Question

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

Answer 1

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

Answer 2

Answer:

Point-slope form: $y+4=\frac{3}{5}(x-2)$

Slope-intercept form: $y=\frac{3}{5}x-\frac{26}{5}$

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

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 $y-y_1=m(x-x_1)$.

Plug in your information.

Again you are given $(x_1,y_1)=(2,-4)$ and $m=\frac{3}{5}$.

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

$y-(-4)=\frac{3}{5}(x-2)$

$y+4=\frac{3}{5}(x-2)$ 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 $y+4=\frac{3}{5}(x-2)$ 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 $y+4=\frac{3}{5}x-\frac{6}{5}$

Subtract 4 on both sides:

$y=\frac{3}{5}x-\frac{6}{5}-4[tex]Combined the like terms:[tex]y=\frac{3}{5}x-\frac{26}{5}$ 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

$5y=5\cdot \frac{3}{5}x-5 \cdot 26/5$

$5y=3x-26$

Now subtract 3x on both sides

$-3x+5y=-26$

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

$3x-5y=26$