Algebra 1 Help Fill In extras?

Best thing to do is first find the slope of the line:

m = (y2 - y1) / (x2 - x1)

m = (-4 + 1) / (7 + 2)

m = -3/9

m = -1/3

Next, we want to do point-slope form:

(y - y1) = m(x - x1)

y + 1 = (-1/3)(x + 2)

Next we want to solve for y to get slope-intercept form:

y + 1 = (-1/3)x - (2/3)

y = (-1/3)x - (5/3)

Now let's find the y-intercept. This is "b" in slope-intercept form so:

y-intercept = (0, -5/3)

Now let's find the x-intercept. This occurs when y=0.

0 = (-1/3)x - (5/3)

(1/3)x = -(5/3)

x = -5

So:

x-intercept = (-5, 0)

Standard form, take your slope-intercept form, and multiply by some clever value so there's no fraction and put in Ax + By = C:

(3)*[y = (-1/3)x - (5/3)]

3y = -1x - 5

x + 3y = -5

(x1, y1) = (-2, -1)

(x2, y2) = (7, -4)

Slope, m = (y2 - y1) / (x2 - x1) = (-4 - -1) / (7 - -2) = (-4 + 1) / (7 + 2) = -3/9 = -1/3

y - y1 = m(x - x1)

y - -1 = (-1/3)(x - -2)

y + 1 = (-1/3)(x + 2)

y + 1 = (-1/3)x - 2/3

y = (-1/3)x - 2/3 - 1

y = (-1/3)x - 2/3 - 3/3

y = (-1/3)x - 5/3 <=== Slope-intercept form of the linear equation

Based on this equation, the y-intercept is -5/3; or point (0, -5/3).

(y + 1) = (-1/3)(x + 2) <=== point-slope form of the linear equation

Cross-multiply:

-3(y + 1) = (x + 2)

-3y - 3 = x + 2

x + 3y = -3 - 2

x + 3y = -5 <=== Standard form of the linear equation

To find the x-intercept, find the value of x when y = 0.

x + 3(0) = -5

x + 0 = -5

x = -5 <=== the x-intercept or point (-5, 0)

x1 = - 2

y1 = - 1

x2 = 7

y2 = - 4

Slope, m = (y2 - y1) / (x2 - x1)

m = [(- 4) - (- 1)] / (7 - (- 2)]

m = (- 4 + 1) / (7 + 2)

m = - 3 / 9

m = - 1/3

b = y - mx

b = - 1 - [(- 1/3)(- 2)]

b = - 1 - (2/3)

b = - 3/3 - 2/3

b = - 5/3

x-int.: y = 0:

- 1/3 x - 5/3 = 0

- 1/3 x = 5/3

x = (5/3) / (- 1/3)

x = (5/3)(- 3)

x = - 15 / 3

x = - 5

x-int.: (- 5, 0)

¯¯¯¯¯¯¯¯¯¯¯

y-int.: x = 0:

y = - 1/3(0) - 5/3

y = - 5/3

y-int. (0, - 5/3)

¯¯¯¯¯¯¯¯¯¯¯¯

Slope-Intercept Form:

y = - 1/3x - 5/3

¯¯¯¯¯¯¯¯¯¯¯¯¯

Standard Form:

1/3x + y + 5/3 = 0

x + 3y + 5 = 0

¯¯¯¯¯¯¯¯¯¯¯¯¯

Point-Slope Form:

(y - y1) = m(x - x1)

(y - (- 1) = - 1/3[x - (- 2)]

(y + 1) = - 1/3(x + 2)

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯

Slope= -1/3

y+1=-1/3(x+2)

y=-1/3x-1/3

y-intercept= -1/3

x-intercept= 1

Hope I helped!

y = -1/3x - 5/3

y-intercept = -5/3

x-intercept = -5

Slope-intercept form: y=mx+b

the slope, m, is found by finding:

(y2-y1)/(x2-x1) = (-4-(-1))/(7-(-2)) = -1/3

So you have:

y = -1/3x + b, where b is the y-intercept.

To find b, plug in one of your point: (-2,-1)

-1 = (-1/3)(-2) + b, which gives b = -5/3

So you have:

y = -1/3x - 5/3

To find the x-intercept, set y=0

0 = -1/3x -5/3

Solving, you get x = -5