Question

write the equation of the locus of a point, the square of whose distance from (-2,-5) is three times its distance from the line 8x+15y=34

Answer 1

well you have a point P(x, y)

and its distance from (-2, -5) is

\[d = \sqrt{x +2)^2 + (y + 5)^2}\]

now square it

\[d^2 = (x + 2)^2 + (y + 5)^2\]

does that make sense...?

Answer 2

x^2 +4x + y^2 +10y +29 = ....

Answer 3

now you need the perpendiculat distance from a point to a line...

point P(x, y) and the line 8x + 15y - 34 = 0

so

\[l = \frac{\left| 8x + 15y - 34 \right|}{\sqrt{8^2 + 15^2}}\]


then it says

d^2 = 3l

so equate them and then keep working on it...

hope it makes sense