Question

If A(a, 0), B(1, 1) and C(0, b) form a triangle, right angled at B when joined, then establish a relation between a and b.

Hint:

Using slopes is often faster and less prone to expanding binomial mistakes:
\[ m_{AB} = \frac{1 - 0}{1 - a} = \frac{1}{1 - a} \] \[ m_{BC} = \frac{b - 1}{0 - 1} = 1 - b \] Since \(AB \perp BC\): \[ \frac{1 - b}{1 - a} = -1 \implies 1 - b = a - 1 \implies a + b = 2 \] Both methods yield the identical, clean result!

Answer 1

Correct Answer: 2

Step 1: Identify the vertices and the right angle.
We have \(A(a, 0)\), \(B(1, 1)\), \(C(0, b)\) with the right angle at \(B\). This means \(BA \perp BC\).
Step 2: Find the slopes of BA and BC.
Slope of BA: \(m_1 = \frac{0 - 1}{a - 1} = \frac{-1}{a-1}\). Slope of BC: \(m_2 = \frac{b - 1}{0 - 1} = 1 - b\).
Step 3: Apply the perpendicularity condition.
\(m_1 \times m_2 = -1\): \(\frac{-1}{a-1} \times (1-b) = -1\).
Step 4: Simplify.
\(\frac{-(1-b)}{a-1} = -1 \Rightarrow 1 - b = a - 1 \Rightarrow a + b = 2\).
Step 5: Alternative - using dot product of vectors.
\(\overrightarrow{BA} = (a-1, -1)\), \(\overrightarrow{BC} = (-1, b-1)\). Dot product \(= (a-1)(-1) + (-1)(b-1) = -a+1-b+1 = 2-a-b = 0\). So \(a + b = 2\).
Step 6: State the relation.
\[ \boxed{a + b = 2} \]