How do you write an equation in standard form of the lines passing through (-1,5) and (0,8)?

2 Answers
Jun 30, 2015

3x-y = -83xy=8

Explanation:

Start with a two point form (based on slope)
color(white)("XXXX")XXXX(y-8)/(x-0) = (8-5)/(0-(-1)y8x0=850(1)
Which simplifies as
color(white)("XXXX")XXXXy-8 = 3xy8=3x

Standard form of a linear equation is
color(white)("XXXX")XXXXAx+By=CAx+By=C with A, B, C epsilon ZZ and A>=0

Converting y-8 = 3x into this form:
color(white)("XXXX")3x-y = -8

Jun 30, 2015

-3x+y=8

Explanation:

The standard form of an equation is given by;
Ax+By=C

To find the equation of line passing through the points (-1,5) and (0,8), we need to used given formula;

(y-y_1)=m(x-x_1)..........equation 1
where m = slope and given by the formula;

m=\frac{y_2-y_1}{x_2-x_1}

Now, Let's assume that (x_1,y_1) is (-1,5) and (x_2,y_2) is (0,8).
First find the slope of line using slope formula, we get;

m=\frac{8-5}{0-(-1)} = \frac{3}{1} = 3

Now, plug (x_1,y_1) is (-1,5) and m = 3 in equation 1, we get
(y-5)=3(x-(-1))
or, y-5=3(x+1)
or, y-5=3x+3

Add 5 on both side, we get,
or, y=3x+3+5
or, y=3x+8

Subtract 3x on both side, we get
or, -3x+y=8
This is our required equation in standard form.