Question

Can you help me with this one: Find a point in first quadrant which belongs to line x+y=8 and which is equally distanced from point (2,8) and line x-3y+2=0.

Answer 1

The equation of the perpendicular to x-3y=2=0 through the point (2,8) is y=-3x+14.
The required point must be on this perpendicular since the distance from a point to a line is on the perpendicular. So equate y=1/3x +2/3 and y =-3x+14

Answer 2

\[\frac{1}{3}x+\frac{2}{3}=-3x+14\]

Answer 3

\[x=4\]

Answer 4

If x = 4, y = 2 and that is the required point.