Question

Find the equations to the straight lines passing through the point of intersection of the straight lines
$Ax+By+C=0$ and $A^{\prime }x+B^{\prime }y+C^{\prime }=0$ and
(1) passing through the origin,
(2) parallel to the axis of $y$,
(3) cutting off a given distance a from the axis of $y$, and
(4) passing through a given point $(x^{\prime },y^{\prime })$.

Answer 1

(1) Equation of line passing through intersection of given lines is

$ax+by+c+\lambda (a^{\prime }x+b^{\prime }y+c^{\prime })=\mathrm{0........}\left(i\right)$

It passes through $(0,0)$

$a\left(0\right)+b\left(0\right)+c+\lambda \left(a^{\prime }\right(0)+b^{\prime }(0)+c^{\prime })=0c+\lambda c^{\prime }=0\Rightarrow \lambda =-\frac{c}{c^{\prime }}$

substituting $\lambda$ in $\left(i\right)$

$ax+by+c+(-\frac{c}{c^{\prime }})(a^{\prime }x+b^{\prime }y+c^{\prime })=0a{c}^{\prime }x+b{c}^{\prime }y+c{c}^{\prime }-a^{\prime }cx-b^{\prime }cy-c{c}^{\prime }=0(a{c}^{\prime }-a^{\prime }c)x+(b{c}^{\prime }-b^{\prime }c)y=0$

${}_{-}-------------------------------------$

(2) Equation of line is

$ax+by+c+\lambda (a^{\prime }x+b^{\prime }y+c^{\prime })=0(a+\lambda a^{\prime })x+(b+\lambda b^{\prime })y+c+\lambda c^{\prime }=\mathrm{0.........}\left(ii\right)$

Slope of line $=m=-\frac{a}{b}=-\frac{a+\lambda a^{\prime }}{b+\lambda b^{\prime }}$

Line is parallel to $y$ axis so $m=\infty$

$\Rightarrow \frac{1}{m}=0-\frac{b+\lambda b^{\prime }}{a+\lambda a^{\prime }}=0b+\lambda b^{\prime }=0\Rightarrow \lambda =-\frac{b}{b^{\prime }}$

substituting $\lambda$ in $\left(ii\right)$

$(a+(-\frac{b}{b^{\prime }})a^{\prime })x+(b+(-\frac{b}{b^{\prime }})b^{\prime })y+c+(-\frac{b}{b^{\prime }})c^{\prime }=0\left(\frac{a{b}^{\prime }-b{a}^{\prime }}{b^{\prime }}\right)x+\left(0\right)y+\frac{c{b}^{\prime }-b{c}^{\prime }}{b^{\prime }}=0(a{b}^{\prime }-b{c}^{\prime })x+(c{b}^{\prime }-b{c}^{\prime })=0$

${}_{-}-------------------------------------$

(3) Equation of line is

$ax+by+c+\lambda (a^{\prime }x+b^{\prime }y+c^{\prime })=0(a+\lambda a^{\prime })x+(b+\lambda b^{\prime })y+c+\lambda c^{\prime }=\mathrm{0......}\left(iii\right)$

It cuts a distance $a$ from $y$ axis , so it passes through $(0,a)$

$(a+\lambda a^{\prime })\left(0\right)+(b+\lambda b^{\prime })\left(a\right)+c+\lambda c^{\prime }=0ab+a{b}^{\prime }\lambda +c+\lambda c^{\prime }=0\Rightarrow \lambda =-\frac{ab+c}{a{b}^{\prime }+c^{\prime }}$

Substituting $\lambda$ in $\left(iii\right)$

$(a+(-\frac{ab+c}{a{b}^{\prime }+c^{\prime }})a^{\prime })x+(b+(-\frac{ab+c}{a{b}^{\prime }+c^{\prime }})b^{\prime })y+c+(-\frac{ab+c}{a{b}^{\prime }+c^{\prime }})c^{\prime }=0(a^2{b}^{\prime }+a{c}^{\prime }-c{a}^{\prime }+ab{a}^{\prime })x+(b^{\prime }c-c^{\prime }b)y+a(b^{\prime }c-c^{\prime }b)=0$

$---------------------------------------$

(4) Equation of line is

$ax+by+c+\lambda (a^{\prime }x+b^{\prime }y+c^{\prime })=\mathrm{0........}\left(iv\right)$

It passes through $(x^{\prime },y^{\prime })$

$a{x}^{\prime }+b{y}^{\prime }+c+\lambda (a^{\prime }{x}^{\prime }+b^{\prime }{y}^{\prime }+c^{\prime })=0\lambda (a^{\prime }{x}^{\prime }+b^{\prime }{y}^{\prime }+c^{\prime })=-(a{x}^{\prime }+b{y}^{\prime }+c)\Rightarrow \lambda =-\frac{a{x}^{\prime }+b{y}^{\prime }+c}{a^{\prime }{x}^{\prime }+b^{\prime }{y}^{\prime }+c^{\prime }}$

substituting $\lambda$ in $\left(iv\right)$

$ax+by+c+(-\frac{a{x}^{\prime }+b{y}^{\prime }+c}{a^{\prime }{x}^{\prime }+b^{\prime }{y}^{\prime }+c^{\prime }})(a^{\prime }x+b^{\prime }y+c^{\prime })=0(a^{\prime }{x}^{\prime }+b^{\prime }{y}^{\prime }+c^{\prime })(ax+by+c)-(a{x}^{\prime }+b{y}^{\prime }+c)(a^{\prime }x+b^{\prime }y+c^{\prime })=0$