Question

The points (3, 3), (h, 0) and (0, k) are collinear if

• A. $\frac{1}{h}+\frac{1}{k}=\frac{1}{3}$
• B. $\frac{1}{h}-\frac{1}{k}=\frac{1}{3}$
• C. $\frac{1}{k}-\frac{1}{h}=\frac{1}{3}$
• D. $\frac{1}{k}-\frac{1}{h}=\frac{1}{7}$

Answer 1

The correct option is A

$\frac{1}{h}+\frac{1}{k}=\frac{1}{3}$

Let A ≡ (3, 3), B ≡ (h, 0) and C ≡ (0, k). The points A, B, C will be collinear if area of $\Delta$ABC = 0

$\Rightarrow \frac{1}{2}$ [3 (0 – k) + h(k – 3) + 0(3 – 0)] = 0

$\Rightarrow \frac{1}{2}$ (–3k + hk – 3h) = 0 or 3k + 3h = hk

$or\frac{3}{h}+\frac{3}{k}=1or\frac{1}{h}+\frac{1}{k}=\frac{1}{3}$