1. ## Help me!

Given that the x-intercept of a line is twice the y-intercept of its y-intercept and that the line passes thru the intersection between the lines 3y+x=3 and 4y-3x=5, find the eqn of the line. i found the intersection of the line and use the point to find the gradient of the the two point at the x-intercept and x-intercept so that i can eqate them and find the intercept. however i got a large fraction and is very confusing. Did i do wrongly? or is there another method to solve this?

2. Originally Posted by helloying
Given that the x-intercept of a line is twice the y-intercept of its y-intercept and that the line passes thru the intersection between the lines 3y+x=3 and 4y-3x=5, find the eqn of the line. i found the intersection of the line and use the point to find the gradient of the the two point at the x-intercept and x-intercept so that i can eqate them and find the intercept. however i got a large fraction and is very confusing. Did i do wrongly? or is there another method to solve this?
Let the line have equation $y = mx + c$.

x-interecept: $x = - \frac{c}{m} \Rightarrow 2c = - \frac{c}{m} \Rightarrow m = - \frac{1}{2}$.

Substitute the known point on the line to find c.

3. Hello, helloying!

Given that the x-intercept of a line is twice the y-intercept,
. . and that the line passes thru the intersection of the lines $\begin{array}{c}3y+x\:=\:3 \\ 4y-3x\:=\:5\end{array}$
find the equation of the line.

The lines intersect at: . $P\left(\frac{3}{13},\;\frac{14}{13}\right)$

The line has y-intercept $Q(0,\,b)$ and x-intercept $R(2b,\,0)$

The slope of line $QR$ is: . $m \:=\:\frac{0-b}{2b-0} \:=\:-\frac{1}{2}$

The line thorugh $P\left(\frac{3}{13},\:\frac{14}{13}\right)$ with slope $-\frac{1}{2}$ has the equation:

. . $y - \frac{14}{13} \:=\:-\frac{1}{2}\left(x - \frac{3}{13}\right) \quad\Rightarrow\quad \boxed{\;y \;=\;-\frac{1}{2}x + \frac{31}{16}\;}$