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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.
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Solution
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)`
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