USING DISTANCE FORMULA TO FIND MISSING COORDINATES

Problem 1 :

If (a, b) is the midpoint of the line segment joining the points A(10, -6) and B(k, 4) and a - 2b = 18, find the value of k and the distance AB.

Solution:

Given, (a, b) is the midpoint of the line segment joining the points A(10, -6) and B(k, 4)

So,

It is given that,

a - 2b = 18

Put b = -1,

a - 2(-1) = 18

a = 18 - 2 

a = 16

Now,

Problem 2 :

The centre of a circle is (2a, a - 7). Find the values of a if the circle passes through the point (11, -9) and has diameter 10√2 units.

Solution:

Given, the length of the diameter is 10√2 units.

So, the radius is 5√2 units.

The centre of the circle be C(2a, a - 7)

It passes through the point P(11, -9)

PC = r

PC² = r²

(11 - 2a)² + (-9 - a + 7)² = (5√2)²

121 + 4a² - 44a + a² + 4 + 4a = 50

5a² - 40a + 75 = 0

5(a² - 8a + 15) = 0

a² - 8a + 15 = 0

(a - 3) (a - 5) = 0

a = 3 or a = 5

Hence, the values of a are 3 or 5.

Problem 3 :

If the distance between the points (2, -2) and (-1, x) is 5, one of the values of x is

(A) -2     (B) 2     (C) -1     (D) 1

Solution:

The distance between (x_1, y₁) and (x₂, y₂) is

Therefore, the distance between the points (2, -2) and (-1, x)

Squaring both sides, we get

25 = 9 + (x + 2)²

25 = 9 + x² + 4 + 4x

x² + 4x + 13 - 25 = 0

x² + 4x - 12 = 0

x² + 6x - 2x - 12 = 0

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

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

x = 2 or x = -6

Hence, the values of x are 2 and -6.

So option (B) is correct.