NCERT Solutions Class 10 Maths Chapter 7 Coordinate Geometry The team of subject matter experts at GFG have made detailed NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry to make sure that every student can understand how to solve Coordinate Geometry problems in a stepwise manner.
Coordination of the point of a line divided into two halves, section formula, the distance between two points, and areas of the triangle are some examples of problems covered in Coordinate Geometry.
Class 10 Maths NCERT Chapter 7 Coordinate Geometry Exercises 





This article provides solutions to all the problems asked in Class 10 Maths Chapter 7 Coordinate Geometry of your NCERT textbook in a stepbystep manner. They are regularly revised to check errors and updated according to the latest CBSE Syllabus 202324 and guidelines.
All the exercises in Class 10 Maths Chapter 7 Coordinate Geometry of your NCERT textbook have been properly covered in NCERT Solutions for Class 10 Maths.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry: Exercise 7.1
Question 1) Find the distance between the following pairs of points:
(i) (2,3), (4,1)
(ii) (5, 7), (1, 3)
(iii) (a, b), (a, b)
Solution:
Formula used in the above question is : √(x_{2 }– x_{1})^{2 }+ (y_{2 }– y_{1})^{2 }(i.e., Distance Formula)
(i) Here, x_{1} = 2, y_{1}= 3, x_{2} = 4, y_{2} = 1
Now, applying the distance formula :
= √(42)^{2} + (13)^{2}
=√(2)^{2} + (2)^{2}
= √8
= 2√2 units
(ii) Here, x_{1}= 5, y_{1}= 7, x_{2} = 1, y_{2} = 3
Now, applying the distance formula :
= √(1 – (5))^{2 }+ (3 – 7)^{2}
= √(4)^{2 }+ (4)^{2}
= 4√2
(iii) Here, x_{1} = a, y_{1} = b . x_{2} = a, y_{2} = b
Now, applying the distance formula :
= √(a – a)^{2} +(b – b)^{2}
= √(2a)^{2} + (2b)^{2}
= √4a^{2} + 4b^{2}
= 2√a^{2} + b^{2}
Question 2) Find the distance between the points (0, 0) and (36, 15). Can you now find the distance between the two towns A and B discussed in section 7.2.
Solution:
Formula used in the above question is : √(x_{2} – x_{1})^{2} + (y_{2} – y_{1})^{2} (i.e, Distance Formula)
Considering, Point A as (0, 0) and Point B as (36, 15) and applying the distance formula we get :
Distance between the two points : √(36 – 0)^{2} + (15 – 0)^{2}
= √(36)^{2} + (15)^{2}
= √1296 + 225
= √1521
= 39 units
Hence, the distance between two towns A and B is 39 units.
Question 3) Determine if the points (1, 5), (2, 3) and (–2, –11) are collinear.
Solution:
Let (1, 5), (2, 3) and (2, 11) be points A, B and C respectively.
Collinear term means that these 3 points lie in the same line. So, to we’ ll check it.
Using distance formula we will find the distance between these points.
AB = √(2 – 1)^{2 }+ (3 – 5)^{2}
=√(1)^{2} + (2)^{2} =√1 + 4 =√5
BC = √(2 – 2)^{2} + (11 – 3)^{2}
= √(4)^{2 }+ (14)^{2 }= √16 + 196 = √212
CA = √(2 – 1)^{2} + (11 – 5)^{2}
= √(3)^{2} + (16)^{2 }= √9 + 256 =√265
As, AB + BC ≠ AC (Since, one distance is not equal to sum of other two distances, we can say that they do not lie in the same line.)
Hence, points A, B and C are not collinear.
Question 4) Check whether (5, – 2), (6, 4) and (7, – 2) are the vertices of an isosceles triangle.
Solution:
Let (5, – 2), (6, 4) and (7, – 2) be the points A, B and C respectively.
Using distance formula :
AB = √(6 – 5)^{2} + (4 – (2))^{2}
= √(1 + 36) = √37
BC = √(7 – 6)^{2} + (2 – 4)^{2}
= √(1 + 36) = √37
AC = √(7 – 5)^{2} + (2 – (2))^{2}
= √(4 + 0) = 2
As, AB = BC ≠ AC (Two distances equal and one distance is not equal to sum of other two)
So, we can say that they are vertices of an isosceles triangle.
Question 5) In a classroom, 4 friends are seated at the points A, B, C and D as shown in Fig. 7.8. Champa and Chameli walk into the class and after observing for a few minutes Champa asks Chameli, “Don’t you think ABCD is a square?” Chameli disagrees. Using distance formula, find which of them is correct.
Solution:
From the given fig, find the coordinates of the points
AB = √(6 – 3) + (7 – 4)
= √9+9 = √18 = 3√2
BC = √(9 – 6) + (4 – 7)
= √9+9 = √18 = 3√2
CD = √(6 – 9) + (1 – 4)
= √9 + 9 = √18 = 3√2
DA = √(6 – 3) + (1 – 4)
= √9+9 =√18 =3√2
AB = BC = CD = DA = 3√2
All sides are of equal length. Therefore, ABCD is a square and hence, Champa was correct.
Question 6) Name the type of quadrilateral formed, if any, by the following points, and give reasons for your answer:
(i) ( 1, – 2), (1, 0), ( 1, 2), ( 3, 0)
(ii) ( 3, 5), (3, 1), (0, 3), ( 1, – 4)
(iii) (4, 5), (7, 6), (4, 3), (1, 2)
Solution:
(i) Here, let the given points are P(1, 2), Q(1, 0), R(1, 2) and S(3, 0) respectively.
PQ = √(1 – (1))² + (0 – (2))²
= √(1+1)²+(0+2)²
= √8 = 2 √2
QR = √(−1−1)²+(2−0)²
= √(−2)²+(2)²
= √8 = 2 √2
RS = √(−3−(−1))²+(0−2)²
= √8 = 2 √2
PS = √((−3−(−1))²+(0−(−2))²
= √8 = 2 √2
Here, we found that the length of all the sides are equal.
Diagonal PR = √(−1−(−1))²+(2−(−2))²
= √ 0+16
= 4
Diagonal QS = √(−3−1)²+(0−0)²
= √ 16 = 4
Finally, we also found that the length of diagonal are also same.
Here, PQ = QR = RS = PS = 2√2
and QS = PR = 4
This is the property of SQUARE. Hence, the given figure is SQUARE.
(ii) Let the points be P(3, 5), Q(3, 1), R(0, 3) and S(1, 4)
PQ = √(3−(−3))²+(1−5)²
= √(3+3)²+(−4)²
= √36+16 = √52 = 2 √13
QR = √(0−3)²+(3−1)²
= √(−3)²+(2)²
= √9 + 4 = √13
RS = √(−1−0)²+(−4−3)²
= √(−1)²+(−7)²
= √1+49 = √50 = 5 √2
PS = √(−1−(−3))²+(−4−5)²
= √(−1+3)²+(−9)²
= √4+81 = √85
Here, All the lengths of sides are unequal.
So, The given points will not create any quadrilateral.
(iii) Let the points be P(4, 5), Q(7, 6), R(4, 3) and S(1, 2)
PQ = √(7−4)²+(6−5)²
= √(3)²+(1)² = √9+1 = √10
QR = √(4−7)²+(3−6)²
= √(−3)²+(−3)² = √9+9 =3 √2
RS = √(1−4)²+(2−3)²
= √(−3)²+(−1)² = √9+1 = √10
PS = √(1−4)²+(2−5)²
= √(−3)²+(−3)² = √9+9 =3 √2
We see that the opposite sides are equal. Lets find the diagonal now.
Diagonal PR = √(4−4)²+(3−5)²
= √0+4 = 2
Diagonal QS = √(1−7)²+(2−6)²
= √36+16
= √52
Here, PQ = RS = √10
and QR = PS = 3√2
We see that the diagonals are not equal.
Hence, the formed quadrilateral is a PARALLELOGRAM.
Question 7) Find the point on the xaxis which is equidistant from (2, – 5) and ( 2, 9).
Solution:
Let the point on the X axis be (x, 0)
Given, distance between the points (2, 5), (x, 0) = distance between points (2, 9), (x, 0)
[ Applying Distance Formula ]
⇒ √(x – 2)²+(0 – (5))² = √(x – (2))²+(0 – 9)²
On squaring both the sides, we get
⇒ (x – 2)² + 5² = (x + 2)² + 9²
⇒ x² – 4x + 4 + 25 = x² + 4x + 4 + 81
⇒ 4x 4x = 85 – 29
⇒ 8x = 56
⇒ x = 7
Question 8) Find the values of y for which the distance between the points P (2, – 3) and Q (10, y) is 10 units.
Solution:
It is given that, the distance b/w two points is 10 units.
So, we’ll find the distance and equate
PQ = √ (10 – 2)^{2} + (y – (3))^{2}
= √ (8)^{2} + (y +3)^{2}
On squaring both the sides, we get :
64 +(y+3)^{2} = (10)^{2}
(y+3)^{2} = 36
y + 3 = ±6
y + 3 = +6 or y + 3 = −6
Hence, y = 3 or 9.
Question 9) If Q (0, 1) is equidistant from P (5, – 3) and R (x, 6), find the values of x. Also find the distance QR and PR.
Solution:
Given, PQ = QR
We will apply distance formula and find the distance between them,
PQ = √(5 – 0)^{2 }+ (3 – 1)^{2}
= √ ( 5)^{2 }+ (4)^{2}
= √ 25 + 16 = √41
QR = √ (0 – x)^{2} + (1 – 6)^{2}
= √ (x)^{2} + (5)^{2}
= √ x^{2} + 25
As they both are equal so on equating them, x^{2 }+ 25 = 41
x^{2 }=16, x = ± 4
So, putting the value of x and obtaining the value of QR and PR through distance formula
For x = +4, PR = √ (4 – 5)^{2} + (6 – (3))^{2}
= √ (1) ^{2}+ (9)^{2}
= √ 82
QR = √ (0 – 4)^{2} + (1 – 6)^{2}
= √ 41
For x = 4, QR = √ (0 – (4))^{2} + (1 – 6)^{2}
= √ 16 + 25 = √ 41
PR = √ (5 + 4)^{2 }+ (3 6)^{2}
= √ 81 + 81 = 9 √ 2
Question 10) Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and ( 3, 4).
Solution:
Let, (x, y) be point P and (3, 6), (3, 4) be pt A and B respectively.
It is given that their distance is equal, so we will equate the equations.
PA = PB (given)
⇒ √(x – 3)^{2 }+(y – 6)^{2} = √(x(3))^{2}+ (y – 4)^{2} [ By applying distance formula ]
On squaring both sides,
(x3)^{2}+(y6)^{2} = (x +3)^{2} +(y4)^{2}
x^{2} +96x+y^{2}+3612y = x^{2} +9+6x+y^{2} +168y
3616 = 6x+6x+12y8y
20 = 12x+4y
3x+y = 5
3x+y5 = 0
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry: Exercise 7.2
Question 1. Find the coordinates of the point which divides the join of (–1, 7) and (4, –3) in the ratio 2 : 3.
Solution:
Let the point P (x,y) divides the line AB in the ratio 2:3
where,
m = 2 and n = 3
x_{1} = 1 and y_{1} = 7
x_{2} = 4 and y_{2} = 3
so, the x coordinate of P will be,
x =
x =
x =
x = 1
and now, the y coordinate of P will be,
y =
y =
y =
y = 3
Hence, the coordinate of P(x,y) is (1,3)
Question 2. Find the coordinates of the points of trisection of the line segment joining (4, –1) and (–2, –3).
Solution:
Let the point P (x_{1},y_{1}) and Q(x_{2},y_{2}) trisects the line.
So, we can conclude that
P divides the line AB in the ratio 1:2.
and Q divides the line AB in the ratio 2:1.
 For P
m = 1 and n = 2
x_{1} = 4 and y_{1} =1
x_{2} = 2 and y_{2} = 3
so, the x coordinate of P will be,
x =
x =
x = 2
and now, the y coordinate of P will be,
y =
y =
y =
Hence, the coordinate of P is (2,).
 For Q
m = 2 and n = 1
x_{1} = 4 and y_{1} =1
x_{2} = 2 and y_{2} = 3
so, the x coordinate of Q will be,
x =
x =
x = 0
and now, the y coordinate of Q will be,
y =
y =
y =
Hence, the coordinate of Q is (0,).
Question 3. To conduct Sports Day activities, in your rectangular shaped school ground ABCD, lines have been drawn with chalk powder at a distance of 1m each. 100 flower pots have been placed at a distance of 1m from each other along AD, as shown in Fig. 7.12. Niharika runs 1/4^{th} the distance AD on the 2nd line and posts a green flag. Preet runs 1/5^{th }the distance AD on the eighth line and posts a red flag. What is the distance between both the flags? If Rashmi has to post a blue flag exactly halfway between the line segment joining the two flags, where should she post her flag?
Solution:
As the given data,
AD = 100 m
Preet posted red flag at of the distance AD
= ( ×100) m
= 20m from the starting point of 8th line.
Therefore, the coordinates of this point will be (8, 20).
Similarly, Niharika posted the green flag at th of the distance AD
= ( ×100) m
= 25m from the starting point of 2nd line.
Therefore, the coordinates of this point will be (2, 25).
Distance between these flags can be calculated by using distance formula,
Distance between two points having coordinates (x_{1},y_{1}) and (x_{2},y_{2}) = √((x_{1}x_{2})^{2} + (y_{1}y_{2})^{2})
Distance between these flags = √((82)^{2} + (2025)^{2})
= √(6^{2} + 5^{2})
Distance between these flags = √61 m
Now as, Rashmi has to post a blue flag exactly halfway between the two flags. Hence, she will post the blue flag in the mid point of the line joining these points. where,
m = n =1
(x_{1},y_{1}) = (8, 20)
(x_{2},y_{2}) = (2, 25)
x =
x =
x =
x = 5
and now, the y coordinate of Q will be,
y =
y =
y =
y = 22.5
Hence, Rashmi should post her blue flag at 22.5m on 5th line.
Question 4. Find the ratio in which the line segment joining the points (– 3, 10) and (6, – 8) is divided by (– 1, 6).
Solution:
Lets consider the ratio in which the line segment joining (3, 10) and (6, 8) is divided by point (1, 6) be k :1.
m = k and n =1
(x_{1},y_{1}) = (3, 10) and (x2,y2) = (6,8)
x = 1
x =
1 =
1(k+1) = 6k+3
k =
Hence, the required ratio is 2:7.
Question 5. Find the ratio in which the line segment joining A(1, – 5) and B(– 4, 5) is divided by the xaxis. Also find the coordinates of the point of division.
Solution:
Let the point P divides the line segment joining A (1, – 5) and B (– 4, 5) in the ratio m : 1.
Therefore, the coordinates of the point of division, say P(x, y) and,
We know that ycoordinate of any point on xaxis is 0.
P(x, 0)
m = m and n = 1
(x_{1},y_{1}) = (1, 5)
(x_{2},y_{2}) = (4,5)
so, as the y coordinate of P is 0,
y =
0 =
5m5=0
m = 1
So, xaxis divides the line segment in the ratio 1:1.
and, x =
x =
x =
Hence, the coordinate of P is (,0).
Question 6. If (1, 2), (4, y), (x, 6), and (3, 5) are the vertices of a parallelogram taken in order, find x and y.
Solution:
Let P, Q, R and S be the points of a parallelogram : P(1,2), Q(4,y), R(x,6) and S(3,5).
Mid point of PR = Mid point of QS (The diagonals of a parallelogram bisect each other, the midpoint O is same)
 Mid point of PR
m = 1 and n = 1
(x_{1},y_{1}) = (1, 2)
(x_{2},y_{2}) = (x,6)
so, the x coordinate of O will be,
x_{o} =
x_{o} =
x_{o} =
and now, the y coordinate of O will be,
y_{o} =
y_{o} =
y_{o} = 4
So, the coordinate of O is ( , 4) ……………..(1)
 For mid point QS
m = 1 and n = 1
(x_{1},y_{1}) = (3,5)
(x_{2},y_{2}) = (4,y)
so, the x coordinate of O will be,
x_{o} =
x_{o} =
x_{o} =
and now, the y coordinate of O will be,
y_{o} =
y_{o} =
y_{o} =
also , the coordinate of O is ……………..(2)
From (1) and (2)
and 4 =
x = 6 and y = 3
Question 7. Find the coordinates of a point A, where AB is the diameter of a circle whose centre is (2, – 3) and B is (1, 4).
Solution:
Let the coordinates of point A be (x, y).
Midpoint of AB is C(2, – 3), which is the centre of the circle.
and, Coordinate of B = (1, 4)
For mid point of two points (x_{1},y_{1}) and (x_{2},y_{2})
x =
y =
By using this formula, we get
(2, 3) = ,
= 2 and = 3
x + 1 = 4 and y + 4 = 6
x = 3 and y = 10
The coordinates of A (3,10).
Question 8. If A and B are (– 2, – 2) and (2, – 4), respectively, find the coordinates of P such that AP = AB and P lies on the line segment AB.
Solution:
The coordinates of point A and B are (2,2) and (2,4) respectively. Since AP = AB
= ——–(1)
subtract 1 from both sides,
– 1 = – 1
Therefore, AP: PB = 3:4
Point P divides the line segment AB in the ratio 3:4.
Here,
m = 3 and n = 4
(x_{1},y_{1}) = (2,2)
(x_{2},y_{2}) = (2,4)
so, the x coordinate of P will be,
x =
x =
x =
and now, the y coordinate of P will be,
y =
y =
y = \frac{20}{7}
Hence, the coordinate of P(x,y) is .
Question 9. Find the coordinates of the points which divide the line segment joining A(– 2, 2) and B(2, 8) into four equal parts.
Solution:
Line segment joining A(– 2, 2) and B(2, 8) divided into four equal parts.
 We can say that Q is the mid point of AB
Here,
m = 1 and n = 1
(x_{1},y_{1}) = (2,2)
(x_{2},y_{2}) = (2,8)
so, the x coordinate of Q will be,
x =
x =
x = 0
and now, the y coordinate of Q will be,
y = \mathbf{\frac{my_2 + ny_1}{m+n}}
y =
y = 5
Hence, the coordinate of Q is (0,5)……………………………..(1)
 We can say that P is the mid point of AQ
Here,
m = 1 and n = 1
(x_{1},y_{1}) = (2,2)
(x_{2},y_{2}) = (0,5)
so, the x coordinate of P will be,
x =
x =
x = 1
and now, the y coordinate of P will be,
y =
y =
y =
Hence, the coordinate of P is (1,)……………………………..(2)
 Now, we can say that R is the mid point of BQ
Here,
m = 1 and n = 1
(x_{1},y_{1}) = (2,8)
(x_{2},y_{2}) = (0,5)
so, the x coordinate of R will be,
x =
x =
x = 1
and now, the y coordinate of R will be,
y =
y =
y =
Hence, the coordinate of R is (1,)……………………………..(3)
From (1), (2) and (3) we conclude that
Three points between A and B are (1,), (0,5) and (1,).
Question 10. Find the area of a rhombus if its vertices are (3, 0), (4, 5), (– 1, 4) and (– 2, – 1) taken in order.[Hint : Area of a rhombus = ½ (product of its diagonals)]
Solution:
Let P(3, 0), Q (4, 5), R(– 1, 4) and S (– 2, – 1) are the vertices of a rhombus PQRS.
Area of a rhombus = ½ (product of its diagonals)
Length of diagonal 1 (PR) = √((3(1))^{2}+(04)^{2}) = √32 = 4√2 units
Length of diagonal 2 (QS) = √((4(2))^{2}+(5(1))^{2}) = √72 = 6√2 units
Area of a rhombus = ½ × 4√2 × 6√2
Area of a rhombus = 24 sq. units
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry: Exercise 7.3
Question 1. Find the area of the triangle whose vertices are:
(i) (2, 3), (1, 0), (2, 4)
Solution:
Area = 1/2 [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
Now by putting all the values in the formula, we will get
Area of triangle = 1/2 [2(0 – (4)) + (1)((4) – (3)) + 2(3 – 0)]
= 1/2 [8 + 7 + 6]
= 21/2
So, the area of triangle is 21/2 square units.
(ii) (5, 1), (3, 5), (5, 2)
Solution:
Area = 1/2 [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
Area of the triangle = 1/2 [(5)((5) – (2)) + 3(2 – (1)) + 5((1) – (5))]
= 1/2[35 + 9 + 20]
= 32
So, the area of the triangle is 32 square units.
Question 2. In each of the following find the value of ‘k’, for which the points are collinear.
(i) (7, 2), (5, 1), (3, k)
Solution:
As we know the result, for collinear points, area of triangle formed by them is always zero.
Let points (7, 2) (5, 1), and (3, k) are vertices of a triangle. (As given in the question)
Area of triangle = 1/2 [7(1 – k) + 5(k – (2)) + 3((2) – 1)] = 0
7 – 7k + 5k +10 9 = 0
2k + 8 = 0
k = 4
(ii) (8, 1), (k, 4), (2, 5)
Solution:
As we know the result, for collinear points, area of triangle formed by them is zero.
So we can say that for points (8, 1), (k, – 4), and (2, – 5), area = 0
1/2 [8((4) – (5)) + k((5) – (1)) + 2(1 (4))] = 0
8 – 6k + 10 = 0
6k = 18
k = 3
Question 3. Find 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 this area to the area of the given triangle.
Solution:
Let us assume that vertices of the triangle be A(0, 1), B(2, 1), C(0, 3).
Let us assume that D, E, F be the midpoints of the sides of triangle.
Coordinates of D, E, and F are
D = ((0 + 2)/2, (1 + 1)/2) = (1, 0)
E = ((0+ 0)/2, (1 + 3)/2) = (0, 1)
F = ((2+0)/2, (1 + 3)/2) = (1, 2)
Area(ΔDEF) = 1/2 [1(2 – 1) + 1(1 – 0) + 0(0 – 2)] = 1/2 (1+1) = 1
Area of ΔDEF is 1 square units
Area(ΔABC) = 1/2 [0(1 – 3) + 2(3 – (1)) + 0((1) – 1)] = 1/2 [8] = 4
Area of ΔABC is 4 square units
So, the required ratio is 1:4.
Question 4. Find the area of the quadrilateral whose vertices, taken in order, are (4, 2), (3, 5), (3, 2), and (2, 3).
Solution:
Let the vertices of the quadrilateral be A(4, 2), B(3, 5), C(3, 2), and D(2, 3).
Here, AC divide quadrilateral into two triangles.
Now, we have two triangles ΔABC and ΔACD.
Area of ΔABC = 1/2 [(4)((5) – (2)) + (3)((2) – (2)) + 3((2) – (5))]
= 1/2 [12 + 0 + 9]
= 21/2 square units
Area of ΔACD = 1/2 [(4)((2) – (3)) + 3((3) – (2)) + 2((2) – (2))]
= 1/2 [20 + 15 + 0]
= 35/2 square units
Now we will add area of both triangle and resultant will give area of quadrilateral
Area of quadrilateral ABCD = Area of ΔABC + Area of ΔACD
= (21/2 + 35/2) square units = 28 square units
Question 5. You have studied in Class IX that a median of a triangle divides it into two triangles of equal areas. Verify this result for ΔABC whose vertices are A (4, 6), B (3, 2), and C (5, 2).
Solution:
Let the vertices of the triangle be A(4, 6), B(3, 2), and C(5, 2).
Let us assume that D be the midpoint of side BC of ΔABC. So, AD is the median in ΔABC.
Coordinates of point D = Midpoint of BC = (4, 0)
Formula: Area = 1/2 [x_{1}(y_{2} – y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1} – y_{2})]
Now,
Area of ΔABD = 1/2 [(4)((2) – (0)) + 3((0) – (6)) + (4)((6) – (2))]
= 1/2 [8 + 18 – 16]
= 3 square units
As we know that, area cannot be negative. So, the area of ΔABD is 3 square units.
Area of ΔACD = 1/2 [(4)(0 – (2)) + 4((2) – (6)) + (5)((6) – (0))]
= 1/2 [8 + 32 – 30]
= 3 square units
As we know that, area cannot be negative. So, the area of ΔACD is 3 square units.
The area of both sides is same.
So we can say that, median AD has divided ΔABC in two triangles of equal areas.
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry: Exercise 7.4
Question 1. Determine the ratio, in which the line 2x + y – 4 = 0 divides the line segment joining the points A(2, 2) and B(3, 7).
Given: 2x + y – 4 = 0 line divides the line segment joining the points A(2, 2) and B(3, 7)
Find: Ratio in which the given line divides the line segment joining the points A and B
So, the coordinates of C is
Now, let us considered the ratio is k:1
=
=
But c lies on 2x+y4=0
So,
2 – 4 = 0
9k = 2
k/1 = 2/9
Hence, the ratio is 2:9
Question 2. Find a relation between x and y, if the points (x, y), (1, 2), and (7, 0) are collinear.
Find: Here, we have to find a relation between x and y, if the points (x, y), (1, 2), and (7, 0) are collinear.
If the given points are collinear then the area of the triangle is 0(created using these points).
So,
Area of triangle = 1/2[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1 }– y_{2})] = 0
=> x(2 – 0) + 1(0 – y) + 7(y – 2) = 0
=> x(2)+ 1(y) + 7y – 14 = 0
=> 2x – y + 7y – 14 = 0
=> 2x + 6y – 14 = 0
On dividing by 2 on both sides, we get
x + 3y – 7 = 0
Hence, the required relation is x + 3y – 7 = 0
Question 3. Find the center of a circle passing through the points (6, 6), (3, 7), and (3, 3).
Let us considered point A(6, 6), B(3, 7), and C(3, 3) and P(x, y) is the center of the circle.
So, AP = BP = CP(radii are equal)
Now first we take
AP = BP
√((x_{2 }– x_{1})^{2 }+ (y_{2 }– y_{1})^{2}) = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
(x – 6)^{2} + (y + 6)^{2} = (x – 3)^{2} + (y + 7)^{2}
x^{2 }+ 36 – 12x + y^{2 }+ 36 + 12y = x^{2 }+ 9 – 6x + y^{2 }+ 49 + 14y
36 – 12x + 36 + 12y = 9 – 6x + 49 + 14y
12x + bx + 12y – 14y + 72 – 58 = 0
6x – 2y + 14 = 0
6x + 2y – 14 = 0
On divided by 2 on both side, we get
3x + y – 7 = 0 (1)
Now, we take
BP = CP
√((x_{2 }– x_{1})^{2 }+ (y_{2 }– y_{1})^{2}) = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
(x – 3)^{2 }+ (y + 7)^{2 }= (x – 3)^{2 }+ (y – 3)^{2}
y^{2 }+ 7^{2 }+ 2(y)(7) = y^{2 }+ 3^{2 }– 2(y)(3)
49 + 14y = 9 – 6y
14y + 6y = 9 – 49
20y = 40
y = 40/20
y = 2
Now, on putting value of y = 2 in eq(1), we get
3x – 2 – 7 = 0
3x – 9 = 0
3x = 9
X = 9/3 = 3
Hence, the center P(x, y) = (3, 2)
Question 4. The two opposite vertices of a square are (1, 2) and (3, 2). Find the coordinates of the other two vertices.
Let us considered ABDC is a square, and its two opposite vertices are A(1, 2) and C(3, 2).
Let point O intersect line AC and BD.
Now, first we will find the coordinate of point O(x, y)
As we know that O is the mid point of line AC,
x = (3 – 1)/2 = 1
y = (2 + 2)/2 = 2
So, the coordinate of point O is (1, 2)
Now we will find the side of the Square
AC = √((3_{ }+ 1)^{2 }+ (2_{ }– 2)^{2})
AC = √16 = 4
So in triangle ACD, using hypotenuse theorem, we get
a = 2√2
So, each side of the square is 2√2
For the coordinate of D:
Let us assume that the coordinate of D is (x_{1}, y_{1}).
As we know that the sides of the squares are equal
so, AD = CD
√((x_{2 }– x_{1})^{2 }+ (y_{2 }– y_{1})^{2}) = √((x_{2} – x_{1})^{2} + (y_{2 }– y_{1})^{2})
√(x_{1 }+ 1)^{2 }+ (y_{1 }– 2)^{2} = √(x_{1} – 3)^{2} + (y_{1 }– 2)^{2}
(x_{1 }+ 1)^{2 }+ (y_{1 }– 2)^{2 }= (x_{1} – 3)^{2} + (y_{1 }– 2)^{2}
(x_{1 }+ 1)^{2 }= (x_{1} – 3)^{2}
x_{1}^{2} + 1 + 2x_{1 }= x_{1}^{2} + 9 – 6x_{1}
1 + 2x_{1 }= 9 – 6x_{1}
2x_{1} + 6x_{1 }= 9 – 1
8x_{1} = 8
x_{1} = 1
Now, CD^{2} = ((x_{2 }– x_{1})^{2 }+ (y_{2 }– y_{1})^{2})
8 = (x_{1} – 3)^{2} + (y_{1 }– 2)^{2}
8 = x_{1}^{2} + 9 – 6x_{1 }+ y_{1}^{2}_{ }+ 4 – 4y_{1}
y_{1 }– 2 =2
y_{1} = 4
So the coordinate of D is (1, 4)
For the coordinate of B:
Let B(x_{2}, y_{2}) and as we know BOD is a line segment so,
1 = x_{2 }+ 1/2
x_{2 }= 1
2 = y_{2 }+ 4/2
y_{2 }= 0
So the coordinate of B = (1, 0)
Question 5. The class X students school in krishnagar have been allotted a rectangular plot of land for their gardening activity. Saplings of Gulmohar are planted on the boundary at a distance of 1 m from each other. There is triangular grassy lawn in the plot as shown in the figure. The students are to sow seeds of flowering plants on the remaining area of the plot.
(i) Taking A as origin, find the coordinates of the vertices of the triangle.
(ii) What will be the coordinates of the vertices of ∆PQR, if C is the origin?
Also, calculate the areas of the triangles in these cases. What do you observe?
(i) Now taking A as an origin the coordinates of the vertices of the triangle are
P = (4, 6)
Q = (3, 2)
R = (6, 5)
(ii) Now taking C as an origin the coordinates of the vertices of the triangle are
P = (12, 2)
Q = (13, 6)
R = (10, 3)
Finding the area of triangle, when A as a origin the coordinates of the vertices
of the triangle are P(4, 6), Q(3, 2), and R(6, 5)
Area1 = 1/2[x_{1}(y_{2 }– y_{3}) + x2(y_{3 }– y_{1}) + x_{3}(y_{1 }– y_{2})]
= [4(2 – 5) + 3(5 – 6) + 6(6 – 2)]/2
= [12 – 3 + 24]/2
= [15 + 24]/2
= 9/2 sq. unit
Finding the area of triangle, when C as a origin the coordinates of the vertices
of the triangle are P(12, 2), Q(13, 6), and R(10, 3)
Area2 = [12(6 – 3) + 13(3 – 2) + 10(2 – 6)]/2
= [12(3) + 13(1) + 10(4)]/2
= [36 + 13 – 40]/2
= [49 – 40]/2
= 9/2 Sq.unit
So, here we observed that the Area1 = Area2
Question 6. The vertices of a ∆ABC are A (4, 6), B (1, 5), and C (7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that AD/AB = AE/AC = 1/4. Calculate the area of the ∆ ADE and compare it with the area of ∆ ABC.
Solution:
Given: AD/AB = AE/AC = 1/4
To find: Area of ∆ADE
Since D and E divides AB and AC is the same ratio 1:4 then DEBC (Using theorem 6.2)
Now, In ∆ADE and ∆ABC
∠A = ∠A (Common angle)
∠D = ∠B (Corresponding angles)
∴∆ADE~∆ABC (AA similarity)
ar.(ADE)/ar.(ABC) = (AD/AB)^{2 }= 1/16
Area of ∆ABC = 1/2[x_{1}(y_{2 }– y_{3}) + x_{2}(y_{3 }– y_{1}) + x_{3}(y_{1 }– y_{2})]
= [4(5 – 2) + 1(2 – 6) + 7(6 – 5)]/2
= [4(3) + 1(4) + 7(1)]/2
= [12 – 4 + 7]/2
= [19 – 4]/2
= 15/2 sq.unit
ar.(ADE)/ar.(ABC) = (1/10)
ar.(ADE) = (1/16) × ar.(ABC)
ar.(ADE) = (1/16) × (15/2) = 15/32 sq.unit
So, we get
ar.(ADE):ar.(ABC) = 1:16
Question 7.Let A(4, 2), B(6, 5) and C(1, 4) be the vertices of ∆ABC.
(i) The median from A meters BC at D. Find the coordinates of the point D.
(ii) Find the coordinates of the point P on the AD, such that AP: PD = 2: 1.
(iii) 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) What do you observe?
[Note: The points which are common to all the three medians is called centroid and this point divides each median in the ratio 2: 1]
(v) If A(x1, y1), B(x2, y2), and C(x3, y3) are the vertices of ∆ABC, find the coordinates of the centroid of the triangles.
Solution:
Given: The vertices of ∆ABC are A(4, 2), B(6, 5) and C(1, 4)
(i) Coordinates of D = Midpoint of BC
=
=
= [7/2, 9/2]
= (3.5, 4.5)
(ii) Coordinates of P =
=
=
= [11/3, 11/3]
(iii) Coordinates E = Midpoint of AC
=
=
= (2.5, 3)
(iv) We observe that the points P, Q and R coincide, i.e the medians of AD, BE, and
concurrent at the point [11/3, 11/3]. This point is known as the centroid of the triangle.
(v) As AD is median, thus D is midpoint of BC
So the coordinates of D:
=
=
Coordinates of G:
= [\frac{x_2m_1+m_2x_1 }{m_1+m_2},\frac{y_2m_1+y_1m_12}{m_1+m_2}]
Question 8. ABCD is a rectangle formed by the points A(1, 1), B(1, 4), C(5, 4) and D(5, 1), P, Q, R and S are the midpoints of AB, BC, CD, and DA respectively. Is the quadrilateral PQRS a square? a rectangle? or a rhombus? Justify your answer.
Solution:
Given: Points A(1, 1), B(1, 4), C(5, 4) and D(5, 1) and P, Q, R and S are the midpoints of AB, BC, CD, and DA
So,
= (2/2, 1, 3/2)
= (4/2, 8/2) = (2, 4)
= (10/2, 8/2) = (5, 3/2)
= (4/2, 2/2) = (2, 1)
So, the length of
= √(61/4)
The length of
= √(61/4)
The length of
= √(61/4)
The length of
= √(61/4)
The length of the diagonal
= √25 = 5
The length of the diagonal
= √36 = 6
Hence, all fours sides are equal(i.e., PQ = QR = RS = SP) but the diagonals are
not equal to each other(i.e. QS ≠ PR).
Therefore, given quadrilateral PQRS is a rhombus.
Key Features of NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry:
 These NCERT solutions are developed by the GfG team, focusing on students’ benefit.
 These solutions are entirely accurate and can be used by students to prepare for their board exams.
 Each solution is presented in a stepbystep format with comprehensive explanations of the intermediate steps.
Also Check:
 NCERT Solutions for Class 9 Maths Chapter 1 Number Systems
 NCERT Solutions for Class 9 Maths Chapter 2 Polynomials
 NCERT Solutions for Class 9 Maths Chapter 3 Coordinate Geometry
 NCERT Solutions for Class 9 Maths Chapter 4 Linear Equations in Two Variables
 NCERT Solutions for Class 9 Maths Chapter 5 Introduction to Euclid’s Geometry
 NCERT Solutions for Class 9 Maths Chapter 6 Lines and Angles
NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry FAQs
Q1: Why is it essential to learn NCERT Class 10 Maths Chapter 7 Coordinate Geometry?
For the purpose of connecting algebra and geometry with the aid of line and curve graphs, coordinate geometry is necessary. Finding points on a plane is a crucial component of mathematics. It also has a number of uses in other scientific fields such dimensional geometry, calculus, and trigonometry.
Q2: What topics are covered in NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry?
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry covers topics such finding points in plane, calculation of area covered by a triangle etc.
Q3: How can NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry help me?
NCERT Solutions for Class 10 Maths Chapter 7 – Coordinate Geometry can help you solve the NCERT exercise without any limitations. If you are stuck on a problem you can find its solution in these solutions and free yourself from the frustration of being stuck on some question.
Q4: How many exercises are there in Class 10 NCERT Maths Chapter 7 Coordinate Geometry?
There are 4 exercises in the Class 10 Maths Chapter 7 – Coordinate Geometry which covers all the important topics and subtopics.
Q5: Where can I find NCERT Solutions for Class 10 Maths Chapter 7 Coordinate Geometry?
You can find these NCERT Solutions in this article created by our team of experts at GeeksforGeeks.