Show that, if the straight lines joining the origin to the points of intersection of $\;$ $3x^2 - xy + 3y^2 + 2x - 3y + 4 = 0$ $\;$ and $\;$ $2x + 3y = k$ $\;$ are at right angles, then $\;$ $6k^2 - 5k + 52 = 0$
Equation of given lines is: $\;$ $3x^2 - xy + 3y^2 + 2x - 3y + 4 = 0$ $\;\;\; \cdots \; (1)$
and $\;$ $2x + 3y = k$ $\;\;\; \cdots \; (2)$
Equation $(2)$ can be written as
$\dfrac{2x + 3y}{k} = 1$ $\;\;\; \cdots \; (2a)$
$\therefore \;$ Making equation $(1)$ homogeneous in $x$ and $y$ using equation $(2a)$ we get,
$3x^2 - xy + 3y^2 + 2 \left(\dfrac{2x + 3y}{k}\right)x - 3 \left(\dfrac{2x + 3y}{k}\right)y + 4 \times \left(\dfrac{2x + 3y}{k}\right)^2 = 0$
i.e. $\;$ $3k^2 x^2 - k^2 xy + 3k^2 y^2 + 4kx^2 + 6kxy -6kxy - 9ky^2 + 16x^2 + 36y^2 + 48xy = 0$
i.e. $\;$ $\left(3k^2 + 4k + 16\right)x^2 + \left(-k^2 + 48\right)xy + \left(3k^2 - 9k + 36\right)y^2 = 0$ $\;\;\; \cdots \; (3)$
Since equation $(3)$ is homogeneous and of the second degree in $x$ and $y$, it represents a pair of straight lines passing through the origin.
Lines represented by equation $(3)$ will be at right angles if
$\text{coefficient of } x^2 + \text{ coefficient of } y^2 = 0$
i.e. $\;$ $3k^2 + 4k + 16 + 3k^2 - 9k + 36 = 0$
i.e. $\;$ $6k^2 - 5k + 52 = 0$