# What is the equation of the line that passes through (0,-1) and is perpendicular to the line that passes through the following points: (-5,11),(10,6) ?

Feb 25, 2017

$y = 3 x - 1$

#### Explanation:

$\text{the equation of a straight line is given by}$

$y = m x + c \text{ where m= the gradient & " c=" the y-intercept}$

$\text{we want the gradient of the line perpendicular to the line}$
$\text{passing through the given points } \left(- 5 , 11\right) , \left(10 , 6\right)$

we will need $\text{ } {m}_{1} {m}_{2} = - 1$

for the line given

${m}_{1} = \frac{\Delta y}{\Delta x} = \frac{{y}_{2} - {y}_{1}}{{x}_{2} - {x}_{1}}$

$\therefore {m}_{1} = \frac{11 - 6}{- 5 - 10} = \frac{5}{-} 15 = - \frac{5}{15} = - \frac{1}{3}$

$\text{ } {m}_{1} {m}_{2} = - 1 \implies - \frac{1}{3} \times {m}_{2} = - 1$

$\therefore {m}_{2} = 3$

so the required eqn. becomes

$y = 3 x + c$

it passes through $\text{ } \left(0 , - 1\right)$

$- 1 = 0 + c \implies c = - 1$

$\therefore y = 3 x - 1$