Question

Find the values of $y$ for which the distance between the points $P\left(2,-3\right)$ and $Q\left(10,y\right)$ is $10$ units.

Answer 1

Evaluate the value of $y$

The distance formula is

$\mathit{d}\mathbf{=}\sqrt{{\left(x_2-x_1\right)}^{\mathbf{2}}\mathbf{+}{\left(y_2-y_1\right)}^{\mathbf{2}}}$

Here,

$d=10$ units

$\left(x_1,x_2\right)=\left(2,10\right)$ and $\left(y_1,y_2\right)=\left(-3,y\right)$.

Substituting in the formula,

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

$\Rightarrow 10=\sqrt{64+y^2+9+6y}$ $\left[\therefore {\left(a+b\right)}^2=a^2+2ab+b^2\right]$

Now, take square roots on both the sides

$\Rightarrow$ $100=73+6y+y^2$

$\Rightarrow$ $y^2+6y-27=0$

$\Rightarrow$$y^2+9y-3y-27=0$

$\Rightarrow$$y(y+9)-3(y+9)=0$

$\Rightarrow$ $(y+9)(y-3)=0$

$\Rightarrow$ $y=3,-9$

Hence, the value of $y$ is $3$ or $-9$.