Question

There is a pair of points, one on each of the lines, whose combined equation is $(4x-3y+5)(6x+8y+5)=0$. If they are such that the distance of the point on one line is $2$ units from the other line then the points are

• A. $(\frac{1}{10},\frac{9}{5})(\frac{1}{2},-1)$
• B. $(\frac{1}{2},-1)(-\frac{23}{10},\frac{7}{3})$
• C. $(\frac{1}{10},\frac{9}{5})(-\frac{23}{10},-\frac{7}{5})$
• D. none of these

Answer 1

The correct option is A $(\frac{1}{10},\frac{9}{5})(\frac{1}{2},-1)$

The lines are $4x-3y+5=0$ and $6x+8y+5=0$
They are mutually perpendicular
Let $X=\frac{6x+8y+5}{\sqrt{6^2+8^2}}$ and $Y=\frac{4x-3y+5}{\sqrt{4^2+{(-3)}^2}}$
Therefore $X,Y$ coordinate the pair of points are
$(2,0),(0,2)$
$(-2,0),(0,2)$
$(2,0),(0,-2)$
$(-2,0),(0,-2)$
Now, For $X=2,Y=0$
$\Rightarrow 6x+8y+5=20,4x-3y+5=0\Rightarrow x=\frac{1}{10},y=\frac{9}{5}$
For $X=0,Y=2$
$\Rightarrow 6x+8y+5=0,4x-3y+5=10\Rightarrow x=\frac{1}{2},y=-1$
For $X=-2,Y=0$
$\Rightarrow 6x+8y+5=-20,4x-3y+5=0\Rightarrow x=-\frac{23}{10},y=-\frac{7}{3}$
And for $X=0,Y=-2$
$\Rightarrow 6x+8y+5=-20,4x-3y+5=-10\Rightarrow x=-\frac{27}{10},y=\frac{7}{5}$