[tex]\bf (\stackrel{x_1}{2}~,~\stackrel{y_1}{-4})~\hspace{10em} slope = m\implies \cfrac{3}{5} \\\\\\ \begin{array}{|c|ll} \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}[/tex]
Answer:
Point-slope form: [tex]y+4=\frac{3}{5}(x-2)[/tex]
Slope-intercept form: [tex]y=\frac{3}{5}x-\frac{26}{5}[/tex]
Standard form: [tex]3x-5y=26[/tex]
Step-by-step 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 [tex]y-y_1=m(x-x_1)[/tex].
Plug in your information.
Again you are given [tex](x_1,y_1)=(2,-4)[/tex] and [tex]m=\frac{3}{5}[/tex].
[tex]y-y1=m(x-x_1)[/tex] with the line before this one gives us:
[tex]y-(-4)=\frac{3}{5}(x-2)[/tex]
[tex]y+4=\frac{3}{5}(x-2)[/tex] 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 [tex]y+4=\frac{3}{5}(x-2)[/tex] 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 [tex]y+4=\frac{3}{5}x-\frac{6}{5}[/tex]
Subtract 4 on both sides:
[tex]y=\frac{3}{5}x-\frac{6}{5}-4[tex]
Combined the like terms:
[tex]y=\frac{3}{5}x-\frac{26}{5}[/tex] 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
[tex]5y=5\cdot \frac{3}{5}x-5 \cdot 26/5[/tex]
[tex]5y=3x-26[/tex]
Now subtract 3x on both sides
[tex]-3x+5y=-26[/tex]
You could also multiply both sides by -1 giving you:
[tex]3x-5y=26[/tex]
