Question

Consider the family of lines $5x+3y-2+{\lambda }_1(3x-y-4)=0$ and $x-3y+1+{\lambda }_2(2x-y-2)=0$, then the equation of a straight line that belongs to both the families, is

• A. $25x-62y+86=0$
• B. $62x-25y+86=0$
• C. $25x-62y=86$
• D. $5x-2y-7=0$

Answer 1

The correct option is D $5x-2y-7=0$

$5x+3y-2+{\lambda }_1(3x-y-4)=0$

concurrent point of this family

of $\mathrm{line}=(1,-1)$

$x-y+1+{\lambda }_2(2x-y-2)=0$

concurrent point of this family of

line $=(3,4)$

now equation belong to both the family of lines have to satisfy

(1,-1) and (3,4)

Hence equation of line passing

through (1,-1) and(3 , 4)

$y-4=\frac{-1-4}{1-3}(x-3)$

$y-4=\frac{-5}{-2}(x-3)$

$2y-8=5x-15$

$5x-2y-7=0$

option $D$ is correct.