Question

What is the equation of the line perpendicular to 2x – 3y = 13 that passes through the point (–6, 5)?

Answer 1

Answer:

$y =- \frac{3}{2}x - 4$

Step-by-step explanation:

Given equation of line is:

2x-3y=13

We will convert the equation of line in point-slope form to find the slope of the line

Let  

m_1  be the slope of the line

So,

2x-3y=13

-3y= -2x+13

Dividing both sides by -3

(-3y)/(-3)=(-2x)/(-3)+13/(-3)

y=(2/3)x-13/3

The co-efficient of x is the slope of the line.

So,

m_1=2/3

Let

m_2  be the slope of second line

As we know that product of slopes of two perpendicular line is -1

m_1 m_2= -1

2/3*m_2= -1  

m_2= -1*3/2

m_2= -3/2

So m2 is the slope of the line perpendicular to given line.

The standard equation of a line is  

y=mx+b

To find the equation of line through (-6,5), put the point and slope in the given form and solve for b

5= -3/2 (-6)+b

5=18/2+b

5=9+b

b=5-9

b= -4  

Putting the values of slope and b, we get

$y =- \frac{3}{2}x - 4$

Answer 2

Answer: $y=-\frac{3}{2}x-4$

Step-by-step explanation:

The equation of the line in Slope-intercept form is:

$y=mx+b$

Where "m" is the slope and "b" is the y-intercept.

To express the given equation of the line in this form, we need to solve for "y":

$2x - 3y = 13\\\\-3y=-2x+13\\\\y=\frac{-2}{-3}x+\frac{13}{-3}\\\\y=\frac{2}{3}x-\frac{13}{3}$

We can identify that the slope of this line is:

$m=\frac{2}{3}$

Since the slopes of perpendicular lines are negative reciprocals, then the slope of the other line is:

$m=-\frac{3}{2}$

Now, we need to substitute the given point  and the slope into  $y=mx+b$ and solve for "b":

$5=-\frac{3}{2}(-6)+b\\\\5=9+b\\\\5-9=b\\\\b=-4$

Substituting values, we get that the equation of this line is:

$y=-\frac{3}{2}x-4$