Question

find the equation of locus of a point which is equidistant from the points A(-3,2)and B(0,4)​

Answer 1

Answer:

Written in image.

Step-by-step explanation:

Written in image.

Answer 2

The locus of the point which is equidistant from the given points is $6x+4y-3=0$.

Given,

A point is equidistant from points A(-3, 2), and B(0, 4).

To find,

Locus of this point.

Solution,

It can be seen that here, a point is given to be equidistant from the points A(-3, 2), and B(0, 4).

Firstly, let this point be P($x,y$).

The distance of above point P($x,y$) from A and B can be determined using the distance formula as follows.

If the two points are $L(x_1,y_1)$ and $M(x_2,y_2)$, then the distance LM is given as

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

• Distance AP

$AP=\sqrt{(-3-x)^2+(2-y)^2}$

$\implies AP=\sqrt{(3+x)^2+(2-y)^2}$

• Distance BP

$BP=\sqrt{(0-x)^2+(4-y)^2}$

$\implies BP=\sqrt{x^2+(4-y)^2}$

The two distances must be equal since P is equidistant from A and B.

Thus,

$AP = BP$

$\implies \sqrt{(3+x)^2+(2-y)^2} =\sqrt{x^2+(4-y)^2}$

Squaring both sides, we get,

$(3+x)^2+(2-y)^2=x^2+(4-y)^2$

Simplifying,

$[9 + 6x+x^2]+[4-4y+y^2]=x^2+[16-8y+y^2]$

$\implies 9 + 6x+4-4y=16-8y$

$\implies 13 + 6x-4y=16-8y$

Rearranging,

$6x-4y+8y+13-16=0$

$\implies 6x+4y-3=0,$ which is the required locus or equation.

Therefore, the locus of the point which is equidistant from the given points is $6x+4y-3=0$.

#SPJ2