Equation of Locus from Point and a Line (1 Viewer)

[Redyapper]

Hi,

The question I'm stuck on is:
Find the equation of the locus of a point so it is equidistant from P(2, -3) and the line y=7. If it was between 2 points, I could've done:
distance formula of Point A = distance formula of Point B and solved it from there, but I'm not sure what to do when I have a line?

Thanks.

 

[gr_111]

you basically had it there, except use perpendicular distance formula = distance from point formula

 

[Redyapper]

you basically had it there, except use perpendicular distance formula = distance from point formula

Ohh right; but what would I put into the perpendicular distance formula?

 

[nightweaver066]

Consider a point (x, y)

If you look at a graph and consider if the point is above the line y = 7 (at A), you notice the distance from the point to the line is (y - 7).

If the point is below the line y = 7 (at B), the distance from the point to the line is (7 - y)

So we can say the distance is |y - 7| (distance cannot be negative).

Now we just apply the distance formula with (x, y) and P(2, -3),



Square both sides, and away you go.

 

[Redyapper]

Consider a point (x, y)

If you look at a graph and consider if the point is above the line y = 7 (at A), you notice the distance from the point to the line is (y - 7).

If the point is below the line y = 7 (at B), the distance from the point to the line is (7 - y)

So we can say the distance is |y - 7| (distance cannot be negative).

Now we just apply the distance formula with (x, y) and P(2, -3),



Square both sides, and away you go.

Thanks a million!