# Use the point on the line and the slope of the line to find three additional points that the line...

## Question:

Use the point on the line and the slope of the line to find three additional points that the line passes through. (There is more than one correct answer.)

Point: (-6,6)

Slope m = 4

## Equation of the Line:

From the point-slope form {eq}y-y_1 = m(x-x_1) {/eq}, the equation of a line requires at least one given point and a slope value ({eq}m {/eq}) to reflect the line's rate of change between points on the line. {eq}x_1 {/eq} and {eq}y_1 {/eq} correspond to given point provided and thus can be replaced by the corresponding components of the point given.

Given: {eq}\text{ Point }: (-6,6) \\ \text{ Slope } = m = 4 {/eq}

To find {eq}3 {/eq} additional points that the line describing the given conditions, the strategy is to create an equation of a line with the formula {eq}y-y_1 = m(x-x_1) {/eq} and substitute any three {eq}x {/eq} inputs into the equation to find the corresponding {eq}y {/eq} values since those values passed by the line.

For the equation of the line:

{eq}\begin{align*} y-y_1 = m(x-x_1) &\Rightarrow y-6 = 4(x-(-6)) \text{ [Given conditions substituted in the equation]} \\ &\Rightarrow y-6 = 4(x+6) \\ &\Rightarrow y-6 = 4x+24 \\ &\Rightarrow y-6+6 = 4x+24+6 \\ &\Rightarrow y = 4x+30 \text{ [Slope-intercept line equation form]} \\ \end{align*} {/eq}

Now three random {eq}x {/eq} inputs can be plugged into the equation of the line to determine the complete points that the line passes through.

When {eq}x = 0 {/eq}:

{eq}\begin{align*} y &= 4(0)+30 \\ &= 0+30 \\ &= 30 \\ \end{align*} {/eq}

When {eq}x = 1 {/eq}:

{eq}\begin{align*} y &= 4(1)+30 \\ &= 4+30 \\ &= 34 \\ \end{align*} {/eq}

When {eq}x = 2 {/eq}:

{eq}\begin{align*} y &= 4(2)+30 \\ &= 8+30 \\ &= 38 \\ \end{align*} {/eq}

Therefore, {eq}3 {/eq} additional points that the line passes through are {eq}(0,30) {/eq}, {eq}(1,34) {/eq}, and {eq}(2,38) {/eq}. 