Question

Find the equations of the straight lines passing through (1, 1) and which are at a distance of 3 units from (-2, 3).

Answer 1

Suppose that the equation of line is

$x+by+c=0.......\left(1\right)$

The line passes through points $(1,1)$

Then,

$1+b+c=0$

$\Rightarrow c=-(b+1)$

The equation of the line is $x+by-(b-1)=0.$

The line is at a distance of $3$ units from

$(-2,3).$

Now,

$\left|\frac{-2+3b-(b+1)}{\sqrt{1+b^2}}\right|=3$

$\Rightarrow \left|\frac{2b-3}{\sqrt{1+b^2}=3}\right|$

On squaring both side and we get,

$\Rightarrow \frac{4b^2-12b+9}{1+b^2}=9$

$\Rightarrow 4b^2-12b+9=9+9b^2$

$\Rightarrow 5b^2+12b=0$

$\Rightarrow b(5b+12)=0$

$\Rightarrow b=0,5b+12=0$

$\Rightarrow b=0,b=-\frac{12}{5}$

Hence, the equation of ,line is $x+by-(b+1)=0$

If b=0 then,

Equation of line is $x-1=0$

If $b=-\frac{12}{5}$

Then equation of line is

$x-\left(\frac{12}{5}\right)+\frac{7}{5}=0$

$5x-12y+7=0$

Hence, this is the answer.