CBSE Class 10 Maths Important Questions: Coordinate Geometry

In mathematics, coordinate geometry is a branch where a specific point on a plane (a flat surface that can infinitely continue in both directions, x-axis & y-axis) can be shown with help of a pair of numbers. It is developed to establish a connection between algebra and geometry with help of graphs, lines, curves, parabolas, etc. It is very useful for locating specific points on a plane. Used extensively in all navigation systems be it on land, sea, or air. In the world of mathematics, it is used for solving problems of trigonometry, calculus, etc.

Important questions for class 10 maths chapter 7 Coordinate Geometry, can be picked from the topic :

• Distance formula description and calculations
• Section formula description and calculations
• Area of a triangle
• Defining  abscissa and ordinate

Important Questions Chapter 7 – Coordinate Geometry

Q1. Find a relation between x and y such that the point (x, y) is equidistant from the points (7, 1) and (3, 5).

Solution:

Let P(x, y) be equidistant from the points A(7, 1) and B(3, 5).

Then, AP = BP

AP^2 = BP^2

Using distance formula,

(x – 7)^2 + (y – 1)^2 = (x – 3)^2 + (y – 5)^2

x^2 – 14x + 49 + y^2 – 2y + 1 = x^2 – 6x + 9 + y^2 – 10y + 25

x – y = 2

Hence, the relation between x and y is x – y = 2.

Q2. Find the area of triangle PQR formed by the points P(-5, 7), Q(-4, -5) and R(4, 5).

Solution:

Given,

P(-5, 7), Q(-4, -5) and R(4, 5)

Let P(-5, 7) = (x1, y1)

Q(-4, -5) = (x2, y2)

R(4, 5) = (x3, y3)

Area of the triangle PQR = (½)|x1(y2 – y3) + x2(y3 – y1) + x3(y1 – y2)|

= (½) |-5(-5 – 5) + (-4)(5 – 7) + 4(7 + 5)|

= (½) |-5(-10) -4(-2) + 4(12)|

= (½) |50 + 8 + 48|

= (½) × 106

= 53

Therefore, the area of triangle PQR is 53 sq. units.

Q3. Write the coordinates of a point on the x-axis which is equidistant from points A(-2, 0) and B(6, 0).

Solution:

Let P(x, 0) be a point on the x-axis.

Given that point, P is equidistant from points A(-2, 0) and B(6, 0).

AP = BP

Squaring on both sides,

(AP)² = (BP)²

Using distance formula,

(x + 2)² + (0 – 0)² = (x – 6)² + (0 – 0)²

x² + 4x + 4 = x² – 12x + 36

4x + 12x = 36 – 4

16x = 32

x = 2

Therefore, the coordinates of a point on the x-axis = (2, 0).

Q4. Find a relation between x and y such that the point P(x, y) is equidistant from the points A (2, 5) and B (-3, 7)

Solution:

Let P (x, y) be equidistant from the points A (2, 5) and B (-3, 7).

∴ AP = BP …[Given

AP^2 = BP^2 …[Squaring both sides

(x – 2)^2 + (y – 5)^2 = (x + 3)^2 + (y – 7)^2

⇒ x^2 – 4x + 4 + y^2 – 10y + 25

⇒ x^2 + 6x + 9 + y^2 – 14y + 49

⇒ -4x – 10y – 6x + 14y = 9 +49 – 4 – 25

⇒ -10x + 4y = 29

∴ 10x + 29 = 4y is the required relation.

Q5. Find the relation between x and y if the points A(x, y), B(-5, 7) and C(-4, 5) are collinear. 

Solution: 

When points are collinear,

∴ Area of ∆ABC = 0

= (x1 (y2 – y3) + x2(y3 – y1) + x3(y1 – y2)) = 0

= x (7 – 5) – 5 (5 – y) -4 (y – 7) = 0

= 2x – 25 + 5y – 4y + 28 = 0

∴ 2x + y + 3 = 0 is the required relation.

Q6. If A(4, 3), B(-1, y) and C(3, 4) are the vertices of a right triangle ABC, right-angled at A, then find the value of y.

Solution:

We have A(4, 3), B(-1, y) and C(3, 4). In right angled triangle ABC,

(BC)^2 = (AB)^2 + (AC)…. [Pythagoras theorem]

⇒ (-1 – 3)^2 + (y – 4)^2 = (4 + 1)^2 + (3 – y)^2 + (4 – 3)^2 + (3 – 4)^2 …(using distance formula

⇒ (-4)^2 + (y^2 – 8y + 16)

⇒ (5)^2 + (9 – 6y + y^2) + (1)^2 + (-1)^2

⇒ y^2 – 8y + 32 = y^2 – 6y + 36 = 0

⇒ -8y + 6y + 32 – 36

⇒ -2y – 4 = 0 ⇒ -2y = 4

∴ y = -2

Q7. Find that value(s) of x for which the distance between the points P(x, 4) and Q(9, 10) is 10 units.

Solution:

PQ = 10 …Given

PQ^2 = 102 = 100 … [Squaring both sides

(9 – x)^2 + (10 – 4)^2 = 100…(using distance formula

(9 – x)^2 + 36 = 100

(9 – x)^2 = 100 – 36 = 64

(9 – x) = ± 8 …[Taking square-root on both sides

9 – x = 8 or 9 – x = -8

9 – 8 = x or 9+ 8 = x

x = 1 or x = 17

Q8. Find the value of y for which the distance between the points A (3,-1) and B (11, y) is 10 units.

Solution:

AB = 10 units … [Given

AB^2 = 10^2 = 100 … [Squaring both sides

(11 – 3)^2 + (y + 1)^2 = 100

8^2 + (y + 1)^2 = 100

(y + 1)^2 = 100 – 64 = 36

y + 1 = ±6 … [Taking square-root on both sides

y = -1 ± 6 ∴ y = -7 or 5

Q9. The point A(3, y) is equidistant from the points P(6, 5) and Q(0, -3). Find the value of y. 

Solution:

PA = QA …[Given

PA^2 = QA^2 … [Squaring both sides

(3 – 6)^2 + (y – 5)^2 = (3 – 0)^2 + (y + 3)^2

9 + (y – 5)^2 = 9 + (y + 3)^2

(y – 5)^2 = (y + 3)^2

y – 5 = ±(y + 3) … [Taking sq. root of both sides

y – 5 = y + 3 y – 5 = -y – 3

0 = 8 … which is not possible ∴ y = 1

Q10. If the point P(k – 1, 2) is equidistant from the points A(3, k) and B(k, 5), find the values of k.

Solution:

PA = PB …Given

PA^2 = PB^2 … [Squaring both sides

⇒ (k – 1 – 3)^2 + (2 – k)^2 = (k – 1 – k)^2 + (2 – 5)^2

⇒ (k – 4)^2 + (2 – k)^2 = (-1)^2 + (-3)^2

k^2 – 8k + 16 + 4 + k^2 – 4k = 1 + 9

2k^2 – 12k + 20 – 10 = 0

2k^2 – 12k + 10 = 0

⇒ k^2 – 6k + 5 = 0 …[Dividing by2]

⇒ k^2 – 5k – k + 5 = 0

⇒ k(k – 5) – 1(k – 5) = 0

⇒ (k – 5) (k – 1) = 0

⇒ k – 5 = 0 or k – 1 = 0

∴ k = 5 or k = 1