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: Let the line be x + y = c:
The perpendicular from the origin makes $45^{\circ}$ with the x-axis, so the line has slope -1, which is $x + y = c$.

Step 2: Area from intercepts:
x-intercept and y-intercept are both $c$. Area $= \frac12 c^2 = 49$, so $c^2 = 98$, $c = 7\sqrt2$.

Step 3: Result:
$x + y = 7\sqrt2$. Distance from origin is $7\sqrt2/\sqrt2 = 7$, consistent with $p = 7$.

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