Find the equation of the locus of a point p (x, y) such that pv = pw, where v= (1, 1) and w = (3, 5)
2x + 2y = 9
2x + 3y = 8
2x + y = 9
x + 2y = 8
Correct answer is D
The locus of a point p(x, y) such that pv = pw where v = (1, 1)
and w = (3, 5). This means that the point p moves so that its distance from v and w are equidistance
\(\sqrt{(x − x_1)^2 + (y − y_1)^2}\) = \(\sqrt{(x − x_2)^2 + (y − y_2)^2}\)
\(\sqrt{(x -1)^2 + (y - 1)^2}\) = \(\sqrt{(x - 3)^2 + (y - 5)^2}\)
square both sides
(x - 1)2 + (y - 1)2 = (x - 3)2 + (y - 5)2
x2 - 2x + 1 + y2 - 2y + 1 = x2 - 6x + 9 + y2 - 10y + 25
x2 + y2 -2x -2y + 2 = x2 + y2 - 6x - 10y + 34
Collecting like terms
x2 - x2 + y2 - y2 - 2x + 6x -2y + 10y = 34 - 2
4x + 8y = 32
Divide through by 4
x + 2y = 8