Question

Help with this one because I do not know how to do perpendicular plz put explanation with steps worth 50!

Answer 1

Answer:

Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line $2x-3y=12$ is $3x+2y=34$  

Step-by-step explanation:

Given:  

$2x-3y=12$  

To Find:  

Equation of line passing through ( 16, -7) and is perpendicular to the line  

$2x-3y=12$  

Solution:  

$2x-3y=12$ ...........Given  

$\therefore y=\dfrac{2}{3}\times x-4$

Comparing with,  

$y=mx+c$  

Where m =slope  

We get  

$Slope = m1 = \dfrac{2}{3}$  

We know that for Perpendicular lines have product slopes = -1.

$m1\times m2=-1$

Substituting m1 we get m2 as

$\dfrac{2}{3}\times m2=-1\\\\m2=-\dfrac{3}{2}$

Therefore the slope of the required line passing through (16 , -7) will have the slope,

$m2=-\dfrac{3}{2}$  

Now the equation of line in slope point form given by  

$(y-y_{1})=m(x-x_{1})$  

Substituting the point (16 , -7) and slope m2 we will get the required equation of the line,  

$(y-(-7))=-\dfrac{3}{2}\times (x-16)\\\\2y+14=-3x+48\\3x+2y=34......Equation\ of\ line$  

Therefore, equation of the line that passes through (16,-7) and is perpendicular to the line  $2x-3y=12$ is

$3x+2y=34$