Find the line segments cut off by the plane $ax+by+cz+d=0$ on the coordinate axes, if $abcd\neq 0$

Question

I'm reading Pogorelov's Geometry.

Find the line segments cut off by the plane $ax+by+cz+d=0$ on the coordinate axes, if $abcd\neq 0$.

Writing the equation as $a(x-x_0)+b(y-y_0)+c(z-z_0)=0$, I think that the line segment is $(x-x_0)$. But when the equation is written as $ax+by+cz+d=0$, I have no clue as to how to think about it.

Answer 1

Condition $abcd\neq 0$ implies that none of $a,b,c,d$ are zero. Since $d\neq 0$ then $ax+by+cz\neq 0$, so the plane $ax+by+cz+d=0$ doesn't 'pass through' (=not including) the origin $(0,0,0)$.

And, because $a,b,c$ are non zero, so the plane cuts $x,y,z$-axes. For example when it cuts $x$-axis this means that in the intersection point of the plane and $x$-axis both $y$ and $z$ are zero (you may more understand it by picture). Thus $y=z=0\implies ax+by+cz+d=ax+d=0\implies x=\dfrac{-d}{a}$.

The others can be found in the same way.

EDIT - In the picture below $OA=\dfrac{-d}{a}$, $OB=\dfrac{-d}{b}$ and $OC=\dfrac{-d}{c}$.

Answer 2

In general the plane: $\color{blue}{ax+by+cz+d=0}$ is expressed in the $\color{blue}{\text{intercept form}}$ as follows $$\frac{x}{\left(-\frac{d}{a}\right)}+\frac{y}{\left(-\frac{d}{b}\right)}+\frac{z}{\left(-\frac{d}{c}\right)}=1$$ Where, $\color{blue}{\left(-\frac{d}{a}\right)}$, $\color{blue}{\left(-\frac{d}{b}\right)}$ & $\color{blue}{\left(-\frac{d}{c}\right)}$ are the intercepts cut by the plane: $\color{blue}{ax+by+cz+d=0}$ on $\color{blue}{\text{x, y & z coordinate axes}}$ respectively.