Find 'p' $3 x + 4 y - 5 = 0, 2 x + 3 y - 4 = 0, p x + 4 y - 6$ are concurent.

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Solution

The correct option is C $2$
$3 x + 4 y - 5 = 0 => x = \frac{5 - 4 y}{3} 2 x + 3 y - 4 = 0 => 2 (\frac{5 - 4 y}{3}) + 3 y - 4 = 0 => 10 - 8 y + 9 y - 12 = 0 => y = 2 x = \frac{5 - 4 y}{3} = \frac{5 - 4 (2)}{3} = - 1 p x + 4 y - 6 = 0 => p (- 1) + 4 (2) - 6 = 0 => - p + 8 - 6 = 0 => p = 2$

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If the area of the parallelogram whose sides are x + 2y + 3 = 0, 3x + 4y - 5 = 0, 2x+4y+5=0 and 3x + 4y - 10 = 0 is 'a' sq. unit. Find the value of 4a

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and

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x + 2 y + 3 = 0, 3 x + 4 y - 5 = 0, 2 x + 4 y + 5 = 0 a n d 3 x + 4 y - 10 = 0

Q. A line through A (-5, -4) meets the lines x + 3y + 2 = 0, 2x + y + 4 = 0, 2x + y + 4 = 0 and x – y – 5 = 0 at B, C and D respectively. If

{(\frac{15}{A B})}^{2} + {(\frac{10}{A C})}^{2} = {(\frac{6}{A D})}^{2},

, then the equation of the line is

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Find 'p' 3x+4y−5=0,2x+3y−4=0,px+4y−6 are concurent.

The correct option is C 23x+4y−5=0=>x=5−4y32x+3y−4=0=>2(5−4y3)+3y−4=0=>10−8y+9y−12=0=>y=2x=5−4y3=5−4(2)3=−1px+4y−6=0=>p(−1)+4(2)−6=0=>−p+8−6=0=>p=2

Find 'p' $3 x + 4 y - 5 = 0, 2 x + 3 y - 4 = 0, p x + 4 y - 6$ are concurent.

The correct option is C $2$
$3 x + 4 y - 5 = 0 => x = \frac{5 - 4 y}{3} 2 x + 3 y - 4 = 0 => 2 (\frac{5 - 4 y}{3}) + 3 y - 4 = 0 => 10 - 8 y + 9 y - 12 = 0 => y = 2 x = \frac{5 - 4 y}{3} = \frac{5 - 4 (2)}{3} = - 1 p x + 4 y - 6 = 0 => p (- 1) + 4 (2) - 6 = 0 => - p + 8 - 6 = 0 => p = 2$