MCQ
The straight lines $4ax + 3by + c = 0$ where $a + b + c = 0$, will be concurrent, if point is
- A$(4,\, 3)$
- ✓$(1/4, \,1/3)$
- C$(1/2, \,1/3)$
- DNone of these
Eliminating $c$, we get $4ax + 3by - (a + b) = 0$
==> $a(4x - 1) + b(3y - 1) = 0$
This passes through the intersection of the lines $4x - 1 = 0$ and $3y - 1 = 0$i.e.$x = \frac{1}{4},\,y = \frac{1}{3}$i.e., $\left( {\frac{1}{4},\,\frac{1}{3}} \right)$.
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($A$) $Q_2 Q_3=12$
($B$) $ R_2 R_3=4 \sqrt{6}$
($C$) area of the triangle $O R_2 R_3$ is $6 \sqrt{2}$
($D$) area of the triangle $P Q_2 Q_3$ is $4 \sqrt{2}$