Question

write the equation for the line that is perpendicular to given line that passes through the given point -3x-6y=17; (6,3)

Answer 1

Slope of the given line = -1/2
the equation is
(y-3)/(x-6)=2
y-3=2x-12
y=2x-9

Answer 2

Answer:

$y=2x-9$

Step-by-step explanation:

you have the line

$-3x-6y=17$

clearing for y:

$-6y=17+3x\\y=\frac{17}{-6} +\frac{3}{-6} x\\y=-\frac{1}{2} x-\frac{17}{6}$

we have an equation of the form

$y=mx+b$

the number that accompanies the x is the slope, and the number alone is the intercept with the y-axis

the slope m is:

$m=-\frac{1}{2}$

i will call the slope of the new line $m_{2}$

so for two perpendiculares lines we must have:$m*m_{2}=-1$

and from this we can find the new slope:

$m_{2}=\frac{-1}{m} \\\\m_{2}=\frac{-1}{\frac{-1}{2} } \\m_{2}=2$

the new slope is 2,

so far we have that the new line is:

$y=2x+b$

so now we have to find the intercept with the y axis ($b$) of the new line, since it passes trough (6,3) ---> x = 6 when y = 3

substituting these x and y values in $y=2x+b$:

$3=2(6)+b\\3=12+b\\b=3-12\\b=-9$

and finally, the equation of the new line that is perpendicular to the original line is:

$y=2x-9$