Question

The area of a triangle formed by a line with the coordinate axes is \(49\) sq. units. If the perpendicular drawn from the origin to this line makes an angle of \(45^{\circ}\) with the positive X-axis, then the equation of line is....

• A. \(x+y = 7\)
• B. \(x+y = 7\sqrt{2}\)
• C. \(x+y = \sqrt{2}\)
• D. \(x+y = 2\)

Hint:

Use the normal form x cos a + y sin a = p with a = 45 degrees.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
If the perpendicular from the origin to a line has length \(p\) and makes angle \(\alpha\) with the positive x-axis, the line is \(x\cos\alpha + y\sin\alpha = p\).

Step 2: Set up:
With \(\alpha = 45^{\circ}\): \(\frac{x}{\sqrt2} + \frac{y}{\sqrt2} = p\), that is \(x + y = p\sqrt{2}\).
Both intercepts equal \(p\sqrt2\).

Step 3: Use the area:
\[ \text{Area} = \frac12 (p\sqrt2)(p\sqrt2) = p^2 = 49 \]
So \(p = 7\) and the line is \(x + y = 7\sqrt{2}\).

Step 4: Why the other options are wrong.
\(x + y = 7\) has intercepts 7 and area 24.5. \(x + y = \sqrt2\) and \(x + y = 2\) have areas 1 and 2, far from 49.

Final Answer:
The line is \(x + y = 7\sqrt2\), option (B). \[ \boxed{x+y=7\sqrt{2}} \]