Question

Statement-1: Two real and distinct straight lines can be drawn passing through the point $M(4,-5)$ whose perpendicular distance from $P(-2,3)$ is equal to 12 .
Statement-2: Perpendicular distance of line $ax+by+c=0$ from $\left(x_1,y_1\right)$ is $\frac{|a_1+b{y}_1+c|}{\sqrt{a^2+b^2}}$.

• A. Statement-1 is true, statement-2 is true and statement-2 is correct explanation for statement-1.
• B. Statement-1 is true, statement-2 is true and statement-2 is NOT the correct explanation for statement-1.
• C. Statement-1 is true, statement-2 is false.
• D. Statement-1 is false, statement-2 is true

Answer 1

Answer: (D)

If possible, let the equation of the line be $(y+5)=m(x-4)$ i.e., $y-mx+4m+5=0$
Then $\left|\frac{3+2m+4m+5}{\sqrt{1+m^2}}\right|=12\Rightarrow (8+6m{)}^2=144\left(1+m^2\right)\Rightarrow 27m^2-24m+20=0\dots$ (i)
Since discriminant of (i) is $(24{)}^2-4\times 27\times 20$, which is negative so there is no real value of $m$ which satisfies (i).
Hence no such line is possible.