The points \(A(2,9),\ B(a,5),\ C(5,5)\) are vertices of a triangle right–angled at \(B\). Find \(a\) and the area of \(\triangle ABC\).
\(a=2\), area \(=6\).
Step 1: Since the triangle is right-angled at \(B\), by Pythagoras theorem:
\(AB^2 + BC^2 = AC^2\).
Step 2: Find \(AB^2\), \(BC^2\), and \(AC^2\) using squared distance (no square roots needed).
Step 3: Compute \(AB^2\).
\(A(2,9),\ B(a,5)\)
\(AB^2 = (2-a)^2 + (9-5)^2 = (2-a)^2 + 16\)
Step 4: Compute \(BC^2\).
\(B(a,5),\ C(5,5)\)
\(BC^2 = (5-a)^2 + (5-5)^2 = (5-a)^2\)
Step 5: Compute \(AC^2\).
\(A(2,9),\ C(5,5)\)
\(AC^2 = (5-2)^2 + (5-9)^2 = 3^2 + (-4)^2 = 9 + 16 = 25\)
Step 6: Apply Pythagoras theorem:
\((2-a)^2 + 16 + (5-a)^2 = 25\)
\((2-a)^2 + (5-a)^2 = 9\)
Step 7: Expand and simplify.
\((2-a)^2 = a^2 - 4a + 4\)
\((5-a)^2 = a^2 - 10a + 25\)
So,
\((a^2 - 4a + 4) + (a^2 - 10a + 25) = 9\)
\(2a^2 - 14a + 29 = 9\)
\(2a^2 - 14a + 20 = 0\)
Divide by 2:
\(a^2 - 7a + 10 = 0\)
\((a-2)(a-5)=0\)
So, \(a=2\) or \(a=5\).
Step 8: Reject \(a=5\) because then \(B(5,5)=C(5,5)\) and no triangle is formed.
Hence, \(a=2\).
Step 9: Find the area of \(\triangle ABC\) (right-angled at \(B\)).
Now \(B=(2,5)\).
\(AB = \sqrt{(2-2)^2 + (9-5)^2} = \sqrt{16} = 4\)
\(BC = \sqrt{(5-2)^2 + (5-5)^2} = \sqrt{9} = 3\)
Area \(= \dfrac{1}{2}\times AB \times BC = \dfrac{1}{2}\times 4 \times 3 = 6\).
Final Answer: \(a=2\) and \(\text{Area}(\triangle ABC)=6\) square units.