A line passing through $A (1, 2)$ is perpendicular to the lines $3 x + 4 y - 2 = 0$ and $3 x + 4 y + 7 = 0$ and intersecting them respectively at the points $P$ and $Q,$ then $\frac{A Q}{A P}$ is equal to

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Solution

$A P = \frac{| 3 (1) + 4 (2) - 2 |}{5}$
$A Q = \frac{| 3 (1) + 4 (2) + 7 |}{5}$
$∴ \frac{A Q}{A P} = \frac{18}{9} = 2$

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A line passing through A(1,2) is perpendicular to the lines 3x+4y−2=0 and 3x+4y+7=0 and intersecting them respectively at the points P and Q, then AQAP is equal to

AP=|3(1)+4(2)−2|5 AQ=|3(1)+4(2)+7|5 ∴AQAP=189=2

A line passing through $A (1, 2)$ is perpendicular to the lines $3 x + 4 y - 2 = 0$ and $3 x + 4 y + 7 = 0$ and intersecting them respectively at the points $P$ and $Q,$ then $\frac{A Q}{A P}$ is equal to

$A P = \frac{| 3 (1) + 4 (2) - 2 |}{5}$
$A Q = \frac{| 3 (1) + 4 (2) + 7 |}{5}$
$∴ \frac{A Q}{A P} = \frac{18}{9} = 2$