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In standard form, find the equation of the line passing through (1,8) and (-2,-1)

0 votes
In standard form, find the equation of the line passing through (1,8) and (-2,-1).
asked Mar 13, 2014 in PRE-ALGEBRA by angel12 Scholar

2 Answers

+1 vote

Let the given points

(x1, y1) = (1, 8)

(x2, y2) = (- 2, - 1)

Line equation through the two points is

image

y - 8 = [(- 1 - 8)/(- 2 - 1)](x - 1)

y - 8 = [- 9/- 3](x - 1)

y - 8 = 9/3(x - 1)

Multiply 3 to each side.

3y - 24 = 9x - 9

Separate variables and constants.

9x - 3y = - 24 + 9

9x - 3y = - 15

Add 15 to each side.

9x - 3y + 15 = 0.

The standard form of the equation is 9x - 3y + 15 =0.

 

answered Mar 13, 2014 by dozey Mentor
0 votes

The standard form line equation is Ax + By = C. A shouldn't be negative, A and B shouldn't both be zero, and A, B and C should be integers.

 

Let the points are (x₁, y₁) = (1, 8) and (x₂, y₂) = (- 2, - 1).

Slope (m) = [(y₂ - y₁)/(x₂ -x₁)]

m = [(-1 - 8)/(- 2 - 1)]

m  = [- 9/- 3]

m = 3.

Now, the line equation is y = 3x + b.

 

Find the y - intercept by substituting any point in the line equation say (x, y) = (1, 8).

8 = (3)(1) + b

b = 8 - 3

b = 5.

The line equation in slope - intercept - form is y = 3x + 5.

 

Write the equation y = 3x + 5 in standard form of line equation.

Subtract y from each side.

3x - y + 5 = 0

Subtract 5 from each side.

3x - y = - 5.

The standard form of line equation is 3x - y = - 5.

answered Jun 12, 2014 by lilly Expert

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