Question

Question 10
Find a relation between x and y such that the point (x, y) is equidistant from the point (3, 6) and (- 3, 4).

Answer 1

Point (x, y) is equidistant from (3, 6) and ( - 3, 4).
And Distance between the points is given by
$\sqrt{{(x_1-x_2)}^2+{(y_1-y_2)}^2}$
$\therefore \sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x-(-3){)}^2+(y-4{)}^2}$
$\sqrt{(x-3{)}^2+(y-6{)}^2}=\sqrt{(x+3{)}^2+(y-4{)}^2}$
$(x-3{)}^2+(y-6{)}^2=(x+3{)}^2+(y-4{)}^2$
$x^2+9-6x+y^2+36-12y=x^2+9+6x+y^2+16-8y$
$36-16=6x+6x+12y-8y$
$20=12x+4y$
$3x+y=5$
$3x+y-5=0$