Straight LineHardBloom L3
Question
The line $(p + 2q)x + (p - 3q)y = p - q$, for different values of $p$ and $q$, passes through the fixed point :
Options
A.$\left( \frac{3}{2},\frac{5}{2} \right)$
B.$\left( \frac{2}{5},\frac{2}{5} \right)$
C.$\left( \frac{3}{5},\frac{2}{5} \right)$
D.$\left( \frac{2}{5},\frac{3}{5} \right)$
Solution
Sol. [Key Concept: Family $(p + 2q)x + (p - 3q)y = p - q$ ⇒ group by $p$ and $q$ to read off the fixed point.]
$${p(x + y - 1) + q(2x - 3y + 1) = 0 }{\Rightarrow (x + y - 1) + \frac{q}{p}(2x - 3y + 1) = 0 }$$Always passes through intersection of $x + y - 1 = 0\ $ and $\ 2x - 3y + 1 = 0$, which is $\left( \frac{2}{5},\frac{3}{5} \right)$.
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