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.

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- Feb 1st 2009, 08:51 PMmathhomeworkEquation 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, 08:57 PMtopher0805
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 $\displaystyle y = mx + b$ :

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

So the slope of line 1 is just:

$\displaystyle -\frac {k}{4}$

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

So the slope of line 2 is just:

$\displaystyle (k-3)$

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

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

Now solve for k.

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

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

$\displaystyle k-3$

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

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.

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

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 $\displaystyle y = mx + b$ with the values ({x intercept},0) and solve for b.

:) - Feb 1st 2009, 09:04 PMred_dog
a) $\displaystyle L_1:kx+4y=1, \ L_2:(3-k)x+y=2$

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

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

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

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

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

Now, the equation of $\displaystyle L_3$ is $\displaystyle y=-\frac{3}{5}\left(x-\frac{22}{3}\right)\Leftrightarrow y=-\frac{3}{5}x+\frac{22}{5}$ - Feb 1st 2009, 09:13 PMstar_tenshi
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 $\displaystyle y = mx +b$, where $\displaystyle m$ is the slope of the line, and $\displaystyle b$ is the y-intercept.

Thus,

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

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

Since both of these must have the same slope, $\displaystyle m$, then: $\displaystyle -\frac{kx}{4} = (k-3)$

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 $\displaystyle y = 0$ into the equation you got with k. The resulting answer should be of the form $\displaystyle x = m$, where $\displaystyle m$ is some number.

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

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