Question

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

Answer 1

According to the question,
A(2,-4) is equidistant from P(3,8) amd Q(-10,y).
i.e.,PA = QA
$\Rightarrow =\sqrt{(2-3{)}^2+(-4-8{)}^2}=\sqrt{(2+10{)}^2+(-4-y{)}^2}\left[\begin{array}{c}\\ \because \text{Distance between the points}(x_1,y_1)and(x_2,y_2),d\\ =\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}\end{array}\right]\Rightarrow \sqrt{(-1{)}^2+(-12{)}^2}=\sqrt{(12{)}^2+(4+y{)}^2}\Rightarrow \sqrt{1+144}=\sqrt{144+16+y^2+8y}\Rightarrow \sqrt{145}=\sqrt{160+y^2+8y}\text{On squaring both the sides, we get}145=160+y^2+8y\Rightarrow y^2+8y+160-145=0\Rightarrow y^2+8y+15=0\Rightarrow y^2+5y+3y+15=0\Rightarrow y(y+5)+3(y+5)=0\Rightarrow (y+5)(y+3)=0Ify+5=0,theny=-5Ify+3=0,theny=-3\therefore y=-3and-5Now,distancebetweenP(3,8)andQ(-10,y),PQ=\sqrt{(-10-3{)}^2+(y-8{)}^2}[puttingy=-3]\Rightarrow =\sqrt{(-13{)}^2+(-3-8{)}^2}=\sqrt{169+121}=\sqrt{290}\text{Again, distance between P(3,8) and (-10,y), PQ}=\sqrt{(-13{)}^2+(-5-8{)}^2}[puttingy=-5]=\sqrt{169+169}=\sqrt{338}$
Hence, the values of y are -3, -5 and corresponding values of PQ are $\sqrt{290}$ and $\sqrt{338}=13\sqrt{2}$, respectively.