Question

The y-coordinate of a point A is thrice its x-
coordinate. If A is equidistant from B(2, - 4)
and C(8, 2), find the coordinates of A.
Coordinates of A =
​

Answer 1

Answer:-

Let the x - Coordinate of the point A be "a".

It's y - Coordinate = 3a.

Given:

A is equidistant from the points B(2 , - 4) and C(8 , 2).

That is,

→ BA = CA

We know that,

$\sf{Distance \: between \: two \: points \: with \: coordinates} \\\\\sf{ A( x_{1} \: , \: y _{1}) \: and \: B( x_{2} \: , \: y _{2}) = \sqrt{ {( x_{2} - x _{1} ) }^{2} + {(y _{2} - y _{1}) }^{2} }}$

$→ \sf {\sqrt{ {(2 - a)}^{2} + {( - 4 - 3a)}^{2} } = \sqrt{ {(8 - a)}^{2} + {(2 - 3a)}^{2} }}$

(Squaring on both sides)

→ (2 - a)² + ( - 4 - 3a)² = (8 - a)² + (2 - 3a)²

Using (a - b)² = a² + b² - 2ab

→ (2)² + a² - 2(2)(a) + ( - 4)² + (3a)² - 2( - 4)

(3a) = (8)² + a² - 2(8)(a) + (2)² + (3a)² - 2(2)

(3a)

→ 4 + a² - 4a + 16 + 9a² + 24a = 64 + a² -

16a + 4 + 9a² - 12a

→ 10a² + 20a + 20 = 10a² - 28a + 68

→ 10a² - 10a² + 20a + 28 = 68 - 20

→ 48a = 48

→ a = 48/48

→ a = 1

→ x - Coordinate of A (a) = 1

→ y - Coordinate of A (3a) = 3(1) = 3

Hence the Coordinates of A are (1 , 3).