SOLUTION: A line that is perpendicular to the line segment connected by (4,-2) and (6,-8) but passes through the point (12,-5) wh at is this is slope and General form

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Question 980907: A line that is perpendicular to the line segment connected by (4,-2) and (6,-8) but passes through the point (12,-5) wh
at is this is slope and General form
Found 2 solutions by josgarithmetic, Cromlix:
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
Find the equation for either form, and convert (algebra steps) into the other form.

slope m, for the points (4,-2) and (6,-8):
m=%28-8-%28-2%29%29%2F%286-4%29=-6%2F2=-3.
The slope perpendicular to this is its negative reciprocal, which is 1%2F3.

The wanted equation using POINT-SLOPE FORM (because this is easier to arrange) is highlight%28y-%28-5%29=%281%2F3%29%28x-12%29%29, and this line passes through (12,-5).

Convert the equation into slope intercept form, and into general form.


--
--
Slope-Intercept Form will look like y=mx+b
and
General Form will look like Ax+By=C.

Answer by Cromlix(4381) About Me  (Show Source):
You can put this solution on YOUR website!
Hi there,
First work out gradient:
Gradient = y2 - y1/x2 - x1
Utilising (4,-2) and (6, -8)
-8 -(-2)/6 - 4
-8 + 2/6 - 4
-6/2
= -3
If two lines are perpendicular to one another
then their gradients multiply together to give -1
m1 x m2 = -1
So, -3 x m2 = - 1
m2 = 1/3
Next use line equation:
y - b = m(x - a)
Utilising m = 1/3 and (12, -5)
y -(-5) = 1/3(x - 12)
y + 5 = 1/3(x - 12)
y + 5 = 1/3x - 4
y = 1/3x -4 - 5
y = 1/3x - 9
or
3y = x - 27
Hope this helps:-)