Question

2. Find the length of the perpendicular drawn from the point of intersection of the lines
3x + 2y + 4 = 0 and 2x + 5y- 1 = 0 to the straight line 7x + 24y - 15 = 0.​

Answer 1

$\huge\tt\blue{Answer}$

.Let P(α, β) be the point of intersection of the lines 3x + 2y + 4 = 0, 2x + 5y – 1 = 0

So,  3α + 2β + 4 = 0 ...... (1)

 

and 2α + 5β – 1 = 0 ..... (2)

Solving (1) and (2) we get

$\sf \: \frac{ \alpha }{ - 2 - 20} \: = \: \frac{ \beta }{8 + 3} \: = \: \frac{1}{15 - 4} \: \dasharrow \: \frac{ \alpha }{ - 22} \: = \: \frac{ \beta }{11} \: = \: \frac{1}{11} \: \dasharrow \alpha \: = \: - 2 \comma \: \beta \: = \: 1 \\ \\ \sf \: \therefore \: p \: = \: ( - 2 \comma \: 1) \\ \\ \sf \: perpendicular \: distance \: from \: p( - 2 \comma \: 1) \: to \: 7x + 24y = 15 \: is \\ \\ \sf \: \frac{ |7( - 2) + 24(1) - 15| }{ \sqrt{7 ^{2} } + \sqrt{24} ^{2} } \: = \: \frac{ | - 14 + 24 - 15| }{ \sqrt{49 + 576} } \\ \\ \sf \: \frac{5}{25} \: = \: \frac{1}{5}$