Straight LineMedium
Question
If a straight line drawn through the point of intersection of the lines $4x + 3y - 1 = 0$ and $3x + 4y - 1 = 0$, meet the co-ordinate axes at the points P and Q , then the locus of the mid point of PQ is:
Options
A.$x + y - 7 = 0$
B.$x + y - 14xy = 0$
C.$2x + y + 14xy = 0$
D.$x + 2y - 14xy = 0$
Solution
Sol. [Key Concept: Find the intersection of the two lines, write the variable line through it in intercept form, and eliminate to get the midpoint locus $x + y - 14xy = 0$.]
Point of intersection of the lines
$3x + 4y = 1$ and $4x + 3y = 1$ is $\left( \frac{1}{7},\frac{1}{7} \right)$
Let the line be $\frac{x}{2h} + \frac{y}{2k} = 1$
Satisfy $\left( \frac{1}{7},\frac{1}{7} \right)$
$${\frac{1}{14\text{ }h} + \frac{1}{14k} = 1 }{\frac{1}{x} + \frac{1}{y} = 14 }$$
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