Question

Distance of line
$y=\left|\begin{array}{ccc}x+1& x& x\\ x& x+2& x\\ x& x& x+3\end{array}\right|$ from the origin is

• A. $\frac{6}{11}$
• B. $\frac{7}{13}$
• C. $\frac{6}{\sqrt{122}}$
• D. $\frac{7}{\sqrt{122}}$

Answer 1

Answer: (C)

$\frac{6}{\sqrt{122}}$

Given,

$y=\left|\begin{array}{ccc}x+1& x& x\\ x& x+2& x\\ x& x& x+3\end{array}\right|$

$=(x+1)det\left(\begin{array}{cc}x+2& x\\ x& x+3\end{array}\right)-xdet\left(\begin{array}{cc}x& x\\ x& x+3\end{array}\right)+xdet\left(\begin{array}{cc}x& x+2\\ x& x\end{array}\right)$

$=(x+1)(5x+6)-x\cdot 3x+x(-2x)$

$\therefore y=11x+6$

Hence, the line equation is given by,

$y=11x+6$ and the point is origin $(0,0)$

Distance between a point and a line is given by,

$D=\frac{|ax+by+c|}{\sqrt{a^2+b^2}}$

$=\frac{11\left(0\right)-\left(0\right)+6}{\sqrt{{11}^2+(-1{)}^2}}$

$=\frac{6}{\sqrt{122}}$