Distance between the parallel lines 3x + 4y + 7 = 0 and 3x + 4y -… - Vidyadip

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
$\frac{3}{5}$

✓

Answer

Correct option: B.

$\frac{{12}}{5}$

b
(b) ${d_1} = $ distance of perpendicular from $(0,\,0)$ to $3x + 4y + 7 = 0$

${d_1} = \frac{{3 \times 0 + 4 \times 0 + 7}}{{\sqrt {{3^2} + {4^2}} }} = \frac{7}{5}$

${d_2} = \frac{{3 \times 0 + 4 \times 0 + ( - 5)}}{{\sqrt {{3^2} + {4^2}} }} = \frac{{ - 5}}{5}$

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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