SOLUTION: Find the value of x so that the line containing points at (x,2) and (-4, 5) is perpendicular to the line containing points at (4,8) and (2,-1).
I have determined that the slope
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Question 121952This question is from textbook Glencoe Mathematics Geometry
: Find the value of x so that the line containing points at (x,2) and (-4, 5) is perpendicular to the line containing points at (4,8) and (2,-1).
I have determined that the slope for the line containing (4,8) and 2,-1 is 9/2 and thus the slope for the perpendicular line would be -2/9. I figured that if I used the y2-y1 over x2-x1 and set that = to -2/9 that I would get the answer and it just is not working. The book says that the answer is 9.5. Can you help?
This question is from textbook Glencoe Mathematics Geometry
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Find the value of x so that the line containing points at (x,2) and (-4, 5) is perpendicular to the line containing points at (4,8) and (2,-1).
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Yes the slope is -2/9
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EQUATION to solve for "x":
(2-5)/(x--4) = -2/9
Cross-multiply:
-2(x+4) = -27
x+4 = 13.5
x = 9.5 or 19/2
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Cheers,
Stan H.
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
The work you have done to find the slope of the perpendicular is spot on. But now you need to use the point-slope form of a line to develop an equation for the perpendicular line.
Now solve for x so that we can answer the question: What must x be when y = 2?
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