If I am given the equation y=2/3x+7 how do i find a line perpendicular that passes through points (0,5)
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The gradient of y=(2/3)x+7 is 2/3.
The gradient of a perpendicular line is -1 ÷ 2/3 = -3/2.
So the equation is y = (-3/2)x + c.
c is the y-intercept and we know the line passes through (0, 5) so c = 5 and the equation is
y = (-3/2)x + 5
The gradient of a perpendicular line is -1 ÷ 2/3 = -3/2.
So the equation is y = (-3/2)x + c.
c is the y-intercept and we know the line passes through (0, 5) so c = 5 and the equation is
y = (-3/2)x + 5
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If a/b is the slope of a line, then -b/a is the slope of the perpendicular line.
The formula point-slope is y - y1 = m(x -x1)
In your problem the slope is 2/3, then the slope of the perpendicular is -3/2 .... the point is (0,5)...
just put in the formula and get : y - 5 = (-3/2)(x-0) ==> y = -3x/2 + 5 OK!
The formula point-slope is y - y1 = m(x -x1)
In your problem the slope is 2/3, then the slope of the perpendicular is -3/2 .... the point is (0,5)...
just put in the formula and get : y - 5 = (-3/2)(x-0) ==> y = -3x/2 + 5 OK!