Question

The angle between lines joining origin and intersection points of line $2x+y=1$ and curve $3x^2+4yx-4x+1=0$

• 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(s)

Find the angle between lines joining origin and intersection points of line.

Given:

$3x^2+4yx-4x+1=0$

$2x+y=1$

Homogenizing the Equation of the curve

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

Angle between lines can be expressed as, by equating coefficient of $x^2$ and coefficient of $y^2$ is equal to $0$.

$$\begin{gathered} LHS: \\ \Rightarrow x^2+y^2 \\ \Rightarrow -1+1 \\ \Rightarrow 0 \\ RHS=0 \\ \therefore LHS=RHS \end{gathered}$$

So that an angle is $90°$ then the line is perpendicular.

Thus, the angle between required line is $\frac{\mathrm{\pi }}{2}$.

Hence, the correct answer is Option A.