Question

A line passing through origin and is perpendicular to two given lines $2x+y+6=0$ and $4x+2y-9=0$. The ratio in which the origin divides points of intersection of given lines and the line perpendicular to given lines, is

• A. $1:2$
• B. $2:1$
• C. $4:2$
• D. $4:3$

Answer 1

The correct option is D

$4:3$

Explanation for the correct option:

Step 1. Find the ratio in which origin divides the line:

Given $2x+y+6=0...\left(i\right)$

The equation of line perpendicular to $2x+y+6=0$ and pass through the point $(0,0)$ is

$(y–y_1)=m(x–x_1)$

Slope of (i) is $-2$

So, the slope of required equation, $m=\frac{1}{2}$ [Since, the lines are perpendicular]

So

$\begin{array}{rcl}y–0& =& \frac{1}{2}(x–0)\end{array}$

$\Rightarrow$ $\begin{array}{rcl}y& =& \frac{1}{2}x\end{array}$

$\Rightarrow$$\begin{array}{rcl}x–2y& =& 0...\left(ii\right)\end{array}$

Step 2. Solving (i) and (ii), to get intersection points

$$\begin{gathered} x=-\frac{12}{5}, \\ y=-\frac{6}{5} \end{gathered}$$

So the point of intersection of line (i) and (ii) is $\left(-\frac{12}{5},-\frac{6}{5}\right)$

Equation of second line is: $4x+2y–9=0...\left(iii\right)$

Point of intersection of (ii) and (iii) is $\left(\frac{9}{5},\frac{9}{10}\right)$

Step 3. Let the origin(0, 0) divides points of intersection of given lines and line $x–2y=0$ in the ratio $\lambda :1,$

Then

$\begin{array}{rcl}x& =& \frac{\left({\displaystyle \frac{9}{5}}\right)\lambda –{\displaystyle \frac{12}{5}}}{\lambda +1}\end{array}$ [Using section formula]

$\Rightarrow$$\begin{array}{rcl}0& =& (9\lambda –12)\end{array}$ [Equating the coordinates]

$\begin{array}{rcl}& \Rightarrow & \lambda =\frac{12}{9}\\ & =& \frac{4}{3}\end{array}$

$\therefore$The ratio is $\mathbf{}\mathbf{4}\mathbf{:}\mathbf{3}$

Hence, option D is the answer.