Question

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

Answer 1

ur answer is in attachment...

Answer 2

Answer:

y=-5,-3

$PQ=\sqrt{290},13\sqrt2$

Step-by-step explanation:

The distance between two points is given by

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Given that point A(2,-4) is equidistant from P(3,8) and Q(-10,y). Hence, distance between A to P is equal to distance between A and Q.

$\sqrt{(3-2)^2+(8+4)^2}=\sqrt{(-10-2)^2+(y+4)^2}$

Simplifying, we get

$\sqrt{1+144}=\sqrt{144+y^2+8y+16}$

Squaring both sides, we get

$1+144=144+y^2+8y+16\\y^2+8y+15=0\\(y+5)(y+3)=0\\y=-5,-3$

Hence, values of y are -5 and -3

For y = -5, the distance PQ is given by

$PQ=\sqrt{(-10-3)^2+(-5-8)^2}\\PQ=\sqrt{169+169}\\PQ=\sqrt{338}\\PQ=13\sqrt2$

For y = -3, the distance PQ is given by

$PQ=\sqrt{(-10-3)^2+(-3-8)^2}\\PQ=\sqrt{169+121}\\PQ=\sqrt{290}$