SOLUTION: I'm don't understanding how to solve this question: The graph of a line passes through the x-intercept of y-20=2/11(x-31) and is perpendicular to the line y=1/7x+7 . The equati

Algebra ->  Linear-equations -> SOLUTION: I'm don't understanding how to solve this question: The graph of a line passes through the x-intercept of y-20=2/11(x-31) and is perpendicular to the line y=1/7x+7 . The equati      Log On


   

Question 957503: I'm don't understanding how to solve this question:
The graph of a line passes through the x-intercept of y-20=2/11(x-31) and is perpendicular to the line y=1/7x+7 . The equation of the line, written in Slope-Point form, is y-p = m(x-q). what is the value of m*q ?
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
First find the x-intercept of
y-20=%282%2F11%29%28x-31%29
So solve for x when y=0
-20=%282%2F11%29%28x-31%29
-110=x-31
x=-79
So the x-intercept is (-79,0).
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Perpendicular lines have slopes that are negative reciprocals.
m%5Bp%5D%2A%281%2F7%29=-1
m%5Bp%5D=-7
So then using the point-slope form,
y-0=-7%28x-%28-79%29%29
y=-7%28x%2B79%29
m=-7
q=-79
m%2Aq=553