A straight line passes through the points (5, | vedclass

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(B) The equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$.Substituting the

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$\sqrt{\frac{17}{2}}$

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(B) The equation of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $\frac{y-y_1}{x-x_1} = \frac{y_2-y_1}{x_2-x_1}$.Substituting the points $(5,0)$ and $(0,3)$,we get $\frac{y-0}{x-5} = \frac{3-0}{0-5} = -\frac{3}{5}$.This simplifies to $5y = -3(x-5)$,which is $3x + 5y - 15 = 0$.The length of the perpendicular from a point $(x_0, y_0)$ to the line $ax + by + c = 0$ is given by $d = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}$.Substituting $(x_0, y_0) = (4,4)$ and the line equation $3x + 5y - 15 = 0$:$d = \frac{|3(4) + 5(4) - 15|}{\sqrt{3^2 + 5^2}} = \frac{|12 + 20 - 15|}{\sqrt{9 + 25}} = \frac{17}{\sqrt{34}}$.Rationalizing the denominator,we get $d = \frac{17\sqrt{34}}{34} = \frac{\sqrt{34}}{2} = \sqrt{\frac{34}{4}} = \sqrt{\frac{17}{2}}$.

$A$ straight line passes through the points $(5,0)$ and $(0,3)$. The length of the perpendicular from the point $(4,4)$ to the line is:

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