How do you find the equation of the line perpendicular to y+5=3(x-2) that passes through the point (6, 2)?

Mar 6, 2018

Let,the equation of line be $y = m x + c$, where, $m$ is its slope and $c$ is the $Y$ intercept.

Now,arranging the given equation in the above mentioned form to get its slope,

Given, $y + 5 = 3 \left(x - 2\right) = y = 3 x - 11$

So,its slope is $3$

Now,for two lines to be mutually perpendicular,their product of slope must be $- 1$

So, $m \cdot 3 = - 1$

or. $m = - \frac{1}{3}$

So,our required line equation becomes, $y = - \frac{1}{3} x + c$

Now,given,that the line passes through $\left(6 , 2\right)$,so putting the values in the equation to get $c$

So, $2 = \left(- \frac{1}{3}\right) \cdot 6 + c$

or, $c = 4$

So,the equation of the line is $y = - \frac{1}{3} x + 4$

or, $3 y + x = 12$ graph{3y+x=12 [-10, 10, -5, 5]}

Mar 6, 2018

$y - 2 = - \frac{1}{3} \left(x - 6\right)$

Or $y = - \frac{1}{3} x + 4$

Or $x + 3 y = 12$

Or $x + 3 y - 12 = 0$

Explanation:

There are several forms for the equation of a straight line:

$y = m x + c , \text{ and " ax +by +c =0" }$ are well known.

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

where $\left({x}_{1} , {y}_{1}\right)$ is a point and $m$ is the slope.

This is exactly the form we have for the given equation:

$y + 5 = \textcolor{b l u e}{3} \left(x - 2\right) \text{ } \rightarrow \therefore \textcolor{b l u e}{m = 3}$

If lines are perpendicular, the product of their slopes is $- 1$

One is the negative reciprocal of the other, so:

${m}_{1} = 3 \text{ "rArr" } {m}_{2} = - \frac{1}{3}$

Using $m = - \frac{1}{3}$ and the point $\left(6 , 2\right)$

$y - {y}_{1} = m \left(x - {x}_{1}\right) \text{ }$ gives$\text{ } y - 2 = - \frac{1}{3} \left(x - 6\right)$

Or in another form:

$y = - \frac{1}{3} x + 2 + 2 \text{ } \rightarrow y = - \frac{1}{3} x + 4$

Or even

$3 y = - x + 12 \text{ "rarr" } x + 3 y = 12$