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

2 Answers

lemjay's profile pic

lemjay | High School Teacher | (Level 3) Senior Educator

Posted on

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:


Substitute value of m.


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.



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




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

jeew-m's profile pic

jeew-m | College Teacher | (Level 1) Educator Emeritus

Posted on

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