Question

Let $P_1,P_2,P_3$ be the perpendicular distances between pair of parallel lines represented by $x^2-3x-4=0$, $y^2-5y+6=0$, $4x^2+20xy+25y^2=0$ respectively then

• A. $P_3<P_2<P_1$
• B. $P_3<P_1<P_2$
• C. $P_2<P_1<P_3$
• D. $P_1<P_2<P_3$

Answer 1

The correct option is A $P_3<P_2<P_1$

$x^2-3x-4=0$
$(x-4)(x+1)=0$

$x=4$ and $x=-1$

Hence the perpendicular distance between these two lines

$P_1=4-(-1)=5$.

$y^2-5y+6=0$
$(y-2)(y-3)=0$
$y=2$ and $y=3$

Hence perpendicular distance between these lines is

$P_2=3-2=1$

Thus $P_2<P_1$

$4x^2+20xy+25y^2=0$

$x=\frac{-20y\pm \sqrt{400y^2-400y^2}}{8}$

$x=\frac{-20y}{8}$

Or

$8x+20y=0$
$2x+5y=0$

Since we get a single line

$P_3=0$
Therefore

$P_3<P_2<P_1$