Question

The straight lines joining the origin to the points of intersection of the line $2x+y=1$ and curve $3x^2+4xy-4x+1=0$include an angle:

• A. $\frac{\mathrm{\pi }}{2}$
• B. $\frac{\mathrm{\pi }}{3}$
• C. $\frac{\mathrm{\pi }}{4}$
• D. $\frac{\mathrm{\pi }}{6}$

Answer 1

The correct option is A

$\frac{\mathrm{\pi }}{2}$

Explanation for the correct option:

Given line : $2x+y=1\dots \left(i\right)$

Equation of curve: $3x^2+4xy–4x+1=0\dots \left(ii\right)$

Substituting $\left(i\right)$ in $\left(ii\right)$ in place of $1$as follows:

$3x^2+4xy–4x\left(1\right)+1{\left(1\right)}^2=0$

$$\begin{gathered} \Rightarrow 3x^2+4xy–4x(2x+y)+1{(2x+y)}^2=0 \\ \Rightarrow 3x^2+4xy–8x^2-4xy+4x^2+4xy+y^2=0 \\ \Rightarrow -x^2+4xy+y^2=0 \end{gathered}$$

Condition that two lines are perpendicular are given by equation: $a{x}^2+2hxy+b{y}^2=0$, where $a$, $b$and $h$are real numbers and not all zero

We get $a=-1,b=1,h=2$

We know the angle between pair of straight lines $a{x}^2+2hxy+b{y}^2=0$ is

$\begin{array}{rcl}\tan \theta & =& \frac{2\sqrt{h^2-ab}}{a+b}\\ \tan \theta & =& \frac{2\sqrt{2^2-(-1)\left(1\right)}}{-1+1}\\ \tan \theta & =& \frac{2\sqrt{4+1}}{0}\\ \tan \theta & =& \infty \\ & \Rightarrow & \theta =\frac{\mathrm{\pi }}{2}\end{array}$

Here $a+b=0$

$\Rightarrow$lines are perpendicular because the angle formed by them is $90°$.

So the angle is $\frac{\mathrm{\pi }}{2}$.

Hence, the correct option is option (A).