Question

What is the equation of the line passing through `(96,72)` and `(19,4)`?

Answer 1

The slope is 0.88311688312.

Explanation:

`(Y_2 - Y_1)/ (X_2 - X_1)` = `m`, the slope

Label your ordered pairs.

(96, 72) `(X_1, Y_1)`
(19, 4) `(X_2, Y_2)`

Plug-in your variables.

`(4 - 72)/(19 - 96)` = `m`

-68/-77 = `m`

Two negatives make a positive, so:

0.88311688312 = `m`

Answer 2

`y = = 68/77x - 984/77`

Explanation:

Recall;

`y = mx + c`

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

`y_2 = 4`

`y_1 = 72`

`x_2 = 19`

`x_1 = 96`

Inputing the values..

`m = (4 - 72)/(19 - 96)`

`m = (-68)/-77`

`m= 68/77`

The new equation is;

`(y - y_1) = m(x - x_1)`

Inputing their values..

`y - 72 = 68/77(x - 96)`

`y - 72 = (68x - 6528)/77`

Cross multiplying..

`77(y - 72) = 68x - 6528`

`77y - 5544 = 68x - 6528`

Collecting like terms..

`77y = 68x - 6528 + 5544`

`77y = 68x - 984`

Dividing through by `77`

`y = = 68/77x - 984/77`

Answer 3

Point-slope form: `y-4=68/77(x-19)`

Slope-intercept form: `y=68/77x-984/77`

Standard form: `68x-77y=984`

Explanation:

First determine the slope using the slope formula and the two points.

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

where `m` is the slope, and `(x_1,y_1)` is one point and `(x_2,y_2)` is the other point.

I'm going to use `(19,4)` as `(x_1,y_1)` and `(96,72)` as `(x_2,y_2)`.

`m=(72-4)/(96-19)`

`m=68/77`

Now use the slope and one of the points to write the equation in point-slope form:

`y-y_1=m(x-x_1)`,

where:

`m` is the slope and `(x_1,y_1)` is one of the points.

I'm going to use `(19,4)` for the point.

`y-4=68/77(x-19)` `larr` point-slope form

Solve the point-slope form for `y` to get the slope-intercept form:

`y=mx+b`,

where:

`m` is the slope and `b` is the y-intercept.

`y-4=68/77(x-19)`

Add `4` to both sides of the equation.

`y=68/77(x-19) +4`

Expand.

`y=68/77x-1292/77 + 4`

Multiply `4` by `77/77` to get an equivalent fraction with `77` as the denominator.

`y=68/77x-1292/77+4xx77/77`

`y=68/77x-1292/77+308/77`

`y=68/77x-984/77` `larr` slope-intercept form

You can convert the slope-intercept form to the standard form:

`Ax+By=C`

`y=68/77x-984/77`

Multiply both sides by `77`.

`77y=68x-984`

Subtract `68x` from both sides.

`-68x+77y=-984`

Multiply both sides by `-1`. This will reverse the signs, but the equation represents the same line.

`68x-77y=984` `larr` standard form

graph{68x-77y=984 [-10, 10, -5, 5]}