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CBSE Important Questions Class 10 Maths Chapter 7 – Coordinate Geometry

Class 10 Maths Chapter 7 Coordinate Geometry is part of the CBSE Class 10 Maths syllabus. It explains how to work with points on the cartesian plane. Students learn the distance formula to find the distance between two points, the section formula to find a point that divides a line segment in a given ratio, the midpoint formula, and the formula for the area of a triangle when the coordinates of its three vertices are known.

These CBSE important questions for Class 10 Maths Chapter 7 cover finding the distance between two points, the distance of a point from the origin, the perimeter of a triangle, distance between points given in terms of sinθ and cosθ, the area of a triangle from its vertices, and a case study on a rectangular plot. Every question has a clear step-by-step solution in simple English with figures, so students can practise on their own, parents can check the working, and teachers can use them for class tests.

Q1. Find the distance between the points A(4, 5) and B(−3, 2).

Answer: The distance formula is:
d = √[(x₂ − x₁)² + (y₂ − y₁)²]
Here x₁ = 4, y₁ = 5, x₂ = −3, y₂ = 2.
d = √[(−3 − 4)² + (2 − 5)²]
= √[(−7)² + (−3)²]
= √(49 + 9)
= √58 units

Q2. What is the perimeter of a triangle with vertices (−6, 0), (0, 0) and (0, −7)?
(a) 13 + √85 units
(b) 7 + √85 units
(c) 6 + √85 units
(d) √85 units

Answer: (a) 13 + √85 units
Distance between (−6, 0) and (0, 0) = 6 units
Distance between (0, 0) and (0, −7) = 7 units
Distance between (−6, 0) and (0, −7) = √[(0 + 6)² + (−7 − 0)²] = √(36 + 49) = √85 units
Perimeter = 6 + 7 + √85 = 13 + √85 units

Q3. Find the distance between the points A(cosθ, sinθ) and B(sinθ, −cosθ).

Answer: Using the distance formula:
d = √[(sinθ − cosθ)² + (−cosθ − sinθ)²]
= √[(sin²θ − 2sinθcosθ + cos²θ) + (cos²θ + 2sinθcosθ + sin²θ)]
The middle terms cancel out:
= √[2sin²θ + 2cos²θ]
= √[2(sin²θ + cos²θ)]
Since sin²θ + cos²θ = 1:
= √2
So the required distance is √2 units.

Q4. Find the distance between the point (3, 4) and the origin.

Answer: The origin is the point (0, 0).
d = √[(3 − 0)² + (4 − 0)²]
= √(9 + 16)
= √25
= 5 units
(In general, the distance of a point (x, y) from the origin is √(x² + y²).)

Q5. Find the distance between the points P(a + b, a − b) and Q(a − b, −a − b).

Answer: Using the distance formula:
d = √{[(a + b) − (a − b)]² + [(a − b) − (−a − b)]²}
First bracket: (a + b − a + b) = 2b
Second bracket: (a − b + a + b) = 2a
d = √[(2b)² + (2a)²]
= √(4a² + 4b²)
= 2√(a² + b²) units

Q6. What is the area of a triangle with vertices (−5, −8), (−9, −5) and (3, −6)?
(a) 32.0 square units
(b) 8.0 square units
(c) 16.0 square units
(d) 24.0 square units

Answer: (c) 16.0 square units
Area = ½ |x₁(y₂ − y₃) + x₂(y₃ − y₁) + x₃(y₁ − y₂)|
= ½ |(−5)(−5 + 6) + (−9)(−6 + 8) + 3(−8 + 5)|
= ½ |(−5)(1) + (−9)(2) + 3(−3)|
= ½ |−5 − 18 − 9|
= ½ × 32
= 16 square units

Q7. What is the area of the triangle formed by the points A(6, 7), B(8, 5) and C(−1, 1)?
(a) 26.0 square units
(b) 6.5 square units
(c) 19.5 square units
(d) 13.0 square units

Answer: (d) 13.0 square units
Area = ½ |6(5 − 1) + 8(1 − 7) + (−1)(7 − 5)|
= ½ |6(4) + 8(−6) + (−1)(2)|
= ½ |24 − 48 − 2|
= ½ × 26
= 13 square units

Q8. AOB is a triangle whose three vertices are A(−3, 0), O(5, 6) and B(−4, 0). What is the area of this triangle?
(a) 4.5 square units
(b) 6.0 square units
(c) 1.5 square units
(d) 3.0 square units

Answer: (d) 3.0 square units
Area = ½ |(−3)(6 − 0) + 5(0 − 0) + (−4)(0 − 6)|
= ½ |−18 + 0 + 24|
= ½ × 6
= 3 square units

Q9. What is the area of the triangle having vertices at (2, 3), (2, 0) and (5, 0)?
(a) 6 unit²
(b) 4.5 unit²
(c) 9 unit²
(d) 18 unit²

Answer: (b) 4.5 unit²
Area = ½ |2(0 − 0) + 2(0 − 3) + 5(3 − 0)|
= ½ |0 − 6 + 15|
= ½ × 9
= 4.5 square units

Q10. The Class X students of a secondary school 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 a triangular grassy lawn in the plot as shown in the figure. (i) Taking A as the 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 both cases. What do you observe?

A rectangular plot with Gulmohar saplings on the boundary and a triangular grassy lawn inside

Answer: (i) Taking A as the origin, with AD along the x-axis and AB along the y-axis, the vertices of the triangle are P(4, 6), Q(3, 2) and R(6, 5).
Area = ½ |4(2 − 5) + 3(5 − 6) + 6(6 − 2)|
= ½ |−12 − 3 + 24| = ½ × 9 = 4.5 square units
(ii) Taking C as the origin, with CB along the x-axis and CD along the y-axis, the vertices become P(12, 2), Q(13, 6) and R(10, 3).
Area = ½ |12(6 − 3) + 13(3 − 2) + 10(2 − 6)|
= ½ |36 + 13 − 40| = ½ × 9 = 4.5 square units
Observation: The coordinates change when we change the origin, but the area of the triangle stays the same. The area of a figure does not depend on where we place the origin.

Q11. Write the section formula and the midpoint formula. Find the coordinates of the point which divides the line segment joining (4, −3) and (8, 5) in the ratio 3 : 1.

Answer: Section formula: The point dividing the line joining (x₁, y₁) and (x₂, y₂) internally in the ratio m : n is
( (mx₂ + nx₁)/(m + n), (my₂ + ny₁)/(m + n) )
Midpoint formula: ( (x₁ + x₂)/2, (y₁ + y₂)/2 )
Here m : n = 3 : 1, (x₁, y₁) = (4, −3) and (x₂, y₂) = (8, 5).
x = [3(8) + 1(4)] / 4 = 28/4 = 7
y = [3(5) + 1(−3)] / 4 = 12/4 = 3
So the required point is (7, 3).

Q12. Show that the points A(1, 7), B(4, 2), C(−1, −1) and D(−4, 4) are the vertices of a square.

Answer: Find the lengths of all four sides and both diagonals.
AB = √[(4 − 1)² + (2 − 7)²] = √(9 + 25) = √34
BC = √[(−1 − 4)² + (−1 − 2)²] = √(25 + 9) = √34
CD = √[(−4 + 1)² + (4 + 1)²] = √(9 + 25) = √34
DA = √[(1 + 4)² + (7 − 4)²] = √(25 + 9) = √34
All four sides are equal.
Now the diagonals:
AC = √[(−1 − 1)² + (−1 − 7)²] = √(4 + 64) = √68
BD = √[(−4 − 4)² + (4 − 2)²] = √(64 + 4) = √68
Both diagonals are also equal.
Since all sides are equal and the diagonals are equal, ABCD is a square.

Related Study Material on Extramarks

Q.1 What is the distance of the point (4, ?5) from the origin?

Marks:1
Ans

Given points are:

$(x_{1}, y_{1}) = (4, - 5) and (x_{2}, y_{2}) = (0, 0)$

? Distance of the point (4, 5) from the origin is

$\begin{array}{l} = \sqrt{{(x_{2} - x_{1})}^{2} + (y_{2} - y_{1})^{2}} \\ = \sqrt{{(4 - 0)}^{2} + (- 5 - 0)^{2}} \\ = \sqrt{{(4)}^{2} + (- 5)^{2}} \\ = \sqrt{16 + 25} = \sqrt{41} units \end{array}$

Q.2 A point P divides the line segment joining the points A(3, –5) and B(–4, 8) such that AP:PB = k:1. If P lies on the line x + y = 0, then find the value of k.

Marks:3
Ans

Given points are: A(3, -5) and B(-4, 8)
Since point P divides AB in the ratio k:1, so by using Section Formula:

$Coordinates of P = (\frac{? 4 k + 3}{k + 1}, \frac{8 k ? 5}{k + 1})$

Since P lies on the line x + y = 0,

$\begin{array}{l} \frac{? 4 k + 3}{k + 1} + \frac{8 k ? 5}{k + 1} = 0 \\ ? 4 k + 3 + 8 k ? 5 = 0 \\ 4 k ? 2 = 0 \\ k = \frac{2}{4} = \frac{1}{2} \end{array}$

Q.3 If A(?5, 7), B(? 4, ?5), C(?1, ?6) and D(4, 5) are the vertices of a quadrilateral, find the area of the quadrilateral ABCD.

Marks:3
Ans

$\begin{array}{l} Now the area of ABD = \frac{1}{2} [5 (5 5) + (4) (5 7) + 4 (7 + 5)] \\ = \frac{1}{2} (50 + 8 + 48) = \frac{106}{2} = 53 square units \\ Also, the area of Î” = \frac{1}{2} [4 (6 5) 1 (5 + 5) + 4 (5 + 6)] \\ = \frac{1}{2} (44 10 + 4) = 19 square units \\ So, the area of quadrilateral ABCD = 53 + 19 = 72 square units \end{array}$

Q.4 Find the value of k if the points A(2, 3), B(4, k) and C(6, ?3) are collinear.

Marks:2
Ans

Since the given points are collinear, the area of the triangle formed by them must be 0,
i.e.,

$\begin{array}{l} \frac{1}{2} [2 (k + 3) + 4 (3 3) + 6 (3 k)] = 0 \end{array}$

$\begin{array}{l} \frac{1}{2} (4 k) = 0 \end{array}$

k=0

Q.5 If the points A(6, 1), B(8, 2), C(7, 4) and D(2p, 3) are the vertices of a parallelogram, taken in order, find the value of p.

Marks:3
Ans

We know that diagonals of a parallelogram bisect each other.

So, the coordinates of the mid point of AC

=coordinates of the mid point of BD

6+22,1+42=8+2p2,2+32
132,52=8+2p2,52
132=8+2p2
8+2p=13
2p=138
p=72

CBSE Important Questions Class 10 Maths Chapter 7 – Coordinate Geometry

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