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., $\left( {\frac{1}{4},\,\frac{1}{3}} \right)$.
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
| Column $I$ | Column $II$ |
| $(A)$ Circle | $(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$ |
| $(B)$ Parabola | $(q)$ Points $z$ in the complex plane satisfying $|z+2|-|z-2|= \pm 3$ |
| $(C)$ Ellipse | $(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$ |
| $(D)$ Hyperbola | $(s)$ The eccentricity of the conic lies in the interval $1 \leq x<\infty$ |
| $(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=|z|^2+1$ |