#### Let $L_1$ be a straight line passing through the origin and$L_2$be the straight line$2x+2y=1$. If the intercepts made by the circle$3 x^2+2 y^2-x+4 y=0$ on$L_1$ and$L_2$ are equal, then which of the following equations can represent$L_1$?Option: 1 $y=(1 / 2) x \\$Option: 2 $y=2 x \\$Option: 3 $y=(1 / 4) x \\$Option: 4 $y=-4 x$

let's consider the general equation of a straight line, which is$y=mx+c$

Since the line passes through the origin (0, 0), the y-intercept (c) is zero.

Therefore, the equation of $L_1$can be written as$y=mx+c$

The intercepts made by the circle equation \begin{aligned} & 3 x^2+2 y^2-x+4 y=0 \\ & \quad 3 x^2+2(m x)^2-x+4(m x)=0 \\ & 3 x^2+2 m^2 x^2-x+4 m x=0 \\ & \left(3+2 m^2\right) x^2+(4 m-1) x=0 \end{aligned}

For the intercepts to be equal, the discriminant of this quadratic equation should be zero.

The discriminant (D) of a quadratic equation $a x^2+b x+c=0$ is given by,$D=b^2-4 a c .$

\begin{aligned} & \quad D=(4 m-1)^2-4\left(3+2 m^2\right)(0) \\ & D=0 \end{aligned}

so,

\begin{aligned} & (4 m-1)^2-4\left(3+2 m^2\right)(0)=0 \\ & (4 m-1)^2=0 \end{aligned}

\begin{aligned} & (4 m-1)^2-4\left(3+2 m^2\right)(0)=0 \\ & (4 m-1)^2=0 \\ & 16 m^2-8 m+1=0 \end{aligned} Quadratic equation for m : \begin{aligned} & 16 m^2-8 m+1=0 \\ m= & \left.\left(-b \pm \sqrt{(} b^2-4 a c\right)\right) /(2 a) \\ m & \left.=\left(-(-8) \pm \sqrt{(}(-8)^2-4(16)(1)\right)\right) /(2(16)) \\ m & =(8 \pm \sqrt{(64-64)) / 32} \\ m= & 8 \pm \sqrt{0} / 32 \\ m & =8 / 32 \\ m & =1 / 4 \end{aligned}

Therefore, the slope of the line $L_1$ is $1/4$

Thus, the equation that represents $L_1$  is $y=(1/4)x$