Question

The area of the triangle formed by the co-ordinate axes and a line \(px+qy = r\) (where p, q and r are positive real numbers) is \(\frac{1}{54}\) sq.units, then ....

• A. p, 3r, q are in G.P.
• B. \(p^2+q^2 = r^2\)
• C. p, \(3\sqrt{3}\) r, q are in G.P.
• D. p, 27 r, q are in G.P.

Hint:

The intercepts of the line are r/p and r/q, so the area is r squared over 2pq. Set it equal to 1/54.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
A line meeting the two axes at \((a, 0)\) and \((0, b)\) forms a right triangle with the axes. Its area is \(\tfrac{1}{2}ab\).

Step 2: Find the intercepts.
Put \(y = 0\): \(x = r/p\). Put \(x = 0\): \(y = r/q\). All of \(p, q, r\) are positive.

Step 3: Write the area.
\[ \frac{1}{2}\cdot\frac{r}{p}\cdot\frac{r}{q} = \frac{r^2}{2pq} = \frac{1}{54} \Rightarrow pq = 27 r^2 \]

Step 4: Test G.P. conditions.
If \(p, m, q\) are in G.P., then \(m^2 = pq = 27r^2\), so \(m = \sqrt{27}\, r = 3\sqrt{3}\, r\).

Step 5: Check the options.
(A) needs \(m = 3r\), which gives \(9r^2\), not \(27r^2\). (D) needs \(m = 27r\), which gives \(729r^2\). (B) \(p^2 + q^2 = r^2\) does not follow from the area. Only (C) with \(m = 3\sqrt{3}\,r\) matches.

Final Answer:
p, \(3\sqrt{3}\,r\), q are in G.P., option (C). \[ \boxed{p,\ 3\sqrt{3}\,r,\ q \text{ in G.P.}} \]