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​

Answer 1

Answer:

Coordinates of point P = (0,2)

Distance = 2√13 units

Step-by-step explanation:

Given:

• Point A (4,8)
• Point B (-6,6)

To Find:

• A point P such that it is equidistant from the points A and B
• The distance

Solution:

Given that the point P is on the Y axis and it is equidistant from the points A and B

Therefore,

Coordinates of P = (0,y)

PA = PB

By distance formula we know that,

$\sf Distance=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Hence,

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

Squaring on both sides,

$\sf 4^2+(8-y)^2=(-6)^2+(6-y)^2$

$\sf 16+64-16y+y^2=36+36-12y+y^2$

$\sf 80-16y=72-12y$

$\sf 4y=8$

$\sf y=2$

Hence the y coordinate is 2.

Therefore coordinates of point P = (0,2)

Now finding the distance by distance formula,

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

$\sf \implies \sqrt{16+36}$

$\sf \implies \sqrt{52}$

$\sf \implies 2\sqrt{13} \: units$

Hence the distance is 2√13 units.