A line has slope m and y-intercept 4. The dis | vedclass

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(C) The equation of a line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$.Given $c = 4$,the equation of the line is $y = mx + 4$,which

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$\frac{4}{\sqrt{m^2 + 1}}$

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(C) The equation of a line with slope $m$ and $y$-intercept $c$ is given by $y = mx + c$.Given $c = 4$,the equation of the line is $y = mx + 4$,which can be rewritten as $mx - y + 4 = 0$.The distance $d$ from a point $(x_1, y_1)$ to a line $Ax + By + C = 0$ is given by $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$.Here,the point is the origin $(0, 0)$,so $x_1 = 0$ and $y_1 = 0$.Substituting these values into the distance formula:$d = \frac{|m(0) - (0) + 4|}{\sqrt{m^2 + (-1)^2}}$$d = \frac{|4|}{\sqrt{m^2 + 1}}$$d = \frac{4}{\sqrt{m^2 + 1}}$.

$A$ line has slope $m$ and $y$-intercept $4$. The distance between the origin and the line is equal to

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