# Equations of lines. Find an equation of the line.?

Find an equation of the line having the given slope and containing the given point.

m = 3/4,(7,-1)

Now I understand that (y-y1)= m(x-x1) and I followed it through but its the solving for y when it comes to fractions that is giving me the hardest time. I think its throwing me off when I subtract or add using fractions. Any help would be greatly appreciated. Thanks in advance.

Relevance

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

Use the point (7, -2) as (x1, y1)

The beauty about fractions is cross multiplication:

(y-(-1))= 3/4 (x-7)

(y+1) = 3/4 (x-7)

Multiply both sides by 4, the denominator of the slope.

4(y+1) = 3(x-7)

4y + 4 = 3x - 21

4y = 3x - 21 - 4

4y = 3x - 25. This is the line you needed.

In slope intercept form:

y = 3x/4 - 25/4

Firstly, just write

y = mx + b,

Then, you know m, so

y = (3/4)x + b

Now, you know a point, so we can solve for b

-1 = (3/4)*7 + b

-1 = 21/4 + b

b = -1 - 21/4 = -6.25 (use a calculator if you don't like fractions)

So, now, we know m, and b, and we just write

y = (3/4)x - 6.25

Thus, that is the equation of your line.

Hope this helps.

y+1=3/4(y-7)

• 4 years ago

Slope intercept for is: y = mx + b considering which you have each and every of the values different than b you are able to fill in and resolve for b. y = mx + b -5 = (4/5)(0) + b -5 = 0 + b b = -5 So, y = (4/5)x - 5

(y + 1) = 3/4 (x- 7) multiply by 4

4y + 4 = 3x - 21

4y = 3x - 25

y = 3/4 x - 25/4

A linear equation is as follows:

y=mx+c

y=3/4x-1