Question

If the point A(2,-4) is equidistant from P(3,8) and Q(-10,y), then find the value of y.

• A. -3
• B. 4
• C. 5
• D. -5

Answer 1

Answer: (A), (D)

The correct options are
A

-3

D

-5

According to the question,

A(2,-4) is equidistant from P(3,8) amd Q(-10,y).

i.e.,PA = QA

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

[$\because$ Distance between the points $(x_1,y_1)$ and $(x_2,y_2),d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$]

$\Rightarrow \sqrt{(-1{)}^2+(-12{)}^2}=\sqrt{(12{)}^2+(4+y{)}^2}$

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

$\Rightarrow \sqrt{145}=\sqrt{160+y^2+8y}$

On squaring both the sides, we get

$145=160+y^2+8y$

$\Rightarrow y^2+8y+160-145=0$

$\Rightarrow y^2+8y+15=0$

$\Rightarrow y^2+5y+3y+15=0$

$\Rightarrow y(y+5)+3(y+5)=0$

$\Rightarrow (y+5)(y+3)=0$

If $y+5=0$, then $y=-5$

If $y+3=0$, then $y=-3$

$\therefore y=-3$ and $-5$

Hence, the values of y are -3, -5.