Discussion Forum

If the  perpendicular distance  between  the point (1,1)  to the line 3x+4y+c=0 is 7, then the possible values of C are

Answer:

Explanation:

We lnow that distance from point $(x_{1},y_{1})$ to the line ax+by+c=0 is 

$|\frac{ax_{1}+by_{1}+c}{\sqrt{a^{2}+b^{2}}}|$

 $\therefore$   Distance from (1,1) to 3x+4y+c=0 is 7

  $\therefore$    $7=|\frac{3(1)+4(1)+c}{\sqrt{3^{2}+4^{2}}}|$

$\Rightarrow$     $7=|\frac{7+c}{5}|$

$\Rightarrow$     $35=|7+c|$

$\Rightarrow$     $7+c=\pm 35$

$\Rightarrow$       $c=-7 \pm 35$

                  c=28,-42