Question

A line passes through `(3 ,4 )` and `(4 ,7 )`. A second line passes through `(7 ,3 )`. What is one other point that the second line may pass through if it is parallel to the first line?

Answer 1

See a solution process below:

Explanation:

First, we need to calculate the slope of the first line: `m_1`. The slope can be found by using the formula: `m = (color(red)(y_2) - color(blue)(y_1))/(color(red)(x_2) - color(blue)(x_1))`

Where `m` is the slope and (`color(blue)(x_1, y_1)`) and (`color(red)(x_2, y_2)`) are the two points on the line.

Substituting the values from the points in the problem gives:

`m_1 = (color(red)(7) - color(blue)(4))/(color(red)(4) - color(blue)(3)) = 3/1`

Because the two lines are parallel, the slope of the second line will also be `3/1`. We can call the slope of the second line `m_2`

Substituting the values from the one point and the slope we calculated for the first line gives:

`m_2 = (color(red)(y) - color(blue)(3))/(color(red)(x) - color(blue)(7))`

`3/1 = (color(red)(y) - color(blue)(3))/(color(red)(x) - color(blue)(7))`

We can now solve these two equations to get another point on the second line:

- Solve for the Numerator:

`y - 3 = 3`

`y - 3 + color(red)(3) = 3 + color(red)(3)`

`y - 0 = 6`

`y = 6`

- Solve for the Denominator:

`x - 7 = 1`

`x - 0 = 8`

One other point on the second line is: `(8, 6)`