Question

WORTH 50!! Answer it with explanation and the answer is not y+6 I have some idea on how to do this.

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,  

$y=mx$

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

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

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 $3x+2y=34$