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

(6, -9)

=( 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

Pardon please?

NB:

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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