Question

Find a point on y axis which is equidistant from a 65 and b - 4, 3

Answer 1

use distance formulae and find it

Answer 2

The coordinates of point P are (0,9).

Step-by-step explanation:

The given points are A(6,5) and B(-4,3).

We need to find the point on y axis which is equidistant from A and B.

Point lie on y-axis, so x-coordinate is 0.

Let the required point is P(0,b).

Distance formula:

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

Using distance formula we get

$AP=\sqrt{(0-6)^2+(b-5)^2}$

$AP=\sqrt{(-6)^2+b^2-2(b)(5)+5^2}$

$AP=\sqrt{36+b^2-10b+25}$

$AP=\sqrt{b^2-10b+61}$

Similarly,

$BP=\sqrt{(0-(-4))^2+(b-3)^2}$

$BP=\sqrt{(4)^2+b^2-2(b)(3)+3^2}$

$BP=\sqrt{16+b^2-6b+9}$

$BP=\sqrt{b^2-6b+25}$

Point P is equidistant from A and B so,

$AP=BP$

$\sqrt{b^2-10b+61}=\sqrt{b^2-6b+25}$

Taking square on both sides.

$b^2-10b+61=b^2-6b+25$

$-10b+61=-6b+25$

Isolate variable terms on left side.

$-10b+6b=-61+25$

$-4b=-36$

Divide both sides by -4.

$b=9$

The value of b is 9. Therefore, the coordinates of point P are (0,9).

#Learn more

Find the distance between the two pairs of point (1,2)and (4,3)​.

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