Question

Find a point on y axis which is at equal distance from p(0,4) and q(-1,-5)

Answer 1

Answer:

Given that the point lies on y-axis, Therefore, its x-coordinate will be 0.

Let the point A(0,y) be equidistant from points.

Given points are P(-6,4) and Q(2,-8).

Given that there distances are equal.

= > PA = QA

= > \sqrt{(0 + 6)^2 + (y - 4)^2} = \sqrt{(0 - 2)^2 + (y + 8)^2}=>(0+6)2+(y−4)2=(0−2)2+(y+8)2

= > \sqrt{36 + y^2 + 16 - 8y} = \sqrt{4 + y^2 + 64 + 16y}=>36+y2+16−8y=4+y2+64+16y

= > \sqrt{y^2 - 8y + 52} = \sqrt{y^2 + 16y + 68}=>y2−8y+52=y2+16y+68

On squaring both sides, we get

= > y^2 - 8y + 52 = y^2 + 16y + 68

= > -24y = 16

= > y = -2/3.

Therefore, the required point is (0, -2/3).

Hope this helps!

Answer 2

Answer:

Let the coordinates of point A on y axis be (0,y).

So,AP=AQ

AP^2=AQ^2

(y-4)^2=1^2+(y+5)^2

y^2+16-8y=1+y^2+25+10y

18y= -8

y= -4/9.

So the point is (0,-4/9).