1. ## Linear function equation help please

I need help understanding the following:

write the equations, in general form, for the lines that pass through point (4,3) and are parallel AND perpendicular to the line y+1=5/7 (x-4). HELP PLEASE?

2. ## Re: Linear function equation help please

Two lines are parallel if they have the same slope. Two lines are perpendicular if the product of their slopes is $-1$. So, find the slope of the line you are given. That is the same slope as the parallel line you want to find. Take its negative reciprocal to get the slope for the perpendicular line. Then, use the Point-Slope form for the equation of a line to find the line with slope equal to the given line that passes through (4,3), and again to find the equation for the line with slope equal to negative the reciprocal of the given line that passes through (4,3).

3. ## Re: Linear function equation help please

Sorry? I'm a 50 year old lawyer trying to assist my 15 year old daughter with my 35 year old math skills. What you have responded with I am sure would make sense to my daughter, but for me to understand it could someone please show me the steps that I would have to follow to find the 2 general form equations? Thanks again.

4. ## Re: Linear function equation help please

Originally Posted by conductor15
in general form, for the lines that pass through point (4,3) and are parallel AND perpendicular to the line y+1=5/7 (x-4).
Originally Posted by conductor15
I'm a 50 year old lawyer trying to assist my 15 year old daughter with my 35 year old math skills.
Write the given line in general form: $5x-7y-27=0$.

The parallel line required will look like $5x-7y+C=0$.

The perpendicular line required will look like $7x+5y+D=0$.

To find the values $C~\&~D$ use $x=4~\&~y=3$.

5. ## Re: Linear function equation help please

A very similar method to what Plato just wrote:

Multiply everything out and simplify: $y+1 = \dfrac{5}{7}(x-4)$ becomes $y = \dfrac{5}{7}x - \dfrac{27}{7}$. Make sure only $y$ is on the left-hand side (LHS) of the equation. The coefficient of $x$ is the number that is multiplied by $x$. In this case, it is $\dfrac{5}{7}$. That is the slope of the line. (If you multiply both sides of the equation by 7 and subtract 7y from both sides, you get the equation Plato wrote in general form). Note that to find the slope of the line as the coefficient of $x$, the coefficient of $y$ must be 1. (Note: If the equation is $y=a$ for some $a$, then the coefficient of x is 0, and the slope is zero. Exception: If the equation is $x=a$ for some $a$ and the coefficient of $y$ is zero, then the line does not have a slope).

A line in point-slope form looks like this: $y-y_0 = m(x-x_0)$ where $m$ is the slope of the line and the line passes through the point $(x_0,y_0)$. So, the slope of the parallel line is $\dfrac{5}{7}$ and the slope of the perpendicular line is $-\dfrac{7}{5}$ (flip the fraction over and change the sign from positive to negative or vise versa).

Since you want both lines to pass through the point $(4,3) = (x_0,y_0)$, the equation for the parallel line is $y-3 = \dfrac{5}{7}(x-4)$ and the equation for the perpendicular line is $y-3 = -\dfrac{7}{5}(x-4)$.

If the original line were horizontal, then the perpendicular line would not have a slope. If it comes up, use Plato's method.