Question

If $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$, find the values of $x$. Also find the distances $QR$ and $PR$.

Answer 1

As given $Q(0,1)$ is equidistant from $P(5,-3)$ and $R(x,6)$

$\Rightarrow PQ=QR$

$\Rightarrow \sqrt{(0-5{)}^2+(1-(-3){)}^2}$$=\sqrt{(x-0{)}^2+(1-6{)}^2}$ [ By using disatnce formula $=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$ ]

$\Rightarrow 25+16=x^2+25$

$\Rightarrow 41=x^2+25$

$\Rightarrow x^2=16$

$\Rightarrow x=4$

Distance between $Q(0,1)$ and $R(4,6)$

$QR=\sqrt{(4-0{)}^2+(6-1{)}^2}=\sqrt{4^2+5^2}=\sqrt{16+25}=\sqrt{41}$

Distance between $P(5,-3)$ and $R(4,6)$

$PR=\sqrt{(4-5{)}^2+(6+3{)}^2}=\sqrt{1+81}=\sqrt{82}$