The equation of the line joining the origin to the point of intersection of the lines `2x^2+xy-y^2+5x-y+2=0 ` is

Options

A. x+y=0
B. x-y=0
C. x-2y=0
D. 2x+y=0

Answer

To solve the problem, we need to find the equation of the line joining the origin (0, 0) to the point of intersection of the lines given by the equation:

\[ 2x^2 + xy - y^2 + 5x - y + 2 = 0 \]

**Step 1: Rewrite the equation in a suitable form.**

We need to express the given equation in a form that reveals the lines it represents. The equation is a quadratic in \(x\) and \(y\).

**Step 2: Identify the coefficients.**

The equation can be rearranged as:

\[ 2x^2 + xy - y^2 + 5x - y + 2 = 0 \]

This is a general conic equation. We will find the point of intersection of the lines represented by this equation.

**Step 3: Find the point of intersection.**

To find the point of intersection, we can use the condition that the discriminant of the quadratic in \(x\) must be zero for the lines to intersect at a single point.

The equation can be treated as a quadratic in \(x\):

\[ 2x^2 + (y + 5)x + (-y^2 - y + 2) = 0 \]

The discriminant \(D\) of this quadratic must be zero for the lines to intersect:

\[ D = (y + 5)^2 - 4 \cdot 2 \cdot (-y^2 - y + 2) = 0 \]

**Step 4: Solve for \(y\).**

Expanding the discriminant:

\[ (y + 5)^2 + 8(y^2 + y - 2) = 0 \]

This simplifies to:

\[ y^2 + 10y + 25 + 8y^2 + 8y - 16 = 0 \]

Combining like terms gives:

\[ 9y^2 + 18y + 9 = 0 \]

Dividing through by 9:

\[ y^2 + 2y + 1 = 0 \]

Factoring gives:

\[ (y + 1)^2 = 0 \implies y = -1 \]

**Step 5: Substitute \(y\) back to find \(x\).**

Now substitute \(y = -1\) back into the original equation to find \(x\):

\[ 2x^2 + (-1 + 5)x - (-1)^2 - (-1) + 2 = 0 \]

This simplifies to:

\[ 2x^2 + 4x + 0 = 0 \]

Factoring out \(2x\):

\[ 2x(x + 2) = 0 \]

This gives us:

\[ x = 0 \quad \text{or} \quad x = -2 \]

**Step 6: Identify the point of intersection.**

The points of intersection are \((0, -1)\) and \((-2, -1)\). We will use one of these points, say \((-2, -1)\).

**Step 7: Find the equation of the line from the origin to the point of intersection.**

The slope \(m\) of the line from the origin \((0, 0)\) to the point \((-2, -1)\) is given by:

\[ m = \frac{-1 - 0}{-2 - 0} = \frac{1}{2} \]

The equation of the line can be expressed as:

\[ y = mx \implies y = \frac{1}{2}x \]

Rearranging gives:

\[ x - 2y = 0 \]

**Final Answer:**

The equation of the line joining the origin to the point of intersection is:

\[ x - 2y = 0 \]

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