Question

Find the points on the y-axis, which are at a distance of $13$ units each from the point $(-5,7).$

Answer 1

Let the point on y-axis be $B(0,b)$ and the given point be $A(-5,7).$ The distance between A and B is $13$ units.
Now,
$AB=13$

$\Rightarrow \sqrt{\{0-(-5){\}}^2+(b-7{)}^2}=13$ [Using Distance formula: $d=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}]$

$\Rightarrow (0+5{)}^2+(b-7{)}^2={13}^2$ [on squaring both sides]

$\Rightarrow 25+b^2+49-14b=169$

$\Rightarrow b^2-14b-95=0$

$\Rightarrow b^2-19b+5b-95=0$

$\Rightarrow b(b-19)+5(b-19)=0$

$\Rightarrow (b-19)(b+5)=0$

$\therefore b=19$ or $-5$
Hence, the points are $(0,19)$ and $(0,-5).$