Question

The lines $ax+(b+c)y=p$, $bx+(c+a)y=p$ and $cx+(a+b)y=p$ intersect at the point $(\frac{3}{4},\frac{3}{4})$ if

• A. $p=3,b+c=4$
• B. $p=3,a+b+c=4$
• C. $p=6,a=b+c=4$
• D. $p=2,a+b+c=3$

Answer 1

Answer: (B)

$p=3,a+b+c=4$

Given lines ,
$ax+(b+c)y=p$, $bx+(a+c)y=p$ and $cx+(a+b)y=p$
By substituting given point in all equations and $p=3$,we get
$a+b+c=4$