# How do you find a standard form equation for the line that is perpendicular to 6x-2y+5=0 and has the same y-intercept as x-y+8=0?

Jun 15, 2017

When given an equation of a line of the form:

$a x - b y + c = 0$

The family of perpendicular lines are of the form:

$b x + a y + k = 0$

Where k is an unknown constant.

#### Explanation:

Given: $6 x - 2 y + 5 = 0$

The family of perpendicular lines are:

$2 x + 6 y + k = 0 \text{ [1]}$

We must find the y intercept of the line $x - y + 8 = 0$:

$0 - y + 8 = 0$

$y = 8$

Substitute the point $\left(0 , 8\right)$ into equation [1] an then solve for k:

$2 \left(0\right) + 6 \left(8\right) + k = 0$

$k = - 48$

Substitute this into equation [1]:

$2 x + 6 y - 48 = 0 \text{ [2]}$

Jun 15, 2017

$y = - \frac{1}{3} x + 8$

This is the same as: $\frac{1}{3} x + y - 8 = 0 \text{ "->" } x + 3 y - 24 = 0$

#### Explanation:

First consider $x - y + 8 = 0 \text{ } \ldots \ldots \ldots \ldots \ldots \ldots E q u a t i o n \left(1\right)$

This can be written as: $y = x + 8$

So the y-intercept for this equation is 8.

Thus the equation we are looking for passes through the point:
$\left(x , y\right) \to \left(0 , 8\right)$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Consider the given equation: $6 x - 2 y + 5 = 0 \text{ } \ldots . . E q u a t i o n \left(2\right)$

Add $2 y$ to both sides

$2 y = 6 x + 5$

Divide both sides by 2

$y = 3 x + \frac{5}{2} \text{ } \ldots \ldots . E q u a t i o n \left({2}_{a}\right)$

Thus the gradient (slope) for $E q u a t i o n \left({2}_{a}\right)$ is 3 from the $3 x$
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Compare $E q u a t i o n \left({2}_{a}\right)$ to the standardised form of$\text{ } y = m x + c$
where $m$ is the gradient (slope).

A line that is perpendicular to it will have the gradient of:
$\left(- 1\right) \times \frac{1}{m} = - \frac{1}{m} \text{ "->" } \left(- 1\right) \times \frac{1}{3} = - \frac{1}{3}$

Thus our standardised form for the new line is:

$y = - \frac{1}{3} x + c$

From $E q u a t i o n \left(1\right)$ we have determined that this line passes through the point $\left(x , y\right) \to \left(0 , 8\right)$

So $y = - \frac{1}{3} x + c \text{ "->" } 8 = - \frac{1}{3} \left(0\right) + c$

Thus $c = 8$

So the equation of the perpendicular line is:
$y = - \frac{1}{3} x + 8$