Question

The straight line through the point of intersection of $ax+by+c=0$ and $a^{\prime }x+b^{\prime }y+c^{\prime }=0$ and parallel to the y axis has the equation?

• A. $x(a{b}^{\prime }-a^{\prime }b)+(c{b}^{\prime }-c^{\prime }b)=0$
• B. $x(a{b}^{\prime }+a^{\prime }b)+(c{b}^{\prime }+c^{\prime }b)$
• C. $y(a^{\prime }b-a{b}^{\prime })+(a^{\prime }c-a{c}^{\prime })=0$
• D. None of the above

Answer 1

The correct option is A $x(a{b}^{\prime }-a^{\prime }b)+(c{b}^{\prime }-c^{\prime }b)=0$

$Given,ax+by+c=0\to \left(i\right)a^{\prime }x+b^{\prime }y+c^{\prime }=0\to \left(ii\right)tofindtheequationparalleltoy-axisandpassingthroughthepointofintersectionofbothequationsavaluatetheyterm.fromequation\left(i\right)y=-\frac{\left(c+gx\right)}{b}onputtingvalueofyinequation\left(ii\right)weget{a}^{\prime }x+b^{\prime }\left[-\frac{\left(c-gx\right)}{b}\right]+c^{\prime }=0b{a}^{\prime }x-b^{\prime }ax-b^{\prime }c+b{c}^{\prime }=0x\left(a^{\prime }b+b^{\prime }a\right)+\left(b{c}^{\prime }-b^{\prime }c\right)=0itcanalsobewrittenas\left[x\left(a{b}^{\prime }-a^{\prime }b\right)+b^{\prime }c-b{c}^{\prime }=0\right]Option\left(A\right)isright$