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(B) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$, where $p$ is the perpendicular distance from the origin and $\alpha$ is the angl

Vedclass · Accepted Answer

$x + y = 7\sqrt{2}$

Vedclass · Answer

(B) The normal form of a line is $x \cos \alpha + y \sin \alpha = p$, where $p$ is the perpendicular distance from the origin and $\alpha$ is the angle the perpendicular makes with the positive $X$-axis.Given $\alpha = 45^\circ$, the equation is $x \cos 45^\circ + y \sin 45^\circ = p$, which simplifies to $\frac{x}{\sqrt{2}} + \frac{y}{\sqrt{2}} = p$, or $x + y = p\sqrt{2}$.The intercepts on the axes are $x = p\sqrt{2}$ and $y = p\sqrt{2}$.The area of the triangle formed with the axes is $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} (p\sqrt{2})(p\sqrt{2}) = p^2$.Given area $= 49$, so $p^2 = 49$, which implies $p = 7$.Substituting $p = 7$ into the equation $x + y = p\sqrt{2}$, we get $x + y = 7\sqrt{2}$.

The area of a triangle formed by a line with the coordinate axes is $49 \text{ sq. units}$. 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 the line is:

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