Let A be the point of intersection of the lines 3x + 2y – 14, 5x – y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, –2) from the line AB is [2024]

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Solution

Q.
Let A be the point of intersection of the lines 3x + 2y = 14, 5x – y = 6 and B be the point of intersection of the lines 4x + 3y = 8, 6x + y = 5. The distance of the point P(5, –2) from the line AB is [2024]

1 8

2 $\frac{5}{2}$

3 $\frac{13}{2}$

4 6

Ans.
(4)

Point of intersection of lines 3x + 2y = 14 and 5x – y = 6 is A(2, 4)

Point of intersection of lines 4x + 3y = 8 and 6x + y = 5 is $B = (\frac{1}{2}, 2)$

So, equation of line AB is, $y - 4 = \frac{2 - 4}{\frac{1}{2} - 2} (x - 2)$

$\Rightarrow y - 4 = \frac{4}{3} (x - 2) \Rightarrow 4 x - 3 y + 4 = 0$

$∴$ Distance of point P(5, –2) from 4x – 3y + 4 is given as

$d =$ $| \frac{4 \times 5 - 3 \times (- 2) + 4}{\sqrt{16 + 9}} |$ $\Rightarrow d = 6 units$

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