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Mice21 [21]
3 years ago
12

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

Mathematics
2 answers:
irga5000 [103]3 years ago
8 0
Slope of the given line = -1/2
the equation is
(y-3)/(x-6)=2
y-3=2x-12
y=2x-9

maxonik [38]3 years ago
6 0

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

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