Algebra Honors ( Periods 5 & 6)

Determining an Equation of a Line 8-5

Write an equation of a line that has slope 2 and y-intercept 3

Easy! remember           y = mx + b

just substitute in what you have… y = 2x + 3

Write an equation of a line that has slope -4 and x-intercept 3.

This is a little different. 

The x-intercept is the x-coordinate of the point where a line crosses the x-axis. So here the point must be (3,0) so now you can substitute into y=mx + b and find b.

First substitute in -4 for m  y = -4x +b

To find b , substitute 3 for x and 0 for y in y = -4x + b

so

0=-4(3) + b

0=-12 + b

12=b

Therefore the equation must be y = -4x + 12

Write an equation of the line passing through the points ( -2, 5) and ( 4, 8)

Since it doesn't give us the slope, we must find it using

   

Substitute  ½ for m in y = mx + b

y =  x/2 + b

Choose one of the points, say (4, 8) and substitute 4 for x and 8 for y.

8 = (1/2)(4) + b

8 = 2 + b

6 = b

Therefore the equation is y = (1/2)x + 6

or y = x/2 + 6

Note: that we could have used the point ( -2, 5) and the resulting equation would have been the same!

Write an equation in STANDARD FORM for the following line described

The line that is parallel to x – 2y + 7 = 0 and contains ( -4, 0)

The first thing we need to do is change the given line into slope-intercept form 

y = mx + b

-2y = -x – 7  becomes   y = x/2 + 7/2

If the line we are trying to find is parallel to that line its slope must also be ½

Now using y = mx + b and the point on the line

0 = (1/2)(-4) + b

0= -2 + b

b = 2

so the line is y = x/2 + 2

BUT that is not in standard form Ax + By = C

Where A, B, and C are integers and A is a  whole number

-x/2 + y = 2

Now multiply everything by -2

x- 2y = -4

Find a line passing through ( -2, 3), (2, 5) and (6, k)  Find k

First find the slope using the two given points

y = (1/2)x +b

Using ( -2,3)

3 = (1/2)(-2) + b

3 = -1 + b

4 = b

y = (1/2)x + 4  is the equation of the line so to solve for k

k = (1/2)(6) + 4

k = 3 + 4 = 7