Question

The length of x-intercept made by the pair of lines \( 2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0 \) is:

• A. \( 2 \)
• B. \( 10 \)
• C. \( 5 \)
• D. \( 20 \)

Hint:

To find the x-intercept of a pair of lines given by \( Ax^2 + 2Hxy + By^2 + 2Gx + 2Fy + C = 0 \), set \( y = 0 \) and solve for \( x \).

Answer 1

The Correct Option is C

To find the length of the x-intercept made by the pair of lines given by the equation \(2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0\), we first determine the points where the lines intersect the x-axis. This happens when \(y = 0\). By substituting \(y = 0\) into the equation, we simplify:

\(2x^2 - 2x - 12 = 0\)

Dividing the entire equation by 2 gives:

\(x^2 - x - 6 = 0\)

This is a quadratic equation in the form \(ax^2 + bx + c = 0\), where \(a = 1\), \(b = -1\), and \(c = -6\). To find the roots, use the quadratic formula:

\(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

Substitute the values of \(a\), \(b\), and \(c\):

\(x = \frac{-(-1) \pm \sqrt{(-1)^2 - 4 \times 1 \times (-6)}}{2 \times 1}\)

\(x = \frac{1 \pm \sqrt{1 + 24}}{2}\)

\(x = \frac{1 \pm \sqrt{25}}{2}\)

\(x = \frac{1 \pm 5}{2}\)

This results in:

\(x = \frac{6}{2} = 3\) and \(x = \frac{-4}{2} = -2\)

The x-intercepts are \(x = 3\) and \(x = -2\). The distance between these intercepts is calculated as the absolute difference:

Distance = \(|3 - (-2)| = |3 + 2| = 5\)

Thus, the length of the x-intercept made by the pair of lines is \(5\).

Answer 2

Step 1: Find the points where the pair of lines intersect the x-axis To determine the x-intercepts, we set \( y = 0 \) in the given equation: \[ 2x^2 + xy - 6y^2 - 2x + 17y - 12 = 0 \] \[ \Rightarrow 2x^2 - 2x - 12 = 0 \]

Step 2: Solve the quadratic equation The quadratic equation simplifies to: \[ 2x^2 - 2x - 12 = 0 \] Dividing throughout by 2: \[ x^2 - x - 6 = 0 \] Factoring: \[ (x - 3)(x + 2) = 0 \] \[ \Rightarrow x = 3 \quad \text{or} \quad x = -2 \]
Step 3: Find the x-intercept length The length of the x-intercept is: \[ |3 - (-2)| = |3 + 2| = 5 \] Thus, the required length is: \[ \boxed{5} \]