The vertices of a triangle are (2, 1),(5, 2) | vedclass

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(D) Let the vertices be $A(2, 1)$,$B(5, 2)$,and $C(4, 4)$.The equation of side $AB$ is $y - 1 = \frac{2 - 1}{5 - 2}(x - 2) \implies y - 1 = \frac{1}{3

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$\frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{10}}$

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(D) Let the vertices be $A(2, 1)$,$B(5, 2)$,and $C(4, 4)$.The equation of side $AB$ is $y - 1 = \frac{2 - 1}{5 - 2}(x - 2) \implies y - 1 = \frac{1}{3}(x - 2) \implies x - 3y + 1 = 0$.The length of the perpendicular from $C(4, 4)$ to $AB$ is $D_C = \frac{|4 - 3(4) + 1|}{\sqrt{1^2 + (-3)^2}} = \frac{|4 - 12 + 1|}{\sqrt{10}} = \frac{7}{\sqrt{10}}$.The equation of side $BC$ is $y - 2 = \frac{4 - 2}{4 - 5}(x - 5) \implies y - 2 = -2(x - 5) \implies 2x + y - 12 = 0$.The length of the perpendicular from $A(2, 1)$ to $BC$ is $D_A = \frac{|2(2) + 1 - 12|}{\sqrt{2^2 + 1^2}} = \frac{|4 + 1 - 12|}{\sqrt{5}} = \frac{7}{\sqrt{5}}$.The equation of side $AC$ is $y - 1 = \frac{4 - 1}{4 - 2}(x - 2) \implies y - 1 = \frac{3}{2}(x - 2) \implies 3x - 2y - 4 = 0$.The length of the perpendicular from $B(5, 2)$ to $AC$ is $D_B = \frac{|3(5) - 2(2) - 4|}{\sqrt{3^2 + (-2)^2}} = \frac{|15 - 4 - 4|}{\sqrt{13}} = \frac{7}{\sqrt{13}}$.Thus,the lengths are $\frac{7}{\sqrt{5}}, \frac{7}{\sqrt{13}}, \frac{7}{\sqrt{10}}$.

The vertices of a triangle are $(2, 1)$,$(5, 2)$ and $(4, 4)$. The lengths of the perpendiculars from these vertices to the opposite sides are

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