If the equations y = mx + c and x cos alpha + | vedclass

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(B) The given equations are $y = mx + c$ (slope-intercept form) and $x \cos \alpha + y \sin \alpha = p$ (normal form).Rewriting the first equation as

Vedclass · Accepted Answer

$c = p \sqrt{1 + m^2}$

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(B) The given equations are $y = mx + c$ (slope-intercept form) and $x \cos \alpha + y \sin \alpha = p$ (normal form).Rewriting the first equation as $mx - y + c = 0$,we can compare it with $x \cos \alpha + y \sin \alpha - p = 0$.Since both represent the same line,the ratio of their coefficients must be equal:$\frac{\cos \alpha}{m} = \frac{\sin \alpha}{-1} = \frac{-p}{-c} = k$From this,$\cos \alpha = mk$ and $\sin \alpha = -k$.Using the identity $\cos^2 \alpha + \sin^2 \alpha = 1$,we get $(mk)^2 + (-k)^2 = 1$,which simplifies to $k^2(m^2 + 1) = 1$,so $k = \frac{1}{\sqrt{1 + m^2}}$.Also,from $\frac{-p}{-c} = k$,we have $\frac{p}{c} = k = \frac{1}{\sqrt{1 + m^2}}$.Therefore,$c = p \sqrt{1 + m^2}$.

If the equations $y = mx + c$ and $x \cos \alpha + y \sin \alpha = p$ represent the same straight line,then:

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