Respuesta :
Answer:
[tex]y =- \frac{3}{2}x - 4[/tex]
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
[tex]y =- \frac{3}{2}x - 4[/tex]
Answer: [tex]y=-\frac{3}{2}x-4[/tex]
Step-by-step explanation:
The equation of the line in Slope-intercept form is:
[tex]y=mx+b[/tex]
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":
[tex]2x - 3y = 13\\\\-3y=-2x+13\\\\y=\frac{-2}{-3}x+\frac{13}{-3}\\\\y=\frac{2}{3}x-\frac{13}{3}[/tex]
We can identify that the slope of this line is:
[tex]m=\frac{2}{3}[/tex]
Since the slopes of perpendicular lines are negative reciprocals, then the slope of the other line is:
[tex]m=-\frac{3}{2}[/tex]
Now, we need to substitute the given point and the slope into [tex]y=mx+b[/tex] and solve for "b":
[tex]5=-\frac{3}{2}(-6)+b\\\\5=9+b\\\\5-9=b\\\\b=-4[/tex]
Substituting values, we get that the equation of this line is:
[tex]y=-\frac{3}{2}x-4[/tex]