Important Questions: Coordinate Geometry - Class 10 MCQ

Let us consider the points as

A = (2, -2)

B = (-1, x)

AB = 5 units

Using the distance formula

AB² = (x₂ - x₁)² + (y₂ - y₁)²

Substituting the values

5² = (-1 - 2)² + (x + 2)²

25 = (-3)² + (x + 2)²

Using the algebraic identity

(a + b)² = a² + b² + 2ab

25 = 9 + x² + 4 + 4x

By further calculation

25 = x² + 4x + 13

x² + 4x + 13 - 25 = 0

x² + 4x - 12 = 0

By splitting the middle term

x² + 6x - 2x - 12 = 0

Taking out the common terms

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

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

So we get

x + 6 = 0

x = -6

And

x - 2 = 0

x = 2

Let A(4, 0), B(-1, -1), C(3, 5)

⇒ AB² + AC² = BC² and AB = AC
Hence, triangle is an isosceles right-angled triangle.

AB= √(9-9)²+(6-0)²=6
BC= √(-9-9)²+(6-6)²=18
CD = √(-9+9)²+(0-6)²=6
DA = √(-9-9)²+(0-0)²=18
so,it's opposite sides are equal,hence it is rectangle

Isosceles triangle.

Since both these points lie on a straight line i.e x axis, distance will be the difference between the respective y coordinates
(0,-2) & (0,6)  =>  6-(-2)) = 6+2 = 8

Distance of  from centre of the circle i. e.

Since both these points lie on a straight line i.e x axis, distance will be the difference between the respective y coordinates

(0,-3)   (0,7)

7-(-3) = 7+3 = 10

Diagonals of parallelogram bisect each other.