The distance between 4x + 3y = 11 and 8x + 6y = 15, is - Vidyadip

MCQ

The distance between $4x + 3y = 11$ and $8x + 6y = 15$, is

A
$\frac{7}{2}$
B
$4$
✓
$\frac{7}{{10}}$
D
None of these

✓

Answer

Correct option: C.

$\frac{7}{{10}}$

c
(c) $4x + 3y = 11$and $4x + 3y = \frac{{15}}{2}$

Therefore, $D = \left| {\,\frac{{11 - \frac{{15}}{2}}}{5}\,} \right| = \frac{7}{{10}}$.

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