If the point A(2, - 4) is equidistant from P(3, 8) and Q( - 10, y), then find the value of y. Also, find distance PQ.
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) =
So, PA = QA
⇒√145= √(160 + y2 + 8y)
On squaring both the 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
If y + 5 = 0, then y = - 5
If y + 3 = 0, then y = - 3
∴ y = - 3, – 5
Now,
Distance P (3, 8) and Q ( - 10, y);
[Putting y = - 3]
→
Again, distance between P (3, 8) and Q (-10, y);
[Putting y = - 5]
→
Hence,
The values of y are - 3, - 5 and corresponding values of PQ are √290 and √338=13√2 respectively.