#### The points A (2, 9), B (a, 5) and C (5, 5) are the vertices of a triangle ABC right angled at B. Find the values of a and hence the area of $\triangle$ABC.

Solution.

$distance\, formula= \sqrt{\left ( x_{2}-x_{1} \right )^{2}+\left ( y_{2}-y_{1} \right )^{2}}$
$\triangle$ABC is a right angle triangle by using Pythagoras theorem we have
(AC)2 = (BC)2 + (AB)2
[(5 – 2)2 + (5 - 9)2] = [(2 – a)2+ (9 – 5)2] + [(a – 5)2 + (5 + 5)2]
(3)2 + (–4)2 = 4 + a2 – 4a + (4)2 + a2 + 25 – 10a
9 + 16 = 4 + a2 – 4a + 16 + a2 + 25 – 10a
25 = 2a2 – 14a + 45
2a2 – 14a + 45 – 25 = 0
2a2 – 14a + 20 = 0
Dividing by 2 we have
a2 – 7a + 10 = 0
a2 – 5a - 2a + 10 = 0
a(a – 5) – 2(a – 5) = 0
(a – 5) (a – 2) = 0
a = 5, a = 2
a = 5 is not possible because if a = 5 then point B and C coincide.
$\therefore$ a = 2
$Area \, of \, triangle = \frac{1}{2}\left [ x_{1} \left ( y_{2} -y_{3}\right )+x_{2} \left ( y_{3} -y_{1}\right )+x_{3} \left ( y_{1} -y_{2}\right )\right ]$
$= \frac{1}{2}\left [ 2\left ( 5-5 \right )+2\left ( 5-9 \right )+5\left ( 9-5 \right )\right ]$
$= \frac{1}{2}\left [ 0-8+20\right ]$
$= \frac{1}{2}\left [ 16\right ]$
= 8 sq. units.