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My teacher told me that if I am given a line $L_1$ and a curve $C_1$ $$L_1 : lx + my + n=0 $$ $$C_1 : ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0$$

If $n \not= 0$ and if the line $L_1$ cuts the curve $C_1$ at $2$ points, then the joint equation of straight lines through these points and passing the origin is given by "homogenizing" the equation of the curve by the equation of the line. Which would give:

$$ax^2 + 2hxy + by^2 + \left(2gx+2fy\right)\left(\frac{lx+my}{-n}\right) + c\left(\frac{lx+my}{-n}\right)^2 = 0$$

She didn't go into much detail about this but told us that it has something to do with making making the degree of every term in the equation the same or making it "homogeneous".

Why does this work and why does it give me the joint equation of the $2$ required straight lines? Are there any sources through which I can do some further reading on this topic?

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You can rewrite the homogenized equation as $Ax^2+Bxy+Cy^2=0$ and it can be factorized as $(p_1x+q_1y)(p_2x+q_2y)=0$. So it represents the line pair $p_1x+q_1y=0$ and $p_2x+q_2y=0$.

Since the points of intersection satisfy both $C_1$ and $L_1$, they satisfy the homogenized equation.

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  • $\begingroup$ But why does it pass through the origin and the two points in which the line $L_1$ and the curve $C_1$ intersect? $\endgroup$ – eem Jul 7 '19 at 2:44
  • $\begingroup$ $(0,0)$ satisfies both $p_1x+q_1y=0$ and $p_2x+q_2y=0$. So it passes through the origin. Take a point of intersection $P(x_0,y_0)$ of $C_1$ and $L_1$. We have $lx_0 + my_0 + n=0$ and $ax_0^2 + 2hx_0y_0 + by_0^2 + 2gx_0 + 2fy_0 + c = 0$. So we have $ax_0^2 + 2hx_0y_0 + by_0^2 + \left(2gx_0+2fy_0\right)\left(\frac{lx_0+my_0}{-n}\right) + c\left(\frac{lx_0+my_0}{-n}\right)^2 = 0$. $P$ satisfies the homogenized equation. $\endgroup$ – CY Aries Jul 7 '19 at 2:49

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