# What is the equation of the line in standard form that passes through the points (8,4) and (4, -7)?

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To solve, determine the slope of the line that passes through (8, 4) and (4,-7).

`m=(y_2 - y_1)/(x_2-x_1)=(-7-4)/(4-8)=(-11)/(-4)=11/4`

So the slope is `11/4` .

Note that the standard form of a line is Ax+By=C and its slope is:

`m=-A/B`

Substitute value of m.

`11/4=-A/B`

So, A=11 and B=-4. Or it could be A=-11 and B=4. In either case, it results to same equation.

Let's use the value A=11 and B=-4. Substitute it to the standard form.

`Ax+By=C`

`11x-4y=C`

Plug-in one of the given points to solve for C. Use (8,4).

`11(8)-4(4)=C`

`88-16=C`

`72=C`

**Hence, the equation of the line in standard form that passes throught the given points is `11x-4y=72` .**

The standard form of a line is Ax+By = C

First let us make the line in the form of y = mx+c

If (x1,y1) and (x2,y2) are two points

`m = (y2-y1)/(x2-x1)`

`m = (-7-4)/(4-8) = (-11)/(-4) = 11/4`

So the gradient m `= 11/4`

So we can write the equation as;

`y = (11/4)x+c`

Now we have to find intercept c.

Let us substitute the first point in to the equation;

`y = (11/4)x+c`

`4 = (11/4)xx8+c`

`c = -18`

So the equation can be written as;

`y = (11/4)x-18`

`4y = 11x-72`

`72 = 11x-4y`

*So the standard form of the line is;*

*11x-4y = 72*