What point on y-axis is at a distance of 5units from the point (-3,4)?

1.	What point on y-axis is at a distance of 5units from the point (-3,4)?
Answer» Given , The point on y axis is at a distance of 5 UNITS from the point (-3 , 4) Let , the point on y axis be (0 , y) We know that , the distance b/w TWO POINTS is given by $\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {( y_{2} - y_{1})}^{2} } }}$ THUS , $\mapsto \tt 5 = \sqrt{ {(0 + 3)}^{2} + { (y - 4) }^{2} }$ $\mapsto \tt 25 = 9 + {y}^{2} + 16 - 8y$ $\mapsto \tt 25 = 25 + y(y - 8)$ $\mapsto \tt y(y - 8) = 0$ $\mapsto \tt y = 0 \: \: or \: \: y = 8$ ★ The point on the y axis will be (0,0) or (0,8)

Answer»

Given ,

The point on y axis is at a distance of 5 UNITS from the point (-3 , 4)

Let , the point on y axis be (0 , y)

We know that , the distance b/w TWO POINTS is given by

$\boxed{ \tt{Distance = \sqrt{ {( x_{2} - x_{1})}^{2} + {( y_{2} - y_{1})}^{2} } }}$