Question

How do you write an equation in standard form for a line passing through (3, 4) and (–3, –8)?

Answer 1

`2x-y-2=0`

Explanation:

The equation of line passing through the points `(x_1, y_1)\equiv(3, 4)` & `(x_2, y_2)\equiv(-3, -8)` is given by following general formula

`y-y_1=\frac{y_2-y_1}{x_2-x_1}(x-x_1)`

`y-4=\frac{-8-4}{-3-3}(x-3)`

`y-4=2(x-3)`

`y-4=2x-6`

`2x-y-2=0`

Answer 2

`2x-y=2`

Explanation:

`"the equation of a line in "color(blue)"standard form"` is.

`color(red)(bar(ul(|color(white)(2/2)color(black)(Ax+By=C)color(white)(2/2)|)))`

`"where A is a positive integer and B, C are integers"`

`"obtain the equation in "color(blue)"slope-intercept form"`

`•color(white)(x)y=mx+b`

`"where m is the slope and b the y-intercept"`

`"to calculate m use the "color(blue)"gradient formula"`

`•color(white)(x)m=(y_2-y_1)/(x_2-x_1)`

`"let "(x_1,y_1)=(3,4)" and "(x_2,y_2)=(-3,-8)`

`m=(-8-4)/(-3-3)=(-12)/(-6)=2`

`y=2x+blarrcolor(blue)"is the partial equation"`

`"to find b substitute either of the 2 given points into"`
`"the partial equation"`

`"using "(3,4)" then"`

`4=6+brArrb=4-6=-2`

`y=2x-2larrcolor(red)"in slope-intercept form"`

`2x-y=2larrcolor(red)"in standard form"`

Answer 3

`2x-y = 2`

Explanation:

If you are given the co-ordinates of two points on a line, here is a good formula to use to get the equation of the line:

`(y-y_1)/(x-x_1) = (y_2-y_1)/(x_2-x_1)`

Use `(3,4)` as `(x_2,y_2)`

`(y-(-8))/(x-(-3)) = (4-(-8))/(3-(-3))`

`(y+8)/(x+3) = (4+8)/(3+3) = 12/6 =2/1" "larr` this is the slope.

`(y+8)/(x+3) = 2/1" "larr` cross multiply

`2x+6 = y+8" "larr`re-arrange into standard form.

`2x-y = 2`