Straight Lines Really Are Straight

h

Let $\displaystyle \alpha$ be the angle between AB and BC. Then $\displaystyle \tan { \alpha } ={ m }_{ 0 }=\left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| ...(1.1)$

Where $\displaystyle { m }_{ 1 }$ and $\displaystyle { m }_{ 2 }$ are slope of BC and AC respectively.

On evaluating (1.1) we get the value of $\displaystyle { m }_{ 0 }=\cfrac { 19 }{ 8 }$ .

Since slope of the line AC is $\displaystyle \cfrac { { m }_{ 1 }-{ m }_{ o } }{ 1+{ m }_{ 1 }{ m }_{ o } } =-\cfrac { 52 }{ 89 }$ . Therefore its equation is $\displaystyle \left( y+7 \right) =-\cfrac { 52 }{ 89 } (x-2)$ , which on simplifications attains the form $\displaystyle 89y+52x+519=0$ , comparing this to the form $\displaystyle ax+by+c=0$ and adding the values of a, b, c; we get the answer as 660.

If you post a comment to this solution, I'll help you with any trouble that you are having understanding my approach. ;)