# Equation of Lines

• Feb 1st 2009, 09:51 PM
mathhomework
Equation of Lines
Given that the straight lines L1: kx+4y=1 and L2: 3x+y=2+kx are parallel.

a. find the value of the constant k.
b. if the line L3 is parallel to L2 and its x-intercept is greater than the x-intercept of L2 by 4, find the equation of L3.
• Feb 1st 2009, 09:57 PM
topher0805
Quote:

Originally Posted by mathhomework
Given that the straight lines L1: kx+4y=1 and L2: 3x+y=2+kx are parallel.

a. find the value of the constant k.
b. if the line L3 is parallel to L2 and its x-intercept is greater than the x-intercept of L2 by 4, find the equation of L3.

For part a, recognize that the slope of the two lines will be equal. I find it helpful to put the equations into the form $y = mx + b$ :

L1: $y = -\frac{kx}{4}+\frac{1}{4}$

So the slope of line 1 is just:

$-\frac {k}{4}$

L2: $y = kx-3x + 2 = (k-3)x + 2$

So the slope of line 2 is just:

$(k-3)$

Since the slopes are equal to each other, we can find k quite easily:

$-\frac {k}{4} = (k-3)$

Now solve for k.

$k = \frac {12}{5}$

For part b, we know that the slope is going to be the same as the slope of line 2:

$k-3$

Where k is equal to $\frac {12}{5}$

And we have been told that the x intercept is 4 greater than that of line 2. So, start by finding the x intercept of line 2.

To find that, simply plug in 0 for y.

$0 = -\frac {3}{5}x + 2$

And solve for x. Add 4 to that number and you have the x intercept for line 3.

Now that you have the x intercept, to find the y intercept you just set up your equation in the form $y = mx + b$ with the values ({x intercept},0) and solve for b.

:)
• Feb 1st 2009, 10:04 PM
red_dog
a) $L_1:kx+4y=1, \ L_2:(3-k)x+y=2$

$L_1\parallel L_2\Rightarrow \frac{k}{3-k}=\frac{4}{1}\Rightarrow k=\frac{12}{5}$

b) For $k=\frac{12}{5}\Rightarrow L_2:\frac{3}{5}x+y=2$

For the x-interception: $y=0\Rightarrow x=\frac{10}{3}$

The x-interception of $L_3$ is $\left(\frac{22}{3},0\right)$

The slope of $L_3$ is $m=-\frac{3}{5}$

Now, the equation of $L_3$ is $y=-\frac{3}{5}\left(x-\frac{22}{3}\right)\Leftrightarrow y=-\frac{3}{5}x+\frac{22}{5}$
• Feb 1st 2009, 10:13 PM
star_tenshi
Quote:

Originally Posted by mathhomework
Given that the straight lines L1: kx+4y=1 and L2: 3x+y=2+kx are parallel.

a. find the value of the constant k.
b. if the line L3 is parallel to L2 and its x-intercept is greater than the x-intercept of L2 by 4, find the equation of L3.

In order for two straight lines to be parallel, they must have the same slope.

a. Given these two equations, you must rearrange them such that they are in the form $y = mx +b$, where $m$ is the slope of the line, and $b$ is the y-intercept.
Thus,
L1: $kx + 4y = 1 \Rightarrow y = -\frac{kx}{4} + \frac{1}{4}$
L2: $3x + y = 2 + kx \Rightarrow y = (k-3)x + 2$
Since both of these must have the same slope, $m$, then: $-\frac{kx}{4} = (k-3)$
You can then solve for k and write out the corresponding equations for L1 & L2.

b. To do this part, find the x-intercept of L2 by plugging in $y = 0$ into the equation you got with k. The resulting answer should be of the form $x = m$, where $m$ is some number.

The x-intercept of L3 is 4 greater than that of L2, thus the x-intercept of L3 would be: $m + 4$

Then you can use the general formula $(y-y_{0}) = m(x-x_{0})$ to find the equation for L3, where $y_{0} = 0$ and $x_{0} = m + 4$.