Help Finding the Equation of a Line with a Given a Point and a Slope

Question

I have been receiving a lot of help today and I really appreciate it, I am nearing the end of my class and need help studying for my final. In my review I came across a problem that i am having difficulty figuring out. I need to write an equation of a line that has a slope of $3/4$ and passes through a point of $(-2,4)$.

So far here is what I have accomplished:

Using the equation $y-y_1= m(x-x_1)$

IDK if this equation is the one I should be using.

$y - 4 = \dfrac{3}{4}[x-(-2)]$

to

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

to

$y - 4 = \dfrac{3x}{4} + \dfrac{6}{4}$

Do I reduce the $\frac{6}{4}$ to $\frac{3}{2}$? Also, is this the correct step I should be doing here?

$y - 4 = 3x + 6$

Since both values on the right have the same denominator, can I just get rid of it?

to

$y = 3x + 10$

Is this correct?

If this is correct, is there a simpler way of doing this? If not, what should I be doing?

Answer 1

You almost have it with $y-4=(3x/4)+(6/4)$. Just get the 4 to the right hand side $y=\frac34x+\frac32+4$. So $$y=\frac34x+\frac{11}2$$

Answer 2

Why not just forget about the different variations of the formula (e.g. point slope, 2 points, slope intercept, etc. ) and just memorize this one.

$$\color{Tomato}{m=\frac{\Delta y}{\Delta x}}$$

If the slope ($m$) and a single point ($x_0, y_0$) are known, then this forumla becomes $$\begin{align} m&=\frac{\Delta y}{\Delta x}\\ m&=\frac{y-y_0}{x-x_0}\\ m(x-x_0)&=y-y_0\\ \end{align}$$

The Slope-intercept form is just a special case where one point is $(0, b)$ $$\begin{align} m&=\frac{\Delta y}{\Delta x}\\ m&=\frac{y-b}{x-0}\\ mx&=y-b\\ y&=mx+b\\ \end{align}$$ And finally, when 2 points are known, we have $$\begin{align} m&=\frac{\Delta_2 y}{\Delta_2 x}=\frac{\Delta_1 y}{\Delta_1 x}\\ m&=\frac{y_1-y_0}{x_1-x_0}=\frac{y-y_0}{x-x_0}\\ y-y_0&=\frac{y_1-y_0}{x_1-x_0}(x-x_0)\\ \end{align}$$