Question

Ndicate the equation of the line through (2, -4) and having slope of 3/5.

Answer 1

Point-slope form:
`y+4=(3)/(5)(x-2)`

Slope-intercept form:
`y=(3)/(5)x-(26)/(5)`

Standard form:
`3x-5y=26`

Explanation:

The easiest form to use here if you know it is point-slope form. I say this because you are given a point and the slope of the equation.

The point-slope form is
`y-y_1=m(x-x_1)`.

Plug in your information.

Again you are given
`(x_1,y_1)=(2,-4)` and
`m=(3)/(5)`.

`y-y1=m(x-x_1)` with the line before this one gives us:

`y-(-4)=(3)/(5)(x-2)`

`y+4=(3)/(5)(x-2)` This is point-slope form.

We can rearrange it for different form.

Another form is the slope-intercept form which is y=mx+b where m is the slope and b is the y-intercept.

So to put
`y+4=(3)/(5)(x-2)` into y=mx+b we will need to distribute and isolate y.

I will first distribute. 3/5(x-2)=3/4 x -6/5.

So now we have
`y+4=(3)/(5)x-(6)/(5)`

Subtract 4 on both sides:

`y=(3)/(5)x-(6)/(5)-4[tex]</p><p>Combined the like terms:</p><p>[tex]y=(3)/(5)x-(26)/(5)` This is slope-intercept form.

We can also do standard form which is ax+by=c. Usually people want a,b, and c to be integers.

So first thing I will do is get rid of the fractions by multiplying both sides by 5.

This gives me

`5y=5\cdot (3)/(5)x-5 \cdot 26/5`

`5y=3x-26`

Now subtract 3x on both sides

`-3x+5y=-26`

You could also multiply both sides by -1 giving you:

`3x-5y=26`

Answer 2

`\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})~\hspace{10em} slope = m\implies \cfrac{3}{5} \\\\\\ \begin{array} \cline{1-1} \textit{point-slope form}\\ \cline{1-1} \\ y-y_1=m(x-x_1) \\\\ \cline{1-1} \end{array}\implies y-(-4)=\cfrac{3}{5}(x-2) \implies y+4=\cfrac{3}{5}x-\cfrac{6}{5} \\\\\\ y=\cfrac{3}{5}x-\cfrac{6}{5}-4\implies \implies y=\cfrac{3}{5}x-\cfrac{26}{5}`