If $l(x,y)=0$ and $l'(x,y)=0$ intersects in $P_0(x_0,y_0)$ then $ \lambda l (x,y) + \lambda ' l'(x,y)=0 , \quad ( \lambda +\lambda ' =1 )$

Question

I'm reading "What is Mathematics?" and found this question.

let $l(x,y)=0$ represents the equation $ax+by+c=0$ of a line $l$,and so does $l'(x,y)=0$ . now let $\lambda + \lambda ' =1 $.
Show that, if $l$ and $l'$ intersect in $P_0(x_0,y_0)$,then every line through $P_0$ has an equation
$$ \lambda l (x,y) + \lambda ' l'(x,y)=0 $$ vice versa. and every such line is uniquely determined by the choice of a pair of values for $\lambda$ and $\lambda '$.

Appreciate for your help.

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updated

i tried to solve the $x_0 = \frac {-cb' +c'b} {ab'-a'b} , y_0 = \frac {c'a-a'c} {a'b-ab'}$ (suppose $l$ is $ax+by+c=0$ and $l'$ is $a'x+b'y+c'=0$ ) and suppose $ m = \frac {\lambda} {\lambda + \lambda'}$ and substitute in the equation $$ y-y_0 = m(x-x_0)$$,but i couldn't go any further to reach the goal.
also tried to suppose an equation $Ax+By+C=0$,and rewrite it in the form $y=mx+D$,and use the $x_0,y_0,m$ to get the relationship between $A,B,C$, also fails to go any further.

Answer 1

First note that $ \lambda l (x,y) + \lambda ' l'(x,y)=0 $ is an equation of a line and that $ \left(\lambda l (x,y) + \lambda ' l'(x,y)\right)_{(x,y)=(x_0,y_0)}=0 $ so $P_0(x_0,y_0)$ is a point of this line $\forall \lambda,\lambda'\in\mathbb{R}$