MCQ
Distance between the parallel lines $3x + 4y + 7 = 0$ and $3x + 4y - 5 = 0$ is
- A$\frac{2}{5}$
- ✓$\frac{{12}}{5}$
- C$\frac{5}{{12}}$
- D$\frac{3}{5}$
${d_1} = \frac{{3 \times 0 + 4 \times 0 + 7}}{{\sqrt {{3^2} + {4^2}} }} = \frac{7}{5}$
${d_2} = \frac{{3 \times 0 + 4 \times 0 + ( - 5)}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{ - 5}}{5}$
Required distance $ = \left| {\,{d_1} - {d_2}} \right|$$ = \left| {\,\frac{7}{5} - \left( {\frac{{ - 5}}{5}} \right)\,} \right| = \frac{{12}}{5}.$
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$1.$ The value of $r$ is
$(A)$ $-\frac{1}{t}$ $(B)$ $\frac{t^2+1}{t}$ $(C)$ $\frac{1}{ t }$ $(D)$ $\frac{t^2-1}{t}$
$2.$ If st $=1$, then the tangent at $P$ and the normal at $S$ to the parabola meet at a point whose ordinate is
$(A)$ $\frac{\left(t^2+1\right)^2}{2 t^3}$ $(B)$ $\frac{a\left(t^2+1\right)^2}{2 t^3}$ $(C)$ $\frac{a\left(t^2+1\right)^2}{t^3}$ $(D)$ $\frac{a\left(t^2+2\right)^2}{t^3}$
Give the answer question $1$ and $2.$