Question

The equation of a straight line upon which the length of the perpendicular from the origin is $5$ and slope of this perpendicular is $3/4$ is

• A. $2x+5y\pm 16=0$
• B. $4x+3y\pm 25=0$
• C. $4x+3y\pm 5=0$
• D. $2x+5y\pm 4=0$

Answer 1

Answer: (B)

Let $y=mx+c$ be the equation of straight line
$\Rightarrow$ $mx-y+c=0$ ..(i)
Now, by condition Perpendicular distance from origin to line (i) = 5
$\Rightarrow$ $\frac{|0-0+c|}{\sqrt{m^2+1}}=5$
$\Rightarrow$ $c=\pm 5\sqrt{m^2\pm 1}$ ..(ii)
Also, given that Slope of perpendicular to line (i) from the origin
$=\frac{3}{4}$
$\Rightarrow$ $-\frac{1}{m}=\frac{3}{4}$
$\Rightarrow$ $m=\frac{-4}{3}$
From Eq. (ii), $c=\pm 5\sqrt{m^2+1}$
$c=\pm 5\sqrt{1+\frac{16}{9}}=\pm 5\times \frac{5}{3}=\pm \frac{25}{3}$
On putting the values of m and c in Eq. (i), we get
$-\frac{4x}{3}-y\pm \frac{25}{3}=0$
$\Rightarrow$ $4x+3y\pm 25=0$
which is required equation of straight line.