Question

How do you find a standard form equation for the line with (-5, -16) with a slope of 2?

Answer 1

`color(red)(2)x - color(blue)(1)y = color(green)(6)`

Explanation:

First, we use the point-slope formula to write an equation for the line:

The point-slope formula states: `(y - color(red)(y_1)) = color(blue)(m)(x - color(red)(x_1))`

Where `color(blue)(m)` is the slope and `color(red)(((x_1, y_1)))` is a point the line passes through.

`(y - color(red)(-16)) = color(blue)(2)(x - color(red)(-5))`

`(y + color(red)(16)) = color(blue)(2)(x + color(red)(5))`

Now we can transform to the standard form.

The standard form of a linear equation is:

`color(red)(A)x + color(blue)(B)y = color(green)(C)`

where, if at all possible, `color(red)(A)`, `color(blue)(B)`, and `color(green)(C)`are integers, and A is non-negative, and, A, B, and C have no common factors other than 1

`y + color(red)(16) = (color(blue)(2) xx x) + (color(blue)(2) xx color(red)(5))`

`y + color(red)(16) = 2x + 10`

`y + color(red)(16) - 16 - color(blue)(2x) = 2x + 10 - 16 - color(blue)(2x)`

`-color(blue)(2x) + y + color(red)(16) - 16 = 2x - color(blue)(2x) + 10 - 16`

`-color(blue)(2x) + y + 0 = 0 + 10 - 16`

`-color(blue)(2x) + y = -6`

`-1(-color(blue)(2x) + y) = -1 xx -6`

`2x - y = 6`

`color(red)(2)x - color(blue)(1)y = color(green)(6)`

Answer 2

`2x-y-6=0`

Explanation:

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

`color(red)(bar(ul(|color(white)(2/2)color(black)(ax+by+c=0)color(white)(2/2)|)))`

Before this, we can write the equation in `color(blue)"point-slope form"`

`color(red)(bar(ul(|color(white)(2/2)color(black)(y-y_1=m(x-x_1))color(white)(2/2)|)))`
where m represents the slope and `(x_1,y_1)" a point on the line"`

here `m=2" and " (x_1,y_1)=(-5,-16)`

`rArry-(-16)=2(x-(-5))`

`rArry+16=2(x+5)`

`rArry+16=2x+10larrcolor(red)" in point-slope form"`

Rearranging in standard form.

`rArr2x-y-6=0larrcolor(red)" in standard form"`