The area of a triangle formed by a line with | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(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:

Similar Questions

The equation of the line passing through the points $(1, 2)$ and $(2, 5)$ is:

If the line $y = mx + c$ passes through the points $(2, 4)$ and $(3, -5)$,then:

The equation of the line passing through $(4, -6)$ and making an angle of $45^\circ$ with the positive $x$-axis is:

If the line $\frac{x}{a} + \frac{y}{b} = 1$ passes through the points $(2, -3)$ and $(4, -5)$,then $(a, b) = $

$A$ line passes through $(-1, -3)$ and is perpendicular to $x + 6y = 5$. Its $x$-intercept is

Vedclass Test Series

Exam Paper Generator

Online Exam Module