Question

What is the equation of a line that passes through the point (1, 8) and is perpendicular to the line whose equation is y=x/2+3 ?

Answer 1

Answer

y = -2x + 10

Explanation

The general equation for a straight line is y = mx + c where m and c are gradient and y-intercept respectively.

y=x/2+3 = y (1/2)x + 3

gradient = 1/2

Gradient of the line perpendicular to y=x/2+3 is;

m × 1/2 = -1

m = -2

Now we find the equation of a line passing through (1,8) and have a gradient of -2.

-2 = (y - 8)/(x - 1)

-2(x - 1) = (y - 8)

2 -2x = y - 8

y = -2x + 10