Question

Line l passes through point (6,0) and line p is the graph of 2x-3y=4. If l is perpendicular to line p, what is the equation of l.

Answer 1

Equation of line p:

`{\sf{2x - 3y = 4}}`

`{\sf{y = (2x)/(3) - (4)/(3)}}`

Slope of line p (m):

`{\sf{(2)/(3)}}`

Since, l is perpendicular to line p, the product of slopes of line l & p should be -1. We assume slope of line l be m2

Hence,

`{\sf{m * m2 = - 1}}`

`{\sf{ (2)/(3) * m2 = - 1}}`

`{\sf{m2 = ( - 3)/(2)}}`

Since, line l passes through points (6, 0).

We apply,

`{\sf{(y - y1) = m2(x - x1)}}`

`{\sf{y - 0 = ( - 3)/(2)(x - 6)}}`

`{\sf{2y - 0 = - 3x + 18}}`

`{\sf{3x + 2y - 18 = 0}}`

The equation of line l:

`{\sf{\red{\boxed{\sf{3x+2y-18=0}}}}}`