The y-intercept made by the line x cos alpha | vedclass

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(B) The given equation of the line is $x \cos \alpha + y \sin \alpha = a$. Dividing both sides by $a$,we get: $\frac{x \cos \alpha}{a} + \frac{y \sin

Vedclass · Accepted Answer

$a \csc \alpha$

Vedclass · Answer

(B) The given equation of the line is $x \cos \alpha + y \sin \alpha = a$. Dividing both sides by $a$,we get: $\frac{x \cos \alpha}{a} + \frac{y \sin \alpha}{a} = 1$ This can be rewritten in the intercept form $\frac{x}{A} + \frac{y}{B} = 1$,where $B$ is the $y$-intercept: $\frac{x}{a \sec \alpha} + \frac{y}{a \csc \alpha} = 1$ Comparing this with the standard intercept form,the $y$-intercept is $a \csc \alpha$.

The $y$-intercept made by the line $x \cos \alpha + y \sin \alpha = a$ is:

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