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.
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 $\displaystyle y = mx + b$ :
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.
a) $\displaystyle L_1:kx+4y=1, \ L_23-k)x+y=2$
$\displaystyle L_1\parallel L_2\Rightarrow \frac{k}{3-k}=\frac{4}{1}\Rightarrow k=\frac{12}{5}$
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 $\displaystyle y=-\frac{3}{5}\left(x-\frac{22}{3}\right)\Leftrightarrow y=-\frac{3}{5}x+\frac{22}{5}$
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 $\displaystyle x_{0} = m + 4$.