Find the equation of the line that passes through the intersection of two lines, $\ 3x-4y=0$ and $\ 2x-5y+7=0$, and form a triangle of area 8 with the coordinate axes.
I know that the intersection point of those lines is $\ (4,3)$. But how can I find the equation of the new line that form the triangle of area 8?
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Let the desired line be$$\frac xa+\frac yb=1$$The intersection point $(4,3)$ lies on it, so$$\frac4a+\frac3b=1$$The area of the triangle formed by the line and the coordinate axes is $\frac12|ab|=8\implies|ab|=16$. Now evaluate the cases $ab>0$ and $ab<0$. The former gives no solution while the latter gives $(a,b)=(8/3,-6),(-8,2)$.
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The equation of any straight line passing through the intersection of $$3x-4y=0$$ and $$2x-5y+7=0$$
is $$0=k(3x-4y)+(2x-5y+7)\iff(3k+2)x-y(4k+5)=-7$$
$$x\cdot\dfrac{3k+2}{-7}+y\cdot\dfrac{4k+5}7=1$$
So, we need $$8=\dfrac{\left|\dfrac{3k+2}{-7}\cdot\dfrac{4k+5}7\right|}2$$
$$\implies16\cdot49|(3k+2)(4k+5)|$$
$$\implies(3k+2)(4k+5)=\pm16\cdot49$$
$k=?$
The vertices of the triangle are $(0,0)$, $(a,0)$ and $(0,b)$ where $a$, $b>0$ such that $ab=16$. Can you (i) find the equation of the line $L$ through $(a,0)$ and $(0,16/a)$ (in terms of $a$) and (ii) the value of $a$ that makes $(4,3)$ lie on $L$?