HOW TO FIND A MISSING COORDINATE USING DISTANCE FORMULA

Problem 1 :

Find the points on the x-axis which are at a distance of 2√5 from the point (7, -4). How many such points are there?

Solution:

Let the coordinates of point on x-axis is (x, 0)

Distance between two points (x₁, y₁) and (x₂, y₂)

The distance between points on x-axis and (7, -4) = 2√5

Squaring on both sides,

(7 - x)² + (-4)² = 4 × 5

49 + x² - 14x + 16 = 20

x² - 14x + 45 = 0

x² - 5x - 9x + 45 = 0

x(x - 5) - 9(x - 5) = 0

(x - 5) (x - 9) = 0

Therefore, the points (5, 0) and (9, 0) are at a distance of 2√5 from the point (7, -4).

Problem 2 :

Find the value of a, if the distance between the points A(-3, -14) and B(a, -5) is 9 units.

Solution:

Given, distance between the points A(-3, -14) and B(a, -5) is 9 units.

By distance formula, 

Squaring on both sides,

Hence, the value of a is -3.

Problem 3 :

If the point A(2, -4) is equidistant from P(3, 8) and Q(-10, y), find the values of y. Also find distance PQ.

Solution:

Given, A(2, -4) is equidistant from P(3, 8) = Q(-10, y) is equidistant from A(2, -4).

By distance formula,

Distance between two points (x₁, y₁) and (x₂, y₂)

Squaring on both sides,

145 - 160 + y² + 8y = 0

y² + 8y + 160 - 145 = 0

y² + 8y + 15 = 0

y² + 5y + 3y + 15 = 0

y(y + 5) + 3(y + 5) = 0

(y + 5) (y + 3) = 0

So, y = (-3, -5)

Distance P(3, 8) and Q(-10, y)

Again, distance P(3, 8) and Q(-10, y)

Hence, the values of y are -3 and -5 and corresponding values of PQ are √290 units and 13√2 units.