Find the equation of the straight line passing through $(−2,−7)$ and having intercept of length $3$ units between straight lines $4𝑥+3𝑦=12$ and $4𝑥+3𝑦=3$

This can be solved by introducing the parametric form of the straight line and doing a bit of substitutions, which gives the slopes, $m=-7/24$ and parallel to y axis.

My Attempt

Let $a$ be the angle made by the line with $4x+3y=3$. Then

img $$ \sin a=3/5\implies \tan a=3/4\\ m'=-4/3\\ \tan a=\frac{3}{4}=|\frac{m+\frac{4}{3}}{1-\frac{4m}{3}}|=|\frac{3m+4}{3-4m}|\\ \implies9-12m=12m+16\quad\text{or}\quad9-12m=-12m-16\\ m=\frac{-7}{24}\quad\text{or}\quad \text{Not possible} $$

In my attempt why am I not getting the slope corresponds to being parallel to y- axis ?

Or the case $9-12m=-12m-16$ means the slope is undefined thus parallel to y - axis ?

This is asked before @Equation Of A Straight Line Passing Through A Point and Having an Intercept but does not address this.


1 Answer 1


Notice that the two parallel lines intersect with $y$-axis at $y=1$ and $y=4$, respectively. So, one of the lines passing through (-2,-7) must be parallel to the $y$-axis in order to have of length of 3 between the lines. So, the line is simply $x= -2$.

If you try to solve for its slope, you should get infinity, which is just the solution to your equation,


In other words, only $m=\pm \infty$ satisfy above equation.

  • $\begingroup$ You mention twice the $y$-axis, whereas I think you mean $x$-axis. $\endgroup$
    – EdG
    Commented Aug 30, 2019 at 2:35
  • $\begingroup$ I do mean $y$, 1st referring to y-intercepts and 2nd stating parallel to y-axis. $\endgroup$
    – Quanto
    Commented Aug 30, 2019 at 2:50
  • $\begingroup$ Sorry, I need to sleep... $\endgroup$
    – EdG
    Commented Aug 30, 2019 at 2:55

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