Question

The slope of the line is -2. The line passes through (t, -1) and (-4,9). How do you find the value of t?

Answer 1

Please see the explanation for steps leading to `t = 1`

Explanation:

Use the formula for the slope:

`m = (y_2 - y_1)/(x_2 - x_1)`

where, `y_2 = 9, y_1 = -1, x_2 - 4 and x_1 = t`:

`-2 = (9 - -1)/(-4 - t)`

Simplify the numerator:

`-2 = 10/(-4 - t)`

Multiply both sides by (-4 - t):

`-2(-4 - t) = 10`

Distribute -2:

`2t + 8 = 10`

Subtract 8 from both sides:

`2t = 2`

`t = 1`

check:

`-2 = (9 - -1)/(-4 - 1) = -2`

This checks

Answer 2

`t=1`

Explanation:

Calculate the slope of the line using the `color(blue)"gradient formula"` and equate to - 2

`color(red)(bar(ul(|color(white)(2/2)color(black)(m=(y_2-y_1)/(x_2-x_1))color(white)(2/2)|)))`
where m represents the slope and `(x_1,y_1),(x_2,y_2)" 2 points on the line"`

Here the 2 points are (t ,-1) and (-4 ,9)

let `(x_1,y_1)=(t,-1)" and " (x_2,y_2)=(-4,9)`

`rArrm=(9-(-1))/(-4-t)=10/(-4-t)`

`rArr10/(-4-t)=-2/1`

cross-multiply.

`rArr-2(-4-t)=10`

`rArr8+2t=10rArr2t=10-8=2`

`(cancel(2) t)/cancel(2)=2/2`

`rArrt=1`