Answer:
y=-[tex]\frac{3}{2}[/tex]x-13
Step-by-step explanation:
You have to convert everything to slope intercept form before starting
2x-3y=5
3y=2x-5
y=2/3x-5/3
Then,
Since Line M is perpendicular Line M's slope will be,
-[tex]\frac{3}{2}[/tex]x
y=-3/2x+b
Enter the given point (2,-10)
-10=-3/2(2)+b
-10=3+b
b=-10-3
b=-13
so the equation for line M is
y=-3/2x-13
Answer:
[tex]y=-\frac{3}{2}x -7[/tex]
Step-by-step explanation:
Line L has the equation:
[tex]2x-3y=5[/tex]
we need to clear for y:
[tex]-3y=-2x+5\\y=\frac{2}{3} x-\frac{5}{3}[/tex]
now we have the form a general line equation
[tex]y=mx+b[/tex]
where [tex]m[/tex] is the slope of the line.
so the slope of the line L is:
[tex]m_{1}=\frac{2}{3}[/tex]
and for two lines to be parallel the following condition must be met
[tex]m_{1}*m_{2}=-1[/tex]
where [tex]m_{2}[/tex] in this case is the slope of line M, substituting the value [tex]m_{1}[/tex] to find [tex]m_{2}[/tex]:
[tex]\frac{2}{3}*m_{2}=-1\\ m_{2}=\frac{-1(3)}{2}\\m_{2}=-\frac{3}{2}[/tex]
This is the slope of line M, and since we also know that it passes through the point (2, -10) where [tex]x_{0}=2[/tex] and [tex]y_{0}=-10[/tex]
we use the point- slope equation and substitute known values to find the equation of the line M:
[tex]y-y_{0}=m(x-x_{0})\\y-(-10)=-\frac{3}{2}(x-2)\\ y+10=-\frac{3}{2}x+3\\y=-\frac{3}{2}x +3-10\\y=-\frac{3}{2}x -7[/tex]
the equation of line M is:
[tex]y=-\frac{3}{2}x -7[/tex]
