Question

How do you find the slope of the line that passes through (-8,-15), (-2,5)?

Answer 1

Question answer: The slope is `10/3`
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Going beyond the question:

`y=10/3x+ 11 2/3`

Explanation:

The two points will give the gradient of a straight line graph.

Consider the standard for of: `y=mx+c`

Where
`y` is the answer

`m` is the gradient (slope)

`c` is the y-axis intercept (point where it crosses the y axis)
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`color(brown)("Determine the gradient")`

Let point 1 `->P_1->(x_1,y_1) = (-8,-15)`

Let point 2 `->P_2->(x_2,y_2)=(-2,5)`

The gradient (slope) is measured moving from left to right on the x-axis. As `" " x_1 < x_2 " "` (`x_1" is less than "x_2`) then we travel from `P_1" to "P_2` for the gradient

So the gradient `->m=("change in up or down")/("change in along") larr "Left to right"`

`m=("change in y-axis")/("change in x-axis") = (y_2-y_1)/(x_2-x_1)`

`m=(5-(-15))/((-2)-(-8)) " "=" "20/6 = 10/3 color(red)( larr" Question answer")`

So the equation now is `y=10/3x+c`

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`color(brown)("Determine y-intercept (crossing point on the y axis)")`

Using any one of the 2 given points substitute to solve for `c`

`P_1->(x_1,y_1)=(color(blue)(-8),color(green)(-15))`
`color(white)(.)`

`=>color(green)(y)=10/3color(blue)(x)+c" "->" "color(green)(-15)=10/3xx(color(blue)(-8))+c`

`-15=-80/3+c`

Add `80/3` to both sides

`11 2/3 = c`
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`color(brown)("Putting it all together")`

`y=mx+c" "->" "y=10/3x+ 11 2/3`

Note that `11 2/3 = 35/3`