Question

Find an equation of the line perpendicular to the graph of 15x-3y = 7that passes through the point at 0,

cy=32 - 4
y=-- + 4
d
= t - 4
3
-
4

Answer 1

5y + x = 5c

Explanation:

Rewriting the equation, 15x - 3y = 7,

3y = 15x - 7

y = 5x - 7/3. The gradient is 5. The gradient of the perpendicular line is -1/5

The equation of the perpendicular to 15x - 3y = 7 passing through (0,c) is

(y - c)/(x - 0) = -1/5

(y - c)/x = -1/5

y - c = -x/5

y = -x/5 + c

5y = -x + 5c

5y + x = 5c