Question

Find the linear relation between x and y such that p(x,y) is equidistant from the point A(1,4)B(-1,2)

Answer 1

use section formula to find
formula is m1x2+m2x1/2 , m1y2+m2y1/2

Answer 2

Answer:

The linear relation between x and y is x+y=3

Step-by-step explanation:

Point P(x,y) is equidistant from the point A(1,4) and B(-1,2)

Formula: Using distance formula

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

First we find the distance of PA

$PA=\sqrt{(x-1)^2+(y-4)^2}$

Now, we find distance of PB

$PB=\sqrt{(x+1)^2+(y-2)^2}$

According to question

PA=PB

$\sqrt{(x-1)^2+(y-4)^2}=\sqrt{(x+1)^2+(y-2)^2}$

Taking square both side to eliminate square root

$(x-1)^2+(y-4)^2=(x+1)^2+(y-2)^2$

$x^2+1-2x+y^2+16-8y=x^2+1+2x+y^2+4-4y$

$-4x-4y+12=0$

$x+y=3$

Hence, The linear relation between x and y is x+y=3