The equation of the line which cuts off the i | vedclass

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

Question

(B) The intercept form of a line is given by $\frac{x}{a'} + \frac{y}{b'} = 1$,where $a'$ and $b'$ are the intercepts on the $x$ and $y$ axes respecti

Vedclass · Accepted Answer

$x \cos 	heta + y \sin 	heta - 2a = 0$

Vedclass · Answer

(B) The intercept form of a line is given by $\frac{x}{a'} + \frac{y}{b'} = 1$,where $a'$ and $b'$ are the intercepts on the $x$ and $y$ axes respectively.Given intercepts are $a' = 2a \sec 	heta$ and $b' = 2a \csc 	heta$.Substituting these into the intercept form:$\frac{x}{2a \sec 	heta} + \frac{y}{2a \csc 	heta} = 1$Since $\frac{1}{\sec 	heta} = \cos 	heta$ and $\frac{1}{\csc 	heta} = \sin 	heta$,we get:$\frac{x \cos 	heta}{2a} + \frac{y \sin 	heta}{2a} = 1$Multiplying both sides by $2a$:$x \cos 	heta + y \sin 	heta = 2a$Rearranging the terms,we get:$x \cos 	heta + y \sin 	heta - 2a = 0$.

The equation of the line which cuts off the intercepts $2a \sec \theta$ and $2a \csc \theta$ on the axes is

Similar Questions

$A$ line is perpendicular to the line $ax + by + c = 0$ and passes through $(a, b)$. The equation of the line is

Find the equation of the line passing through the point of intersection of the lines $4x + 7y - 3 = 0$ and $2x - 3y + 1 = 0$ that has equal intercepts on the axes.

The straight line $L$ is perpendicular to the line $5x - y = 1$. If the area of the triangle formed by the line $L$ and the coordinate axes is $5$,find the equation of the line $L$.

The line $(3x - y + 5) + \lambda (2x - 3y - 4) = 0$ will be parallel to the $y$-axis,if $\lambda$ =

Find the slope of the line passing through the points $(3, -2)$ and $(-1, 4)$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module