Question

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.

Answer 1

Given points are A(2, – 4), P(3, 8) and Q(–10, y)

According to the question,

PA = QA

`sqrt((2 - 3)^2 + (-4 - 8)^2) = sqrt((2 + 10)^2 + (-4 - y)^2)`

`sqrt((-1)^2 + (-12)^2) = sqrt((12)^2 + (4 + y)^2)`

`sqrt(1 + 144) = sqrt(144 + 16 + y^2 + 8y)`

`sqrt(145) = sqrt(160 + y^2 + 8y)`

On squaring both sides, we get

145 = 160 + y² + 8y

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

⇒ y + 5 = 0

⇒ y = –5

And y + 3 = 0

⇒ y = –3

∴ y = – 3, – 5

Now, PQ = `sqrt((-10 - 3)^2 + (y - 8)^2`

For y = – 3 

PQ = `sqrt((-13)^2 + (-3 - 8)^2`

= `sqrt(169 + 121)`

= `sqrt(290)` units

And for y = – 5 

PQ = `sqrt((-13)^2 + (-5 - 8)^2`

= `sqrt(169 + 169)`

= `sqrt(338)` units

Hence, values of y are – 3 and – 5, PQ = `sqrt(290)` and `sqrt(338)`