# How do you find a standard form equation for the line with (8, -2) and perpendicular to the line whose equation is x-5y-7=0?

Jul 8, 2017

$5 x + y + 2 = 0$

#### Explanation:

the basic equation of a straight line with known gradient $\text{ "m" }$and one known set of co-ordinates" (x_1,y_1)

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

we are given a line perpendicular to the required line so we use the fact that if the gradient of the required line is$\text{ "m_1" }$and the gradient of a line perpendicular to it $\text{ "m_2" }$ then we have

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

that is the product of the gradients of perpendicular lines is $- 1$

so we use the line given to find the gradient

$x - 5 y - 7 = 0$

rearrange to the form

$y = m x + c$

$x - 7 = 5 y$

$y = \frac{1}{5} x - \frac{7}{5}$

$\implies m = \frac{1}{5}$

this is the perpendicular gradient ${m}_{2}$

so

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

$\implies {m}_{1} \times \frac{1}{5} = - 1$

$\therefore {m}_{1} = - 5$

we also have

" (x_1,y_1)=(8,-2)

$y - {y}_{1} = m \left(x - {x}_{1}\right)$

becomes

$y - 8 = - 5 \left(x - - 2\right)$

$y - 8 = - 5 x - 10$

$5 x + y + 2 = 0$

If the reader compares this with the original lines $x - 5 y - 7 = 0$

a pattern may be seen .