Question

4. A line segment has two end-points M(3, 7) and
N(11, -6). Find the coordinates of the point W
that lies on the y-axis such that W is equidistant
from M and from N.
Hint: The term 'equidistant' means 'same distance'.​

Answer 1

Given:-

M(3,7)

N(11,-6)

WM=WN

W(0,y)

Required to find:-

w (0,y)

Procedure:-

WM=WN

Squaring on both sides

$(WM)^2=(WN)^2$

So,

${(x_{w} -x_{M} )^{2}+(y_{w}-y_m)^2 } = (x_{w} -x_{n} )^{2}+(y_{w}-y_n)^2 }$

Plugging in the values, we get

${(0-3)^2+(y-7)^2}= (0-11)^2+(y+6)^2}$

$9+y^{2} -14y+49=121+y^2+12y+36$

$26y-99=0$

$26y=99$$y=\frac{99}{26}$

∴ The point W is on $(0,\frac{99}{26} )$.