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write the slope intercept equation for the line that pass through (-12,10) and is perpendicular to 4x+6y =3Asked by Jackie
Write the slope-intercept equation for the line that passes through (-3,-15)and is perpendicular to -6x+8y=3
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First, let's solve the equation for y to determine the slope of the line.
-6x + 8y = 3
8y = 6x + 3
y = (6/8)x + 3/8 = (3/4)x + 3/8
Slope = 3/4
The slope of a line perpendicular to this one must be its negative reciprocal, or -4/3.
Now, use the point-slope form to find the other line:
y - y1 = m(x - x1)
y - (-15) = -4/3 [x - (-3)]
y + 15 = -4/3 (x + 3)
y + 15 = (-4/3)x - 4
y = (-4/3)x - 4 - 15
y = (-4/3)x - 19
And there you have it! By showing you the steps, I hope this will help with other problems of this type.
-6x + 8y = 3
8y = 6x + 3
y = (6/8)x + 3/8 = (3/4)x + 3/8
Slope = 3/4
The slope of a line perpendicular to this one must be its negative reciprocal, or -4/3.
Now, use the point-slope form to find the other line:
y - y1 = m(x - x1)
y - (-15) = -4/3 [x - (-3)]
y + 15 = -4/3 (x + 3)
y + 15 = (-4/3)x - 4
y = (-4/3)x - 4 - 15
y = (-4/3)x - 19
And there you have it! By showing you the steps, I hope this will help with other problems of this type.
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