Question

Find the ordinate of point whose abscissa is $10$ and which is at a distance of $10$ units from the point $P(2,-3)$.

Answer 1

Given

$x_1=10y_1=y$

$x_2=2y_2=-3$

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

Or $d^2=(x_1-x_2{)}^2+(y_1-y_2{)}^2$

$100=(10-2{)}^2+(y+3{)}^2$

$(y+3{)}^2=100-64$

$(y+3{)}^2=36$

$y+3=6\Rightarrow y=3$

$y+3=-6\Rightarrow y=-9$