I have no idea how to do this at all, I am trying to study before I take calculus again.

I am supposed to find equations for the line that passes through the point $(2, -5)$ and:

  1. Has slope $-3$.
  2. is parallel to the x-axis.
  3. is parallel to y axis
  4. parallel to the line $2x-4y = 3$

Is this kind of information important to calculus? I am not familiar with any of the terms I am seeing and I don't remember doing anything similar to this in class. It seems like it is so rarely used that it is near impossible to remember all these small things.

  • $\begingroup$ You are actually trying to find four equations for four different lines. You do realize that, right? $\endgroup$ – Arturo Magidin Jan 1 '12 at 20:37
  • $\begingroup$ Yes, I will edit to make that more clear. $\endgroup$ – user138246 Jan 1 '12 at 20:42
  • $\begingroup$ You might find Khan Academy useful for seeing problems explained as they are solved. $\endgroup$ – Austin Mohr Jan 1 '12 at 21:00

What formulas do you know for finding equations of lines? There are a couple of standard ones:

  • Point-slope. If you know a point $(a,b)$ through which the line goes and the slope $m$ of the line, then the equation of the line is given by $y-b = m(x-a)$.

  • Two-points. If you know two points $(a,b)$ and $(c,d)$ that are on the line, then:

    • If $a=b$, the line is vertical, and the equation is $x=a$.
    • If $a\neq b$, then the slope of the line is $$m = \frac{\text{rise}}{\text{run}} = \frac{d-b}{c-a}.$$ Now use the point-slope formula with $(a,b)$ and $m$.
  • Slope-intercept. If you know the slope $m$ and the $y$-intercept $(0,b)$, then you are actually in a "point-slope" situation, with the point $(0,b)$. So the equation is just $y-b=mx$, or $y=mx+b$.

The first problem, you know a point on the line and the slope of the line. The point-slope formula gives you exactly what you want.

In the second problem, you know a point, and you are implicitly told the slope: "parallel to the $x$-axis" means that the line has to be horizontal. What is the slope of a horizontal line?

Same in the third problem: "parallel to the $y$-axis" means the line has to be vertical. The equations of vertical lines are always of the form $x=k$ for some constant $k$. If the line has to go through $(2,-5)$ and be vertical, what is the equation?

And in the fourth problem, you are again given the slope implicitly: "parallel to $2x-4y=3$" means "having the same slope as $2x-4y=3$". Find the slope of the line $2x-4y=3$ and proceed form there.

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  • $\begingroup$ How can I find the slope of an equation if I don't know what the (a,b) and stuff are supposed to be in their equation? $\endgroup$ – user138246 Jan 1 '12 at 20:47
  • $\begingroup$ @Jordan: Equations don't have slopes. Lines (and, in calculus, graphs of functions at specific points) have slopes. I don't understand your question: you are given the point, and you are being given enough information to find the slope. What is the problem? $\endgroup$ – Arturo Magidin Jan 1 '12 at 20:49
  • $\begingroup$ I don't know how to manipulate the equation $2x-4y=3$ in a way to make it into a line of the form y=mx+b I am getting nonsensical answers that do not match what the book has. $\endgroup$ – user138246 Jan 1 '12 at 20:50
  • $\begingroup$ @Jordan: $2x-4y = 3$ is the same as $2x=3+4y$, which is the same as $2x-3=4y$, which is the same as $4y=2x-3$, which is the same as $y = \frac{1}{4}(2x-3)$, which is the same as $y = \frac{2}{4}x - \frac{3}{4}$, which is the same as $y=\frac{1}{2}x - \frac{3}{4}$. So $m=\frac{1}{2}$. $\endgroup$ – Arturo Magidin Jan 1 '12 at 20:52
  • $\begingroup$ The book gets $y=1/2x-6$ and I can't figure out how. $\endgroup$ – user138246 Jan 1 '12 at 20:54

Hint: Every line can be written as $y = mx + b$, where $m$ denotes the line's slope and $b$ denotes its $y$-intercept. For each of your conditions (which are four separate problems), you are given the value of $m$ (the slope). You can plug in the other point $(2, -5)$ and solve for $b$.

Here is how you might work the first one. Since the slope of the line is $-3$, you know the equation looks like $$ y = -3x + b. $$ To determine $b$, plug in the point you know about. $$ -5 = -3 \cdot 2 + b. $$ Some algebra reveals that $b = 1$. Thus, the equation for the line is $$ y = -3x + 1. $$

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  • $\begingroup$ y=mx+b is all I have memorized for this but I couldn't make it work. $\endgroup$ – user138246 Jan 1 '12 at 20:48

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