrt((5 - 2)^2 + (3 - (-1))^2) = sqrt(9 + 16) = sqrt(25) = 5.
BC = sqrt((11 - 5)^2 + (11 - 3)^2) = sqrt(36 + 64) = sqrt(100) = 10.
AC = sqrt((11 - 2)^2 + (11 - (-1))^2) = sqrt(81 + 144) = sqrt(225) = 15.
Since AB + BC = 5 + 10 = 15 = AC, the points A, B, C are collinear.
Find the coordinates of the point which divides the line segment joining the points (4, -3) and (8, 5) in the ratio 3:1 internally.
Using section formula with m:n = 3:1, A(4, -3), B(8, 5):
x = (3(8) + 1(4))/(3 + 1) = (24 + 4)/4 = 28/4 = 7
y = (3(5) + 1(-3))/(3 + 1) = (15 - 3)/4 = 12/4 = 3
The required point is (7, 3).
Find the area of the triangle whose vertices are (2, 3), (4, 7) and (6, 2).
Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
= (1/2)|2(7 - 2) + 4(2 - 3) + 6(3 - 7)|
= (1/2)|2(5) + 4(-1) + 6(-4)|
= (1/2)|10 - 4 - 24|
= (1/2)|-18| = (1/2)(18) = 9 sq. units
Show that the points A(-1, -1), B(1, 1) and C(-sqrt(3), sqrt(3)) are the vertices of an equilateral triangle.
AB = sqrt((1 - (-1))^2 + (1 - (-1))^2) = sqrt(4 + 4) = sqrt(8) = 2sqrt(2)
BC = sqrt((-sqrt(3) - 1)^2 + (sqrt(3) - 1)^2) = sqrt(3 + 2sqrt(3) + 1 + 3 - 2sqrt(3) + 1) = sqrt(8) = 2sqrt(2)
AC = sqrt((-sqrt(3) - (-1))^2 + (sqrt(3) - (-1))^2) = sqrt((1 - sqrt(3))^2 + (sqrt(3) + 1)^2) = sqrt(1 - 2sqrt(3) + 3 + 3 + 2sqrt(3) + 1) = sqrt(8) = 2sqrt(2)
Since AB = BC = AC = 2sqrt(2), triangle ABC is equilateral.
If the point C(-1, 2) divides internally the line segment joining A(2, 5) and B in the ratio 3:4, find the coordinates of B.
Let B = (x, y). C(-1, 2) divides AB in ratio 3:4.
-1 = (3x + 4(2))/(3 + 4) => -7 = 3x + 8 => 3x = -15 => x = -5
2 = (3y + 4(5))/(3 + 4) => 14 = 3y + 20 => 3y = -6 => y = -2
B = (-5, -2)
If the distance between the points (3, -5) and (x, -5) is 15 units, find the values of x.
Distance = sqrt((x - 3)^2 + (-5 - (-5))^2) = 15
sqrt((x - 3)^2 + 0) = 15
|x - 3| = 15
x - 3 = 15 or x - 3 = -15
x = 18 or x = -12
Find the coordinates of the point which divides the line segment joining the points (1, -2) and (4, 7) in the ratio 1:2.
Using section formula with m:n = 1:2, A(1, -2), B(4, 7):
x = (1(4) + 2(1))/(1 + 2) = (4 + 2)/3 = 6/3 = 2
y = (1(7) + 2(-2))/(1 + 2) = (7 - 4)/3 = 3/3 = 1
The required point is (2, 1).
Find the value of a for which the points A(a, -1), B(5, 1) and C(a, 4) are collinear.
For collinear points, area of triangle = 0.
(1/2)|a(1 - 4) + 5(4 - (-1)) + a((-1) - 1)| = 0
|a(-3) + 5(5) + a(-2)| = 0
|-3a + 25 - 2a| = 0
|-5a + 25| = 0
-5a + 25 = 0
a = 5
Find the point on the y-axis which is equidistant from the points (5, -2) and (-3, 2).
Any point on the y-axis is of the form (0, y).
Distance from (0, y) to (5, -2) = Distance from (0, y) to (-3, 2)
sqrt(25 + (y + 2)^2) = sqrt(9 + (y - 2)^2)
Squaring: 25 + y^2 + 4y + 4 = 9 + y^2 - 4y + 4
25 + 4y + 4 = 9 - 4y + 4
8y = 9 + 4 - 25 - 4 = -16
y = -2
The point is (0, -2).
The vertices of a triangle are A(1, 2), B(4, 5) and C(6, 1). Find the area of the triangle ABC.
Area = (1/2)|1(5 - 1) + 4(1 - 2) + 6(2 - 5)|
= (1/2)|1(4) + 4(-1) + 6(-3)|
= (1/2)|4 - 4 - 18|
= (1/2)|-18| = 9 sq. units
Find the centroid of the triangle whose vertices are (3, -5), (-7, 4) and (10, -2).
Centroid G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3)
= ((3 + (-7) + 10)/3, ((-5) + 4 + (-2))/3)
= (6/3, -3/3)
= (2, -1)
Find the distance between the points A(5, -3) and B(-7, -2).
Using the distance formula d = sqrt((x2 - x1)^2 + (y2 - y1)^2), we get:
AB = sqrt((-7 - 5)^2 + (-2 - (-3))^2) = sqrt((-12)^2 + (1)^2) = sqrt(144 + 1) = sqrt(145).
Therefore the distance between points A(5, -3) and B(-7, -2) is sqrt(145) units.
Find the ratio in which the line segment joining A(1, -5) and B(-4, 5) is divided by the x-axis. Also find the coordinates of the point of division.
Let the x-axis divide AB in ratio k:1. Any point on x-axis has y = 0.
y-coordinate: (k(5) + 1(-5))/(k + 1) = 0
5k - 5 = 0
k = 1
Ratio = 1:1 (i.e., the x-axis bisects the segment).
x-coordinate: (1(-4) + 1(1))/(1 + 1) = (-4 + 1)/2 = -3/2
Ratio is 1:1 and the point is (-3/2, 0).
If the point P(x, y) is equidistant from the points A(5, 1) and B(-1, 5), prove that 3x = 2y.
PA = PB
sqrt((x - 5)^2 + (y - 1)^2) = sqrt((x + 1)^2 + (y - 5)^2)
Squaring: (x - 5)^2 + (y - 1)^2 = (x + 1)^2 + (y - 5)^2
x^2 - 10x + 25 + y^2 - 2y + 1 = x^2 + 2x + 1 + y^2 - 10y + 25
-10x - 2y + 26 = 2x - 10y + 26
-12x + 8y = 0
-12x = -8y
12x = 8y
3x = 2y (dividing by 4). Hence proved.
Find the area of the triangle formed by the points (1, -1), (-4, 6), and (-3, -5).
Area = (1/2)|1(6 - (-5)) + (-4)((-5) - (-1)) + (-3)((-1) - 6)|
= (1/2)|1(11) + (-4)(-4) + (-3)(-7)|
= (1/2)|11 + 16 + 21|
= (1/2)(48) = 24 sq. units
Want more PYQ-style practice? AI generates unlimited similar questions from this chapter.
Practice PYQ PatternsAssertion-Reason Questions
(A) Both true, Reason explains Assertion · (B) Both true, Reason doesn't explain · (C) Assertion true, Reason false · (D) Assertion false, Reason true
AR1
Assertion (A): The distance between the points (3, 4) and (0, 0) is 5 units.
Reason (R): The distance formula is d = sqrt((x2 - x1)^2 + (y2 - y1)^2).
Answer: (A)
Both are true. d = sqrt((3-0)^2 + (4-0)^2) = sqrt(9+16) = sqrt(25) = 5. The reason (distance formula) correctly explains why the assertion is true.
AR2
Assertion (A): The midpoint of the line segment joining (2, 3) and (4, 7) is (3, 5).
Reason (R): The midpoint formula is M = ((x1+x2)/2, (y1+y2)/2).
Answer: (A)
Both are true. M = ((2+4)/2, (3+7)/2) = (3, 5). The reason (midpoint formula) correctly explains the assertion.
AR3
Assertion (A): The points (1, 1), (2, 3), and (3, 5) are collinear.
Reason (R): Three points are collinear if the area of the triangle formed by them is zero.
Answer: (A)
Area = (1/2)|1(3-5) + 2(5-1) + 3(1-3)| = (1/2)|(-2) + 8 + (-6)| = (1/2)|0| = 0. Since area is 0, points are collinear. Both assertion and reason are true, and the reason explains the assertion.
AR4
Assertion (A): The point (-2, 5) lies on the perpendicular bisector of the line segment joining (1, 2) and (5, 8).
Reason (R): A point on the perpendicular bisector is equidistant from the endpoints of the segment.
Answer: (D)
Reason is TRUE. For the assertion, check distances: from (-2,5) to (1,2) = sqrt(9+9) = sqrt(18); from (-2,5) to (5,8) = sqrt(49+9) = sqrt(58). Since sqrt(18) is not equal to sqrt(58), the point is NOT equidistant. Assertion is FALSE.
AR5
Assertion (A): The centroid of a triangle with vertices (0, 0), (6, 0), and (0, 6) is (2, 2).
Reason (R): The centroid divides each median in the ratio 1:2 from the vertex.
Answer: (A)
Centroid = ((0+6+0)/3, (0+0+6)/3) = (2, 2). The assertion is true. The reason is also true and explains the centroid property. Both true, reason explains assertion.
AR6
Assertion (A): The point (3, 3) divides the segment joining (1, 1) and (5, 5) in the ratio 1:1.
Reason (R): The section formula gives the coordinates of a point dividing a line segment in a given ratio.
Answer: (A)
Midpoint of (1,1) and (5,5) = ((1+5)/2, (1+5)/2) = (3,3). The ratio 1:1 means the midpoint. Both assertion and reason are true, and the reason (section formula) explains the assertion.
Case Study Questions
A city corporation planned to build a triangular park in a residential area. The coordinates of the three corners of the park on a map are A(2, 1), B(6, 5) and C(10, 1). The corporation wants to install a fountain at the centroid of the park and place streetlights at the midpoints of each side.
(i) Find the distance between vertices A and B. (1 mark)
AB = sqrt((6-2)^2 + (5-1)^2) = sqrt(16 + 16) = sqrt(32) = 4sqrt(2) units
(ii)(a) Find the coordinates of the centroid where the fountain is to be placed. (2 marks)
Centroid G = ((2 + 6 + 10)/3, (1 + 5 + 1)/3) = (18/3, 7/3) = (6, 7/3)
(ii)(b) Find the coordinates of the midpoint of side BC where one streetlight is placed. (1 mark)
Midpoint of BC = ((6+10)/2, (5+1)/2) = (16/2, 6/2) = (8, 3)
(iii) Find the area of the triangular park. (1 mark)
Area = (1/2)|2(5-1) + 6(1-1) + 10(1-5)|
= (1/2)|2(4) + 6(0) + 10(-4)|
= (1/2)|8 + 0 - 40|
= (1/2)(32) = 16 sq. units
During a Sports Day event, a relay race was organized on a coordinate grid painted on the school ground. Four relay stations were set up at points P(2, -3), Q(4, 1), R(0, 5) and S(-2, 1). Each runner runs from one station to the next.
(i) Find the distance from station P to station Q. (1 mark)
PQ = sqrt((4-2)^2 + (1-(-3))^2) = sqrt(4 + 16) = sqrt(20) = 2sqrt(5) units
(ii) Check whether PQRS is a parallelogram by finding the midpoints of its diagonals. (2 marks)
Midpoint of PR = ((2+0)/2, (-3+5)/2) = (1, 1)
Midpoint of QS = ((4+(-2))/2, (1+1)/2) = (1, 1)
Since both diagonals have the same midpoint (1, 1), the diagonals bisect each other.
PQRS is a parallelogram.
(iii) Find the total distance covered by a runner who runs from P to Q to R to S and back to P. (2 marks)
PQ = sqrt((4-2)^2 + (1+3)^2) = sqrt(4+16) = sqrt(20) = 2sqrt(5)
QR = sqrt((0-4)^2 + (5-1)^2) = sqrt(16+16) = sqrt(32) = 4sqrt(2)
RS = sqrt((-2-0)^2 + (1-5)^2) = sqrt(4+16) = sqrt(20) = 2sqrt(5)
SP = sqrt((2-(-2))^2 + (-3-1)^2) = sqrt(16+16) = sqrt(32) = 4sqrt(2)
Total = 2sqrt(5) + 4sqrt(2) + 2sqrt(5) + 4sqrt(2) = 4sqrt(5) + 8sqrt(2) units
MCQ Practice
MCQ1. The distance of the point P(2, 3) from the x-axis is:
(A) 2
(B) 3
(C) 1
(D) 5
Answer: (B) - The distance of any point (x, y) from the x-axis equals the absolute value of its y-coordinate. Distance = |3| = 3.
MCQ2. The point which divides the line segment joining (2, -5) and (5, 2) in the ratio 2:3 is:
(A) (0, -1)
(B) (16/5, -11/5)
(C) (3, -1)
(D) (16/5, 11/5)
Answer: (B) - x = (2(5) + 3(2))/(2+3) = (10+6)/5 = 16/5. y = (2(2) + 3(-5))/(2+3) = (4-15)/5 = -11/5. Point is (16/5, -11/5).
MCQ3. The distance between the points (a, b) and (-a, -b) is:
(A) sqrt(a^2 + b^2)
(B) a^2 + b^2
(C) 2sqrt(a^2 + b^2)
(D) 4(a^2 + b^2)
Answer: (C) - d = sqrt((-a-a)^2 + (-b-b)^2) = sqrt(4a^2 + 4b^2) = 2sqrt(a^2 + b^2).
MCQ4. If the centroid of a triangle formed by (7, x), (y, -6) and (9, 10) is (6, 3), then (x, y) is:
(A) (5, 2)
(B) (2, 5)
(C) (4, 2)
(D) (2, 4)
Answer: (A) - Centroid = ((7+y+9)/3, (x-6+10)/3) = (6, 3). So (16+y)/3 = 6 => y = 2. And (x+4)/3 = 3 => x = 5. So (x, y) = (5, 2).
MCQ5. If the distance between (4, p) and (1, 0) is 5 units, then the value of p is:
(A) 4 only
(B) -4 only
(C) +4 or -4
(D) 0
Answer: (C) - sqrt((4-1)^2 + (p-0)^2) = 5 => 9 + p^2 = 25 => p^2 = 16 => p = +4 or -4.
MCQ6. The area of the triangle with vertices (0, 0), (3, 0) and (0, 4) is:
(A) 12 sq. units
(B) 6 sq. units
(C) 7 sq. units
(D) 24 sq. units
Answer: (B) - Area = (1/2)|0(0-4) + 3(4-0) + 0(0-0)| = (1/2)|0 + 12 + 0| = 6 sq. units.
MCQ7. The perimeter of a triangle with vertices (0, 4), (0, 0) and (3, 0) is:
(A) 5 units
(B) 12 units
(C) 11 units
(D) 7 + 5 = 12 units
Answer: (B) - Side 1 (0,4 to 0,0) = 4. Side 2 (0,0 to 3,0) = 3. Side 3 (3,0 to 0,4) = sqrt(9+16) = 5. Perimeter = 4 + 3 + 5 = 12 units.
MCQ8. The midpoint of the line segment joining (a, b) and (a + 2b, 2a - b) is:
(A) (a + b, a)
(B) (a, a + b)
(C) (2a + b, 2a + b)
(D) (a + b, (3a - b)/2)
Answer: (A) - Midpoint = ((a + a + 2b)/2, (b + 2a - b)/2) = ((2a + 2b)/2, 2a/2) = (a + b, a).
MCQ9. If the points (a, 0), (0, b) and (1, 1) are collinear, then 1/a + 1/b equals:
(A) 0
(B) 1
(C) 2
(D) -1
Answer: (B) - For collinearity: a(b - 1) + 0(1 - 0) + 1(0 - b) = 0 => ab - a - b = 0 => ab = a + b. Dividing by ab: 1 = 1/b + 1/a. So 1/a + 1/b = 1.
MCQ10. The coordinates of the point which is equidistant from the three vertices of a right triangle with vertices O(0, 0), A(0, 2y) and B(2x, 0) is:
(A) (x, y)
(B) (y, x)
(C) (x/2, y/2)
(D) (2x, 2y)
Answer: (A) - The circumcentre of a right triangle lies at the midpoint of the hypotenuse. Midpoint of AB = ((0+2x)/2, (2y+0)/2) = (x, y).
MCQ11. If the distance between (8, p) and (4, 3) is 5 units, then the value of p is:
(A) 6
(B) 0
(C) 6 or 0
(D) 3
Answer: (C) - sqrt((8-4)^2 + (p-3)^2) = 5 => 16 + (p-3)^2 = 25 => (p-3)^2 = 9 => p - 3 = +3 or -3 => p = 6 or 0.
MCQ12. If the point P(k, 0) divides the line segment joining A(2, -2) and B(-7, 4) in the ratio 1:2, then the value of k is:
(A) 1
(B) -1
(C) 2
(D) -2
Answer: (B) - k = (1(-7) + 2(2))/(1+2) = (-7 + 4)/3 = -3/3 = -1. Verification: y = (1(4) + 2(-2))/3 = (4-4)/3 = 0. So point is (-1, 0). k = -1.
Try Quiz mode - 10 MCQs per topic with instant feedback and scoring.
Start Quiz ModeImportant Questions - Mark-wise
1 Mark Questions
Find the distance between the points (3, 4) and (0, 0).
d = sqrt((3-0)^2 + (4-0)^2) = sqrt(9 + 16) = sqrt(25) = 5 units.
Find the midpoint of the line segment joining (2, 7) and (6, 1).
Midpoint = ((2+6)/2, (7+1)/2) = (4, 4).
What are the coordinates of the centroid of a triangle with vertices (1, 3), (5, 7) and (3, 2)?
Centroid = ((1+5+3)/3, (3+7+2)/3) = (9/3, 12/3) = (3, 4).
Find the distance of the point (8, -6) from the origin.
d = sqrt(8^2 + (-6)^2) = sqrt(64 + 36) = sqrt(100) = 10 units.
What is the distance between (2, 0) and (0, 3)?
d = sqrt((2-0)^2 + (0-3)^2) = sqrt(4 + 9) = sqrt(13) units.
2-3 Mark Questions
Find the ratio in which the y-axis divides the line segment joining the points (5, -6) and (-1, -4).
Let the y-axis divide the segment in ratio k:1. The x-coordinate of the point on y-axis is 0.
0 = (k(-1) + 1(5))/(k+1)
0 = -k + 5
k = 5
Ratio = 5:1.
Find the coordinates of the point which divides the join of (4, -3) and (8, 5) in the ratio 3:1.
x = (3(8) + 1(4))/(3+1) = (24+4)/4 = 7
y = (3(5) + 1(-3))/(3+1) = (15-3)/4 = 3
Point = (7, 3).
Find the point on the x-axis which is equidistant from (2, -5) and (-2, 9).
Let the point be (x, 0).
sqrt((x-2)^2 + 25) = sqrt((x+2)^2 + 81)
Squaring: x^2 - 4x + 4 + 25 = x^2 + 4x + 4 + 81
-8x = 56
x = -7
Point = (-7, 0).
Prove that the points (1, -2), (3, 0), (1, 2) and (-1, 0) are the vertices of a square.
A(1,-2), B(3,0), C(1,2), D(-1,0).
AB = sqrt(4+4) = 2sqrt(2). BC = sqrt(4+4) = 2sqrt(2). CD = sqrt(4+4) = 2sqrt(2). DA = sqrt(4+4) = 2sqrt(2).
All sides equal. AC = sqrt(0+16) = 4. BD = sqrt(16+0) = 4. Diagonals equal.
ABCD is a square.
3-4 Mark Questions
If A(1, 2), B(4, 3) and C(6, 6) are the three vertices of a parallelogram ABCD, find the coordinates of the fourth vertex D.
In a parallelogram, diagonals bisect each other. Let D = (x, y).
Midpoint of AC = ((1+6)/2, (2+6)/2) = (7/2, 4)
Midpoint of BD = ((4+x)/2, (3+y)/2)
Equating: (4+x)/2 = 7/2 => x = 3. (3+y)/2 = 4 => y = 5.
D = (3, 5).
Find the area of the quadrilateral whose vertices taken in order are (-4, -2), (-3, -5), (3, -2) and (2, 3).
Divide into triangles ABC and ACD where A(-4,-2), B(-3,-5), C(3,-2), D(2,3).
Area ABC = (1/2)|(-4)(-5+2) + (-3)(-2+2) + 3(-2+5)| = (1/2)|12 + 0 + 9| = 10.5
Area ACD = (1/2)|(-4)(-2-3) + 3(3+2) + 2(-2+2)| = (1/2)|20 + 15 + 0| = 17.5
Total area = 28 sq. units.
The vertices of a triangle are A(2, -4), B(6, -2) and C(5, 3). Find the area of the triangle. Also find the length of the median through vertex A.
Area = (1/2)|2(-2-3) + 6(3-(-4)) + 5((-4)-(-2))|
= (1/2)|2(-5) + 6(7) + 5(-2)|
= (1/2)|-10 + 42 - 10| = (1/2)(22) = 11 sq. units
Midpoint of BC (let it be D) = ((6+5)/2, (-2+3)/2) = (11/2, 1/2)
AD = sqrt((11/2 - 2)^2 + (1/2 + 4)^2) = sqrt((7/2)^2 + (9/2)^2) = sqrt(49/4 + 81/4) = sqrt(130/4) = sqrt(130)/2 units.
Key Concepts
Important Formulas
Study Plan - Coordinate Geometry
Distance Formula
Read NCERT theory on Cartesian plane and distance formula. Solve Exercise 7.1 Q1-Q5. Focus on computing sqrt values accurately.
Distance Formula Applications
Complete Exercise 7.1 Q6-Q10. Practice identifying type of triangle and quadrilateral using distances. Focus on equidistant point problems.
Section Formula and Midpoint
Learn section formula and midpoint formula. Solve Exercise 7.2 Q1-Q5. Practice trisection and ratio-finding problems.
Area of Triangle and Collinearity
Learn the area formula. Solve Exercise 7.2 Q6-Q10 and all of Exercise 7.3. Practice collinearity checks and area of quadrilateral.
PYQs, MCQs, and Full Revision
Solve all 16 PYQs, all MCQs, AR questions, and case study questions under timed conditions. Review common mistakes and exam tips.
Common Mistakes to Avoid
- ✗Swapping (x1, y1) and (x2, y2) in the section formula: remember the ratio m:n means m is towards the second point and n is towards the first point
- ✗Forgetting to take the absolute value when computing area of a triangle using the coordinate formula, which can give a negative result
- ✗Using the section formula with the wrong ratio direction: if point P divides AB in ratio k:1, then P = (k*B + 1*A)/(k+1), not the other way around
- ✗Making sign errors when substituting negative coordinates into the distance or section formula: always use brackets around negative values
- ✗Confusing collinearity (area = 0) with the distance condition (AB + BC = AC): both methods work, but the area method is faster and less error-prone
- ✗Forgetting to verify the y-coordinate after finding the ratio using the x-coordinate in section formula problems: examiners expect verification for full marks
Tips for Board Exam
- ✓Always draw a rough figure on the coordinate plane before solving: this helps identify which formula to use and catches sign errors early
- ✓For section formula problems, clearly state the ratio m:n and label which point is A (first) and which is B (second): the order matters
- ✓In area of triangle questions, arrange the formula neatly as (1/2)|x1(y2-y3) + x2(y3-y1) + x3(y1-y2)| and compute each term separately to avoid errors
- ✓For collinearity questions, use the area formula (set area = 0) rather than the distance method: it involves less computation and fewer chances of error
- ✓MCQ shortcut: if a question asks for the type of quadrilateral, compute all 4 sides first, then diagonals only if needed. Equal sides + equal diagonals = square, equal sides + unequal diagonals = rhombus
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Start Coordinate Geometry - Choose Your ModeAbout Coordinate Geometry - Class 10 Mathematics
Chapter 7 Coordinate Geometry is one of the most important chapters in CBSE Class 10 Mathematics. It belongs to the Coordinate Geometry unit which carries 6 marks in the board exam. The chapter builds on the basic concepts of plotting points on the Cartesian plane (learned in Class 9) and introduces three powerful formulas: the distance formula, the section formula, and the area of a triangle formula.
What is Coordinate Geometry?
Coordinate Geometry (also called Analytical Geometry) is the branch of mathematics that uses algebraic methods to solve geometric problems. Every point in a plane can be represented by an ordered pair (x, y) called its coordinates. The horizontal axis is the x-axis and the vertical axis is the y-axis, and they intersect at the origin O(0, 0). This system was introduced by the French mathematician Rene Descartes.
The Distance Formula
The distance between two points P(x1, y1) and Q(x2, y2) in the Cartesian plane is given by:
d = sqrt((x2 - x1)^2 + (y2 - y1)^2)
This formula is derived from the Pythagorean theorem. If you draw a horizontal line from P and a vertical line from Q, they form a right triangle, and the distance PQ is the hypotenuse. The distance of any point P(x, y) from the origin O(0, 0) is simply sqrt(x^2 + y^2).
The distance formula has several important applications in Class 10:
- Finding the distance between two given points
- Checking whether three points are collinear (if AB + BC = AC, the points are collinear)
- Determining the type of triangle formed by three points (equilateral, isosceles, scalene, or right-angled)
- Determining the type of quadrilateral formed by four points (square, rectangle, rhombus, or parallelogram)
- Finding a point equidistant from two or more given points
The Section Formula
If a point P(x, y) divides the line segment joining A(x1, y1) and B(x2, y2) internally in the ratio m:n, then:
x = (mx2 + nx1)/(m + n) y = (my2 + ny1)/(m + n)
This is called the section formula. The key thing to remember is that m is the ratio towards B (the second point) and n is towards A (the first point). A common mistake is to reverse this order.
When the ratio is 1:1, the section formula simplifies to the midpoint formula:
M = ((x1 + x2)/2, (y1 + y2)/2)
The section formula is used for:
- Finding the coordinates of a point that divides a line segment in a given ratio
- Finding the ratio in which a given point divides a line segment (assume ratio k:1 and solve)
- Finding the midpoint of a line segment
- Finding the points of trisection (ratios 1:2 and 2:1)
- Proving that diagonals of a parallelogram bisect each other (midpoints of both diagonals are equal)
- Finding the centroid of a triangle: G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3). The centroid is the point where all three medians intersect.
Area of a Triangle
The area of a triangle with vertices A(x1, y1), B(x2, y2) and C(x3, y3) is:
Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
The absolute value is important because the formula can give a negative result depending on the order of vertices, but area is always positive. If the area equals zero, the three points are collinear (they lie on the same straight line).
To find the area of a quadrilateral with vertices A, B, C, D (taken in order), divide it into two triangles using one diagonal. For example, diagonal AC divides quadrilateral ABCD into triangles ABC and ACD. The total area is the sum of the areas of these two triangles.
Collinearity of Points
Three points are said to be collinear if they all lie on the same straight line. There are two ways to check collinearity in coordinate geometry:
- Area method: If the area of the triangle formed by the three points is zero, the points are collinear. This is the preferred method in board exams.
- Distance method: If for points A, B, C, the sum AB + BC = AC (or any permutation), the points are collinear.
Exercise Structure for CBSE 2026-27
Exercise 7.1 contains 10 questions on the distance formula. Students practice finding distances, checking collinearity using the distance method, identifying types of triangles and quadrilaterals, and finding points satisfying given distance conditions.
Exercise 7.2 contains 10 questions on the section formula. Problems include finding a point dividing a segment in a given ratio, trisection points, finding the ratio when the dividing point is known, midpoint problems, centroid problems, and parallelogram diagonal properties.
Exercise 7.3 contains 5 questions on the area of a triangle. Students find areas, check collinearity using the area method, and compute the area of quadrilaterals by dividing them into triangles.
The older Exercise 7.4 has been removed from the CBSE 2026-27 syllabus but is still relevant for state board students.
CBSE Exam Pattern and Weightage
Coordinate Geometry typically contributes questions worth 5-6 marks. The typical breakdown is:
- 1 MCQ on distance formula or section formula (1 mark)
- 1 short answer on section formula, midpoint, or collinearity (2-3 marks)
- 1 question on area of triangle or proving a geometric property (3 marks)
Since 2023, case study questions have been introduced that combine multiple concepts from this chapter. A single case study might ask students to find distances, midpoints, and areas within a real-world context.
Connection to Other Chapters
Coordinate Geometry connects strongly with Chapter 6 (Triangles) for proving properties of geometric figures using coordinates, with Chapter 3 (Pair of Linear Equations) for understanding graphical representation, and with Chapter 10 (Circles) for problems involving the distance of a point from the centre of a circle. For students preparing for competitive exams, coordinate geometry extends to straight lines, circles, and conic sections in Classes 11-12.
Important Problem Types
The most frequently asked problem types in board exams and competitive exams from this chapter are:
- Finding the distance between two points: Direct application of the distance formula. Always simplify the square root at the end.
- Finding a point equidistant from two given points: Set the two distances equal, square both sides, and simplify to get the required coordinates.
- Type of triangle or quadrilateral: Compute all side lengths using the distance formula. For quadrilaterals, also compute diagonal lengths. Equal sides with equal diagonals make a square. Equal sides with unequal diagonals make a rhombus. Equal diagonals with unequal sides make a rectangle.
- Section formula ratio problems: Assume the ratio as k:1, substitute into the section formula, and solve for k using one coordinate. Always verify with the other coordinate.
- Trisection problems: Use ratios 1:2 and 2:1 to find the two trisection points.
- Area of triangle and collinearity: Apply the area formula. For collinearity, set area equal to zero and solve.
- Area of quadrilateral: Divide into two triangles and add areas.
Key Points to Remember
- The distance formula is derived from the Pythagorean theorem
- In the section formula, the ratio m:n means m parts towards B and n parts towards A
- The centroid always divides each median in the ratio 2:1 from the vertex
- Area of a triangle using coordinates will be zero if and only if the points are collinear
- Always take the absolute value of the area formula result
- To find the type of quadrilateral: compute all 4 sides and both diagonals, then classify
- The midpoint formula is a special case of the section formula with m = n = 1
- For ratio-finding problems, assume ratio k:1 and solve for k using one coordinate, then verify using the other
- When computing areas, arrange vertices either clockwise or anticlockwise consistently
- The distance of a point (x, y) from the x-axis is |y| and from the y-axis is |x|
- For problems involving a point on the x-axis, its y-coordinate is 0 and vice versa
Frequently Asked Questions
How many exercises are in NCERT Class 10 Maths Chapter 7 for CBSE 2026-27?
For CBSE 2026-27, Chapter 7 Coordinate Geometry has 3 exercises: Exercise 7.1 (10 questions on distance formula), Exercise 7.2 (10 questions on section formula), and Exercise 7.3 (5 questions on area of triangle). The older Exercise 7.4 has been removed from the CBSE syllabus.
What is the distance formula in Coordinate Geometry?
The distance between two points P(x1, y1) and Q(x2, y2) is d = sqrt((x2 - x1)^2 + (y2 - y1)^2). This formula is derived from the Pythagorean theorem by forming a right triangle with horizontal and vertical sides.
How many marks does Coordinate Geometry carry in the CBSE Class 10 board exam?
Coordinate Geometry is part of its own unit which carries 6 marks in the CBSE Class 10 board exam. This typically includes 1 MCQ (1 mark), 1 short answer question (2-3 marks), and 1 problem on area or section formula (3 marks).
What is the section formula and when is it used?
The section formula gives the coordinates of a point that divides a line segment joining (x1, y1) and (x2, y2) internally in the ratio m:n. The formula is: P = ((mx2 + nx1)/(m + n), (my2 + ny1)/(m + n)). It is used for finding dividing points, midpoints, trisection points, and proving parallelogram properties.
How do you check if three points are collinear using coordinates?
Use the area formula: Area = (1/2)|x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|. If the area is 0, the three points are collinear (they lie on the same line). Alternatively, you can use the distance method: if AB + BC = AC for points A, B, C in order.
What is the centroid formula for a triangle?
The centroid of a triangle with vertices (x1, y1), (x2, y2), (x3, y3) is G = ((x1 + x2 + x3)/3, (y1 + y2 + y3)/3). The centroid is the point of intersection of all three medians and divides each median in the ratio 2:1 from the vertex.
Is Exercise 7.4 in the CBSE 2026-27 syllabus?
No. Exercise 7.4 has been removed from the CBSE 2026-27 syllabus. Only Exercises 7.1, 7.2, and 7.3 are part of the current CBSE syllabus. However, Exercise 7.4 is still relevant for state board students (UP Board, Bihar Board, MP Board, Rajasthan Board).
How do you find the area of a quadrilateral using coordinates?
Divide the quadrilateral into two triangles using one diagonal. Find the area of each triangle using the coordinate area formula, then add both areas. For example, for quadrilateral ABCD, use diagonal AC to get triangles ABC and ACD.