Question

Find a point P on the y-axis which is equidistant from the points A(4, 8) and B(– 6, 6). Also find the distance AP.

Answer 1

(0-4)sq+(y-8)sq=(0+6)sq+(y-6)sq
16-36=(y-6+y-8)(y-6-y+8)
-20=(2y-14)(2)
-5=y-7
y=2
thus point on y axis will be (0,y)=(0,2).

Answer 2

The correct answer is $2\sqrt{13}$.

Given: Points = A(4, 8) and B(– 6, 6).

To Find: Distance AP.

Solution:

If point P is on y-axis, its x coordinate will be 0.

Points = A(4, 8) and B(– 6, 6).

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

For length AP.

Point A = (4, 8)

Point P = (0, y)

For length BP.

Point B = (– 6, 6)

Point P = (0, y)

If point P is equidistant.

AP = BP

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

$\sqrt{16 +(y-8)^{2}}=\sqrt{36 +(y-6)^{2}}$

Squaring both the sides.

$16+(y-8)^{2} =36+(y-6)^{2}$

$y^{2}-16y+64 =20+y^{2} -12y+36$

$4y=8$

y = 2

The point P is (0, 2).

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

= $\sqrt{(0-4 )^{2} +(2-8 )^{2}}$

= $\sqrt{16+36}$

= $\sqrt{52}$

= $2\sqrt{13}$

Hence, the length AP is $2\sqrt{13}$.

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