Question

Equation of all lines passing through the intersection of lines $2x+3y=5$ and $x+y=2$ is given by $(2x+3y-5)+\lambda (x+y-2)=0$, of all these lines one line is farthest from point $(7,4)$, the value of $\lambda$ is :

• A. $2$
• B. $4$
• C. $-4$
• D. None of the above

Answer 1

Answer: (C)

$-4$

Intersection point of given lines is $(1,1)$

The farthest line from $(7,4)$ will be perpendicular to the line joining $(1,1)$ and $(7,4)$

The line joining $(1,1)$ and $(7,4)$ is: $\frac{y-1}{x-1}=\frac{3}{6}⟹2y=x+1$

The slope of the line which is perpendicular to $2y=x+1$ is $-2$

From the variable equation of the line, the slope of the line is given by:

$\frac{-(2+\lambda )}{3+\lambda }=-2⟹-2-\lambda =-6-2\lambda$

$\lambda =-6+2=-4$

Hence, option C is correct.