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Write an equation (a) in slope-intercept form and (b) in standard form for the line passing through (1,6) and perpendicular to 3 x + 7y = 1.a) The equation of the line in slope-intercept form is   enter your response here

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The equation


3x+7y=1
7y=-3x+1
y=(-3x)/(7)+(1)/(7)

has slope -3/7.

The product of perpendicular lines must be -1, so the slope of the new equation must be:


m\cdot-(3)/(7)=-1
m=(-1\cdot7)/(-3)=(7)/(3)

The line with that slope and that passes throught (1,6) is:


(y-6)=m(x-1)
(y-6)=(7)/(3)(x-1)
y-6=(7)/(3)x-(7)/(3)
y=(7)/(3)x-(7)/(3)+6
y=(7)/(3)x-(7)/(3)+(18)/(3)
y=(7)/(3)x+(11)/(3)

The following graph shows both equations:


y=(7)/(3)x+(11)/(3)
3\cdot y=((7)/(3)x+(11)/(3))\cdot3
3y=7x+11
-7x+3y=11

Write an equation (a) in slope-intercept form and (b) in standard form for the line-example-1
User ChrisRob
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