Write an equation for the line in point slope form and general form
Passing through (6, -9) and perpendicular to the line whose equation is x-8y-3=0
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1.Find the gradient of the line
for perpendicular lines G1 x G2 = -1
x - 8y -3 = 0
-8y = - x + 3
y = 1/8 x -3
G1 = 1/8
1/8 G2 = -1
G2 = - 8
The "point-slope" form of the equation of a straight line is:
y - y1 = m(x - x1)
The coordinates are
=( y + 9) = -8(x - 6)
I am submitting the general form in my next submission due to time.
asks for the equation of the function ?
The "General Form" of the equation of a straight line is:
Ax + By + C = 0
A or B can be zero, but not both at the same time.
y + 9 = - 8x + 48
8x + y + 9 - 48 = 0
8x + y -39 = 0
The Slope-Intercept Form of the equation of a straight line:
y = mx + b
The Point-Slope Form of the equation of a straight line:
y − y1 = m(x − x1)
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