Suppose we have a joint equation of planes $8x^2-3y^2-10z^2+10xy+17yz+2xz=0$.Suppose we put $z=0$ we get a joint equation of pair of straight lines. Now which particular pair of straight lines does this equation represent on putting $z=0$.
I'm finding it really difficult to visualize. Please help!
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By putting $z=0$ we get a homogeneous equation of second degree in $x$ and $y$: $$8x^2-3y^2+10xy=0$$
Divide the entire equation by $x^2$ and let $y=mx$:
$$3m^2-10m-8=0$$
Solving the quadratic we get two values of $m$ both of which correspond to a line in X-Y plane passing through origin.
The visualization is really simple. Each of the lines is the result of the intersection of one of the planes represented by the given equation with the plane $z=0$ .
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Putting $z=0$, you get from your equation, $$8x^2-3y^2+10xy=0$$ or, $$8x^2+12xy-2xy-3y^2=0$$ or, $$4x(2x+3y)-y(2x+3y)=0$$ or, $$(3y+2x)(-y+4x)=0$$ or, $$(y-4x)(y+\frac{2}{3}x)=0$$
So you have $2$ lines, $y=4x$ and $y=-\frac{2}{3}x$
The equation in $x,y,z$ you supplied is the plane on which these $2$ straight lines lie.
Notice, setting $z=0$ in the joint equation of the planes: $8x^2-3y^2-10z^2+10xy+17yz+2xz=0$ one should get $$8x^2-3y^2+10xy=0$$ setting $y=mx+c$, one should get $$8x^2-3(mx+c)^2+10x(mx+c)=0$$ $$(8+10m-3m^2)x^2+(10c-6mc)x-3c^2=0$$ comparing the corresponding coefficients on both the sides, one should get $c=0$ & $$3m^2-10m-8=0$$$$\implies m=4, \ -\frac{2}{3} $$ hence, setting values of $m$ & $c$, the pair of lines is $\color{red}{y=4x}$ & $\color{red}{y=-\frac{2}{3}x}$