Question 102449This question is from textbook glencoe geometry concepts and applications
: Find the equation of the line satisfying the given conditions: perpendicular to the graph of y + 6 = -1 / 3x and passes through the point at (-4, -5).
Find the equation of the line satisfying the given conditions: parallel to the x-axos and passes through the point (-5, -3).
Use slope to show that (triangle symble)ABC with vertices A(0,2), B(2,3) and C(1,5) is a right triangle.
This question is from textbook glencoe geometry concepts and applications
Answer by stanbon(75887)   (Show Source):
You can put this solution on YOUR website! Find the equation of the line satisfying the given conditions: perpendicular to the graph of y + 6 = (-1/3)x and passes through the point at (-4, -5).
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The given line has a slope of -1/3; so every perpendicular line must
have a slope of +3
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The equation you want is y = mx+b; you need to find m and b.
m= + 3
-5 = 3(-4)+ b
-5 = -12 + b
b = 7
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Equation is y = 3x+7
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Find the equation of the line satisfying the given conditions: parallel to the x-axis and passes through the point (-5, -3).
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Parallel to the x-axis means the slope = zero
Passing thru (-5,-3) means
-3 = 0(-5)+ b
b = -3
EQUATION:
y = -3
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Use slope to show that (triangle symbol)ABC with vertices A(0,2), B(2,3) and C(1,5) is a right triangle.
Check the slope of each side:
slope of AB: [3-2]/[2-0] = 1/2
slope of AC: [5-2]/[1-0] = 3
slope of BC: [5-3]/[1-2] = -2
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AB and BC are perpendicular and therefore form a right triangle at B.
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Cheers,
Stan H.
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