Slope Intercept Equation

Algebra
Tutor: None Selected Time limit: 1 Day

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

Jun 7th, 2015

Thank you for the opportunity to help you with your question!

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)

 so (y - -9) = -8(x-6)

=( y + 9) = -8(x - 6)

I am submitting the general form in my next submission due to time.

Please let me know if you need any clarification. I'm always happy to answer your questions.
Jun 7th, 2015

asks for the equation of the function ?

Jun 7th, 2015

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) = -8(x - 6)

y + 9 = - 8x + 48

8x + y + 9 - 48 = 0

8x + y -39 = 0


Jun 7th, 2015

Pardon please?

Jun 7th, 2015

NB:

The "General Form" of the equation of a straight line is:

Ax + By + C = 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)


Jun 7th, 2015

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