Question

A straight line through origin O meets the parallel lines $4x+2y=9$ and $2x+y+6=0$ at point P and Q respectively. Then the point O divides the segment PQ in the ratio.
$\begin{array}{rl}& a)1:2\\ & b)3:4\\ & c)2:1\\ & d)4:3\end{array}$

Answer 1

Hint: Now we are given that a straight line through origin O meets the parallel lines $4x+2y=9$ and $2x+y+6=0$ at point P and Q respectively. We know that equation of line passing through origin is in $y=mx$ hence we will use this equation to solve with both lines to get the coordinates P and Q. Now we have coordinates of P, Q and O. hence we can use section formula which says if (x, y) divides the line joining $(x_1,x_2)$ and $(y_1,y_2)$ in ration m : n. then we have.
$(x,y)=(\left({\displaystyle \frac{m{x}_1+n{x}_2}{m+n}}\right),\left({\displaystyle \frac{m{y}_1+n{y}_2}{m+n}}\right))$ . hence we can find the ratio m : n.

Complete step-by-step answer:

Now consider the equation of line passing through origin
We know that the equation of line passing through origin is $y=mx$
Now let us say the equation of line PQ is $y=mx$
Now Let us first find the point of intersection P.
P lies on line $4x+2y=9$ and the line $y=mx$
Now consider the line $4x+2y=9$
Dividing throughout by 2 we get
$2x+y={\displaystyle \frac{9}{2}}$
Hence we get $y={\displaystyle \frac{9}{2}}-2x$
Now at intersection point of line $4x+2y=9$ and the line $y=mx$ which is P we have
$$\begin{array}{rl}& mx={\displaystyle \frac{9}{2}}-2x\\ & mx+2x={\displaystyle \frac{9}{2}}\\ & x(m+2)={\displaystyle \frac{9}{2}}\\ & x={\displaystyle \frac{9}{2(m+2)}}\end{array}$$
Now substituting $$x={\displaystyle \frac{9}{2(m+2)}}$$ in $y=mx$ we get.
$y={\displaystyle \frac{9m}{2(m+2)}}$
Hence the coordinates of P is $({\displaystyle \frac{9}{2(m+2)}},\left({\displaystyle \frac{9m}{2(m+2)}}\right)).............(1)$
Now line $2x+y+6=0$ and $y=mx$ intersects at Q.
Rearranging the terms of $2x+y+6=0$ we get $y=-2x-6$
Now at interaction point Q of line $2x+y+6=0$ and $y=mx$we will have
$mx=-2x-6$
Hence we get
$\begin{array}{rl}& mx+2x=6\\ & \Rightarrow x={\displaystyle \frac{6}{(m+2)}}\end{array}$
Now substituting $x={\displaystyle \frac{6}{(m+2)}}$ in $y=mx$ we get $y={\displaystyle \frac{6m}{(m+2)}}$
Hence the coordinates of Q are $({\displaystyle \frac{6}{m+2}},{\displaystyle \frac{6m}{m+2}}).....................(2)$
Now we have coordinates of P is $({\displaystyle \frac{9}{2(m+2)}},\left({\displaystyle \frac{9m}{2(m+2)}}\right))$ and coordinates of Q are $({\displaystyle \frac{6}{m+2}},{\displaystyle \frac{6m}{m+2}})$
Now we have O = (0, 0) divides the line PQ internally.
Let us say that that the point O divides the line PQ in ratio λ : 1.
Then we know by section formula if (x, y) divides the line joining $(x_1,x_2)$ and $(y_1,y_2)$ in ration m : n. then we have.
$(x,y)=(\left({\displaystyle \frac{m{x}_1+n{x}_2}{m+n}}\right),\left({\displaystyle \frac{m{y}_1+n{y}_2}{m+n}}\right))$
Hence for the line PQ we have.
$(0,0)=({\displaystyle \frac{({\displaystyle \frac{\lambda 9}{2(m+2)}}-{\displaystyle \frac{6}{m+2}})}{\lambda +1}},{\displaystyle \frac{({\displaystyle \frac{\lambda 9m}{2(m+2)}}-{\displaystyle \frac{6m}{m+2}})}{\lambda +1}})$
Now first equating x coordinate we get
$$\begin{array}{rl}& {\displaystyle \frac{({\displaystyle \frac{\lambda 9}{2(m+2)}}-{\displaystyle \frac{6}{m+2}})}{\lambda +1}}=0\\ & \Rightarrow ({\displaystyle \frac{\lambda 9}{2(m+2)}}-{\displaystyle \frac{6}{m+2}})=0\\ & \Rightarrow {\displaystyle \frac{\lambda 9}{2(m+2)}}={\displaystyle \frac{6}{m+2}}\\ & \Rightarrow {\displaystyle \frac{3\lambda }{2}}={\displaystyle \frac{2}{1}}\\ & \Rightarrow \lambda ={\displaystyle \frac{4}{3}}\end{array}$$
Hence the value of λ is ${\displaystyle \frac{4}{3}}$ .
Now we have point O divides the line PQ in ratio λ : 1.
Hence O divides PQ in ${\displaystyle \frac{4}{3}}:1=4:3$
Hence we have O divides line PQ in 4 : 3.
Option d is the correct answer.

Note: In section formula we use the ratio as m : n, and to solve we have assumed the ratio to be λ : 1 for simplicity. The answer through both the methods will be the same.