Question

Find the point on Y axis each of which is a distance of 13 units from a point (-5,7)

Answer 1

The answer is (0,19) or (0,-5).
By distance formula,
$\sqrt{(7 - y) {}^{2} + 5 {}^{2} } = 13$
after squaring on both sides we and solving we get,
$y {}^{2} - 14y - 95 = 0$
so, y=19 or -5

Answer 2

Answer:

The required point is either (0,-5) and (0,19).

Step-by-step explanation:

Let the point on the y-axis be (0,y), which is 13 units from a point (-5,7).

The distance formula is

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

$13=\sqrt{(-5-0)^2+(y-7)^2}$

Squaring both sides.

$169=25+(y-7)^2$

$144=(y-7)^2$

Taking square root both sides.

$\pm 12=y-7$

$y-7=12\Rightarrow y=19$

$y-7=-12\Rightarrow y=-5$

Therefore the required point is either (0,-5) and (0,19).