Question

A straight line L passes through the point of intersection of the lines \(x-y+1 = 0\) and \(2x+y-7 = 0\). If L intersects the positive x-axis at \(A(a,0)\) and the positive y-axis at \(B(0,b)\), then the minimum area of the triangle \(OAB\) (where \(O\) is the origin) is ....

• A. \(6\) square units.
• B. \(12\) square units.
• C. \(24\) square units.
• D. \(48\) square units.

Hint:

Find the point of intersection, write the intercept form, and minimise ab with AM-GM.

Answer 1

The Correct Option is B

Step 1: Find the Fixed Point:
Adding \(x-y+1=0\) and \(2x+y-7=0\) gives \(3x-6=0\), so \(x=2\) and then \(y=3\). The line L passes through \((2,3)\).

Step 2: Write the Line:
With intercepts \(a\) and \(b\) on the positive axes, \(\dfrac{x}{a}+\dfrac{y}{b}=1\). Passing through \((2,3)\):
\[ \frac{2}{a}+\frac{3}{b}=1 \]

Step 3: Minimise the Area:
Area \(=\tfrac12ab\). By AM-GM,
\[ 1=\frac2a+\frac3b\geq 2\sqrt{\frac{6}{ab}} \Rightarrow \sqrt{ab}\geq 2\sqrt6 \Rightarrow ab\geq 24 \]
So the minimum area is \(\tfrac12(24)=12\) square units. Equality needs \(2/a=3/b=1/2\), i.e. \(a=4,\ b=6\).

Step 4: Check:
For \(a=4,\ b=6\): \(2/4+3/6=1\), valid, and area \(=12\). Options 6 is impossible because \(ab\geq24\) forces area at least 12.

Final Answer:
The minimum area is 12 square units, option (B). \[ \boxed{\text{(B) } 12\ \text{square units}} \]