If a straight line drawn through the point of intersection of the lines 4x+3y-1=0 and 3x+4…

Question

If a straight line drawn through the point of intersection of the lines $4x+3y-1=0$ and $3x+4y-1=0$ meets the co-ordinate axes at the points $P$ and $Q$, then the locus of the mid point of $PQ$ is:

Accepted Answer

Subtracting the two given lines: $x-y=0\implies x=y$. Substituting: $7x=1\implies x=y=\dfrac17$. So all lines through the fixed point $\left(\dfrac17,\dfrac17\right)$ meeting the axes at $P(a,0),Q(0,b)$ must satisfy
$$\frac1{7a}+\frac1{7b}=1 \implies \frac1a+\frac1b=7$$

Let the midpoint of $PQ$ be $(h,k)=\left(\dfrac a2,\dfrac b2\right)$, so $a=2h,\ b=2k$:
$$\frac1{2h}+\frac1{2k}=7 \implies h+k = 14hk$$

Locus: $x+y-14xy=0$.

Solutions · 1