Graphing Parallel Lines with a Missing Value..HELP?

Question

Find k so that the line containing the points (-3,k) and (4,8) is parallel to the line containing the points (5,3) and (1, -6).

Answer 1

Parallel lines have the same slope.

 

For line 1:

 

Point 1 = (-3, k)

Point 2 = (4, 8)

 

 

For line 2:

 

Point 3 = (1, -6)

Point 4 = (5, 3)

 

Slope = change in y / change in x

 

Slope of line 1 = Slope of line 2

 

(y₂ - y₁) / (x₂ - x₁) = (y₄ - y₃) / (x₄ - x₃)

 

(8 - k) / (4 - (-3)) = (3 - (-6)) / (5 - 1)

 

(8 - k) / 7 = 9 / 4

 

 

Cross multiply.

 

4(8 - k) = 7(9)

 

4(8 - k) = 63

 

 

Divide both sides of the equation by 4.

 

8 - k = 15.75

 

 

Subtract 8 on both sides of the equation.

 

-k = 7.75

 

 

Divide both sides of the equation by -1.

 

 k = -7.75

Answer 2

Hi Mario.

Parallel lines have slopes that are the same. So the first step is to find the slope of the line between the two points that we know. Between these two points, x goes from 1 to 5, so 5-1=4. In that span, y goes from -6 six to 3, where 3-(-6)=9.

So (y_2-y_1)/(x_2-x_1)=9/4.

Now, we want the second line to also have a slope of 9/4. The change in x from -3 to 4 is 7.

so what we want is that (8-k)/7=9/4.

If we multiply this whole equation by 7, we get.

8-k=(9*7)/4       subtract 8 from both sides:

 

-k= (9*7)/4-8     multiply by -1

 

k= 8- (9*7)/4

 

k= 9-63/4

k= -6.75  (or negative 6 and 3 quarters)