Question

If P(x, y) is equidistant from the points A(2, 1) and B(-2, 4), the relation between x and y is .

• A. 8x - 6y + 15 = 0
• B. 4x - 7y + 5 = 0
• C. x - 6y + 25 = 0

Answer 1

The correct option is A 8x - 6y + 15 = 0

Given: P(x, y), A(2, 1), and B(-2, 4)

The distance between two points $(x_1,y_1)\text{and}(x_2,y_2)$ is $\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}.$

P is equidistant from A and B, so PA = PB.

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

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

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

$-4x-2y+5=4x-8y+20$

8x - 6y + 15 = 0