Respuesta :
When some line named l1 is perpendicular to some other line named l2 ,
then there is slope coefficijent
S1= - 1/ S2
In our case we must first transform given line from standard form to slope form
2x+3y= -5 => 3y = - 2x-5 => y = (-2/3)x -5
The slope coefficijent od the given line is S= - 2/3
Then the slope coefficijent of the requested line is
S1= - 1/(-2/3) => I suppose that you know to solve double fraction
S1=3/2
The equation for the line which passes through one point is
y-y1 = S1 (x-x1) => y-1 = 3/2 (x-(-6)) => y-1 =3/2 (x+6)
We will multiply both sides of the equation with number 2 and get
2y-2=3x+18 => 2y=3x+18+2 => 2y = 3x+20
When we divide the both sides with number 2 we finally get
y= (3/2)x+10.
Good luck!!!!
then there is slope coefficijent
S1= - 1/ S2
In our case we must first transform given line from standard form to slope form
2x+3y= -5 => 3y = - 2x-5 => y = (-2/3)x -5
The slope coefficijent od the given line is S= - 2/3
Then the slope coefficijent of the requested line is
S1= - 1/(-2/3) => I suppose that you know to solve double fraction
S1=3/2
The equation for the line which passes through one point is
y-y1 = S1 (x-x1) => y-1 = 3/2 (x-(-6)) => y-1 =3/2 (x+6)
We will multiply both sides of the equation with number 2 and get
2y-2=3x+18 => 2y=3x+18+2 => 2y = 3x+20
When we divide the both sides with number 2 we finally get
y= (3/2)x+10.
Good luck!!!!
Answer:
[tex]y=\frac{3}{2}x+10[/tex]
Step-by-step explanation:
Given equation of line : [tex]2x + 3y = -5[/tex]
Standard form of equation of line = [tex]y=mx+c[/tex] ---A
Where m is the slope
Convert the given equation in standard form
[tex]2x + 3y = -5[/tex]
[tex] 3y = -5-2x[/tex]
[tex] y =\frac{-5}{3}-\frac{2}{3}x[/tex]
So, slope =[tex]m = -\frac{2}{3}[/tex]
If the two lines are perpendicular then the product of their slopes is -1
Let n be the slope of required equation of line
So, [tex]m \times n = -1[/tex]
[tex]-\frac{2}{3}\times n = -1[/tex]
[tex]n=\frac{3}{2}[/tex]
Substitute this value in A
[tex]y=\frac{3}{2}x+c[/tex] --B
Now we are given that the required perpendicular line passes through the point (–6, 1)
So, substitute (–6, 1) in B
[tex]1=\frac{3}{2}(-6)+c[/tex]
[tex]1-(\frac{3}{2}(-6))=c[/tex]
[tex]10=c[/tex]
Substitute the value of c in B
[tex]y=\frac{3}{2}x+10[/tex]
Hence the slope-intercept form of the equation of the line that passes through the point (–6, 1) and is perpendicular to the graph of 2x + 3y = –5? is [tex]y=\frac{3}{2}x+10[/tex]
