What is the slope of a line that has an X-Intercept of 8 and a Y-Intercept of 11?

dourtuntellorvl
2022-06-26
Answered

What is the slope of a line that has an X-Intercept of 8 and a Y-Intercept of 11?

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Reagan Madden

Answered 2022-06-27
Author has **15** answers

If we know the X-Intercept and the Y-Intercept, then we have two points. With two points, we can define the slope of the line and, indeed, an equation for the line through those two points.

Using your example, suppose the X-Intercept is 8 and Y-Intercept is 11. Then we know that $(8,0)$ and $(0,11)$ are points on this line. Therefore, the slope of the line is $\frac{\mathrm{\Delta}y}{\mathrm{\Delta}x}$, or the change in $y$ over the change in $x$. Therefore, $slope=\frac{0-11}{8-0}=\frac{-11}{8}$. Now we can use point-slope form for a line through a point to give a formula for the line.

Point-slope form is defined by $y-{y}_{1}=m(x-{x}_{1})$) where ${x}_{1},{y}_{1}$ are the x and y values from one of your points, and m is the slope. I will use the point $(8,0)$, although we can very easily choose the other point and get the same formula, so that the formula for this line is $y-0=\frac{-11}{8}(x-8)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y=\frac{-11}{8}x+11$

Using your example, suppose the X-Intercept is 8 and Y-Intercept is 11. Then we know that $(8,0)$ and $(0,11)$ are points on this line. Therefore, the slope of the line is $\frac{\mathrm{\Delta}y}{\mathrm{\Delta}x}$, or the change in $y$ over the change in $x$. Therefore, $slope=\frac{0-11}{8-0}=\frac{-11}{8}$. Now we can use point-slope form for a line through a point to give a formula for the line.

Point-slope form is defined by $y-{y}_{1}=m(x-{x}_{1})$) where ${x}_{1},{y}_{1}$ are the x and y values from one of your points, and m is the slope. I will use the point $(8,0)$, although we can very easily choose the other point and get the same formula, so that the formula for this line is $y-0=\frac{-11}{8}(x-8)\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}y=\frac{-11}{8}x+11$

gnatopoditw

Answered 2022-06-28
Author has **5** answers

The intercepts can be treated as special points. In your case you have $(8,0)$ and $(0,11)$.

You can get the gradient from that by using gradient

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{11-0}{0-8}=-\frac{11}{8}$

You can get the gradient from that by using gradient

$m=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}=\frac{11-0}{0-8}=-\frac{11}{8}$

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