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

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6 votes

Answer:


y=2x-9

Explanation:

you have the line


-3x-6y=17

clearing for y:


-6y=17+3x\\y=(17)/(-6) +(3)/(-6) x\\y=-(1)/(2) x-(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=-(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)=(-1)/(m) \\\\m_(2)=(-1)/((-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

User Zamnuts
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