Question

find the coordinates of the points on the y-axis which are at a distance of 17 units from the point (-8,3)

Answer 1

Answer:

(0,18) or (0,-12)

Step-by-step explanation:

Steps are explained in the picture

Answer 2

Given that ,

The distance between the points on the y-axis and (-8,3) is 17 units

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

We know that ,

The distance between two points is given by :

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

Thus ,

$\sf \mapsto 17 = \sqrt{ {(0 - ( - 8))}^{2} + {(y - 3)}^{2} } \\ \\ \sf Squaring \: both \: on \: sides \: , \: we \: get \\ \\ \sf \mapsto 289 = 64 + {(y)}^{2} + 9 - 6y \\ \\ \sf \mapsto {(y)}^{2} - 6y - 216 = 0 \\ \\ \sf \mapsto {(y)}^{2} - 18y + 12y - 216 = 0 \\ \\ \sf \mapsto y(y - 18) + 12(y - 18) \\ \\ \sf \mapsto (y + 12)(y - 18) \\ \\ \sf \mapsto y = - 12 \: \: or \: \: y = 18$

Therefore , the point on the y - axis will be (0,-12) or (0,18)