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  1. #1
    Member helloying's Avatar
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    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?
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  2. #2
    Flow Master
    mr fantastic's Avatar
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    Quote Originally Posted by helloying View Post
    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 $\displaystyle y = mx + c$.

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

    Substitute the known point on the line to find c.
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  3. #3
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    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 $\displaystyle \begin{array}{c}3y+x\:=\:3 \\ 4y-3x\:=\:5\end{array}$
    find the equation of the line.

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


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

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


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

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

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