If the straight lines joining the origin and the points of intersection of the curves $5{x^2} + 12xy - 6{y^2} + 4x - 2y + 3 = 0$ and $x + ky - 1 = 0$ are equally inclined to the co-ordinate axis, then find the value of $\left| k \right|$

Hint: Whenever we have the curve and the line and we need to find their point of intersection, we need to homogenise the two equations by making their each term’s degree equal. Now after homogenising we will get the equation in the form of $A{x^2} + 2Hxy + B{y^2} = 0$
For the equal inclination to the co-ordinate axis, we can say that ${\text{coefficient of }}xy = 0$.

Complete step-by-step answer:
Here we are given that the straight lines joining the origin and the points of intersection of the curves $5{x^2} + 12xy - 6{y^2} + 4x - 2y + 3 = 0$ and $x + ky - 1 = 0$ are equally inclined to the co-ordinate axis, then we need to find the value of $\left| k \right|$
We can draw the figure as follows:


So first of all we will find the value of $k$ and then we know that $\left| k \right| = k{\text{ if }}k > 0$ and $\left| k \right| = - k{\text{ if }}k < 0$
Whenever we have the curve and the line and we need to find their point of intersection, we need to homogenise the two equations by making their each term’s degree equal. Now we are given the curve
$5{x^2} + 12xy - 6{y^2} + 4x - 2y + 3 = 0$$ - - - - (1)$
Line equation is given as
$x + ky - 1 = 0$$ - - - - (2)$
So we can homogenise by putting the value of $1 = x + ky$ from the equation (2) in the equation (1) to make the degree equal to $2$ because degree of ${x^2},{y^2},xy$ are all equal to $2$
So $5{x^2} + 12xy - 6{y^2} + 4x - 2y + 3 = 0$
$5{x^2} + 12xy - 6{y^2} + 4x.1 - 2y.1 + {3.1^2} = 0$
As degree of $x{\text{ and }}y$ is $1$ so we will put the same value of $1 = x + ky$ but here we also have the constant which is $3$ whose degree is $0$ so we need to multiply it by ${1^2}$ to make the degree equal to $2$
So putting the value we get
$\Rightarrow$ $5{x^2} + 12xy - 6{y^2} + 4x.(x + ky) - 2y.(x + ky) + 3.{(x + ky)^2} = 0$
Now simplifying it we get:
$\Rightarrow$ $5{x^2} + 12xy - 6{y^2} + 4{x^2} + 4kxy - 2xy + 2k{y^2} + 3({x^2} + {k^2}{y^2} + 2kxy) = 0$
$\Rightarrow$ $5{x^2} + 12xy - 6{y^2} + 4{x^2} + 4kxy - 2xy + 2k{y^2} + 3{x^2} + 3{k^2}{y^2} + 6kxy = 0$
Now separating coefficients of the terms ${x^2},{y^2},xy$ we get
$\Rightarrow$ $5{x^2} + 4{x^2} + 3{x^2} + 12xy + 4kxy - 2xy + 6kxy - 6{y^2} + 2k{y^2} + 3{k^2}{y^2} = 0$
$\Rightarrow$ $(5 + 4 + 3){x^2} + (12 + 4k - 2 + 6k)xy + ( - 6 + 2k + 3{k^2}){y^2} = 0$
$\Rightarrow$ $12{x^2} + (10 + 10k)xy + (3{k^2} + 2k - 6) = 0$$ - - - - - - (3)$
So now we have got the equation in the form $A{x^2} + 2Hxy + B{y^2} = 0$
So we know that ${m_1} + {m_2} = - \dfrac{{2H}}{B}$ and ${m_1}{m_2} = \dfrac{A}{B}$
Here ${m_1}$ is the slope of the line joining origin to the point of intersection of the curves with $x - {\text{axis}}$ and ${m_2}$ with the $y - {\text{axis}}$
But here we are told that there is equal inclination to the co-ordinate axis. Therefore we can say that if ${m_1} = \tan \theta $ then ${m_2} = - \tan \theta $
So we can say that ${m_1} + {m_2} = - \dfrac{{2H}}{B} = 0$
So we get that $ - \dfrac{{2H}}{B} = 0$
On solving we get $H = 0$
So we can say that the coefficient of $xy = 0$ for the equal inclination
Equating it to zero we get
$\Rightarrow$ $10 + 10k = 0$
$\Rightarrow$ $10k = - 10$
$\Rightarrow$ $k = - \dfrac{{10}}{{10}} = - 1$
$\Rightarrow$ So $k = - 1$
We need to find the value of $\left| k \right|$
 We know that $\left| k \right| = k{\text{ if }}k > 0$ and $\left| k \right| = - k{\text{ if }}k < 0$
As $k = - 1 < 0$
So $\left| k \right| = - k = - ( - 1) = 1$
So we get $\left| k \right| = 1$

Note: Here a student must know that whenever we are given the curve and the line equation and we are told about the curve or line joining the point of intersection of the curve and the line, we need to homogenise the curve and make the degree the same. Then we get the equation of the form $A{x^2} + 2Hxy + B{y^2} = 0$
So we must know that ${m_1} + {m_2} = - \dfrac{{2H}}{B}$ and ${m_1}{m_2} = \dfrac{A}{B}$
Here ${m_1}$ is the slope of the line joining origin to the point of intersection of the curves with $x - {\text{axis}}$ and ${m_2}$ with the $y - {\text{axis}}$