Question

If the equation \(kx^2-5xy+6y^2+x-3y = 0\) represents a pair of straight lines, then the co-ordinates of their point of intersection are

• A. \((-3,-1)\)
• B. \((-3,1)\)
• C. \((3,-1)\)
• D. \((3,1)\)

Hint:

Find \(k\) from the pair-of-lines condition, factorise, and solve the two lines.

Answer 1

The Correct Option is A

Step 1: Understanding the Question:
For \(ax^2+2hxy+by^2+2gx+2fy+c = 0\) to be a pair of lines, \(abc + 2fgh - af^2 - bg^2 - ch^2 = 0\).

Step 2: Find k:
Here \(a = k\), \(b = 6\), \(h = -\frac52\), \(g = \frac12\), \(f = -\frac32\), \(c = 0\).
\[ 0 + 2\left(-\frac32\right)\left(\frac12\right)\left(-\frac52\right) - k\cdot\frac94 - 6\cdot\frac14 - 0 = 0 \]
\(\frac{15}{4} - \frac{9k}{4} - \frac{6}{4} = 0\), so \(k = 1\).

Step 3: Factorise:
With \(k=1\): \(x^2 - 5xy + 6y^2 + x - 3y = (x-2y)(x-3y) + (x-3y) = (x-3y)(x-2y+1)\).
The lines are \(x-3y = 0\) and \(x-2y+1 = 0\).

Step 4: Intersect:
From the first, \(x = 3y\). In the second: \(3y - 2y + 1 = 0\), so \(y = -1\) and \(x = -3\). The point is \((-3,-1)\).

Final Answer:
The lines meet at \((-3,-1)\), option (A). \[ \boxed{(-3,-1)} \]