Question

If the lines 3y+4x=1, y=x+5 and 5y+bx=3 are concurrent then b=?

Please explain step by step.
Points :-

Answer 1

Given,
4x + 3y -1 = 0
x - y + 5 = 0
bx + 5y -3 = 0 are concurrent .
so,
P( 4x + 3y -1) + Q (x - y + 5) + R (bx + 5y -3) = 0

x ( 4P + Q + bR) + y(3P -Q + 5R) + (-P + 5Q -3R) = 0

this is possible when, we think ,
det[{ 4, 1 , b } {3 , -1 , 5 } { -1 , 5 , -3 }] = 0
4(3 - 25) - (-9+5) + b( 15 -1 ) = 0
-88 +4 + 14b = 0
14b - 84 = 0
b = 6

Answer 2

$\huge\boxed{\texttt{\fcolorbox{Red}{aqua}{Hey Mate!!!!}}}$

$<b>$ Given that:
$4x + 3y - 1 = 0 \\ x - y + 5 = 0 \\ bx + 5y - 3 = 0 \:$
are concurrent.

So,
$p(4x + 3y - 1) + q(x - y + 5) + r(bx + 5y - 3) = 0 \\ \\ x(4p + q + br) + y(3p - q + 5r) + ( - p + 5q - 3r) = 0$
It can be possible if we think

[{ 4, 1, b } { 3, -1, 5 } { -1, 5, -3 }] = 0
$4(3 - 25) - ( - 9 + 5) + b(15 - 1) = 0 \\ - 88 + 4 + 14b = 0 \\ 14b - 84 = 0 \\ b = 6$

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