# What is the equation of the line perpendicular to y=-3/11x  that passes through  (8,7) ?

Feb 1, 2016

3y - 11x +67 = 0

#### Explanation:

The equation of the line is of the form : y - b = m(x - a )

where m represents the gradient and (a,b) a point on the line.

Here (a , b) =(8 , 7) is given but require m.

When 2 lines are perpendicular to each other, the product of

their gradients is - 1 .

${m}_{1.} {m}_{2} = - 1$

let ${m}_{1} = - \frac{3}{11} \textcolor{b l a c k}{\text{ the gradient of given line }}$

then ${m}_{2} \textcolor{b l a c k}{\text{ is gradient of perpendicular line}}$

hence ${m}_{2} = - \frac{1}{m} _ 1 = \frac{- 1}{- \frac{3}{11}} = \frac{11}{3}$

equation : y - 7 $= \frac{11}{3} \left(x - 8\right)$

(multiply by 3 to eliminate fraction )

hence 3 y - 21 = 11x - 88 $\Rightarrow 3 y - 11 x + 67 = 0$