If a variable line drawn through the intersection of the lines x3+ y4=1 and x4+ y3=1, meet…

Question

If a variable line drawn through the intersection of the lines $\dfrac x3+\dfrac y4=1$ and $\dfrac x4+\dfrac y3=1$, meets the coordinate axes at $A$ and $B$, ($A
e B$), then the locus of the midpoint of $AB$ is:

Accepted Answer

The two given lines are $4x+3y-12=0$ and $3x+4y-12=0$. All lines through their intersection: $(4x+3y-12)+\lambda(3x+4y-12)=0$
$$x(4+3\lambda)+y(3+4\lambda) = 12(1+\lambda)$$

$x$-intercept: $\dfrac{12(1+\lambda)}{4+3\lambda}$, $y$-intercept: $\dfrac{12(1+\lambda)}{3+4\lambda}$.

Midpoint $(h,k)$: $h=\dfrac{6(1+\lambda)}{4+3\lambda}$, $k=\dfrac{6(1+\lambda)}{3+4\lambda}$.

Eliminating $\lambda$ (as in the source) gives:
$6(h+k)=7hk$

Locus: $7xy=6(x+y)$.

Solutions · 1