Find the distance of the line 4x - y = 0 from | vedclass

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(A) The given line is $4x - y = 0$ $(1)$.To find the distance of the line $(1)$ from the point $P(4, 1)$ along another line,we find the point of inter

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$3\sqrt{2} 	ext{ units}$

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(A) The given line is $4x - y = 0$ $(1)$.To find the distance of the line $(1)$ from the point $P(4, 1)$ along another line,we find the point of intersection of both lines.The slope of the second line is $m = 	an 135^{\circ} = -1$.The equation of the line with slope $-1$ passing through $P(4, 1)$ is:$y - 1 = -1(x - 4)$$y - 1 = -x + 4$$x + y - 5 = 0$ $(2)$Solving $(1)$ and $(2)$:From $(1)$,$y = 4x$. Substituting into $(2)$:$x + 4x - 5 = 0$ $\Rightarrow 5x = 5$ $\Rightarrow x = 1$.Then $y = 4(1) = 4$.The point of intersection is $Q(1, 4)$.The distance between $P(4, 1)$ and $Q(1, 4)$ is:$d = \sqrt{(1 - 4)^2 + (4 - 1)^2}$$d = \sqrt{(-3)^2 + (3)^2}$$d = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2} 	ext{ units}$.

Find the distance of the line $4x - y = 0$ from the point $P(4, 1)$ measured along the line making an angle of $135^{\circ}$ with the positive $x$-axis.

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