The length of the altitude through A of the t | vedclass

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(D) In $\Delta ABC$,the vertices are $A \equiv (-3, 0)$,$B \equiv (4, -1)$,and $C \equiv (5, 2)$.First,we calculate the length of the base $BC$ using

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$\frac{22}{\sqrt{10}}$

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(D) In $\Delta ABC$,the vertices are $A \equiv (-3, 0)$,$B \equiv (4, -1)$,and $C \equiv (5, 2)$.First,we calculate the length of the base $BC$ using the distance formula:$BC = \sqrt{(5 - 4)^2 + (2 - (-1))^2} = \sqrt{1^2 + 3^2} = \sqrt{1 + 9} = \sqrt{10}$.Next,we calculate the area of $\Delta ABC$ using the coordinate formula:$	ext{Area} = \frac{1}{2} |x_A(y_B - y_C) + x_B(y_C - y_A) + x_C(y_A - y_B)|$$= \frac{1}{2} |(-3)(-1 - 2) + 4(2 - 0) + 5(0 - (-1))|$$= \frac{1}{2} |(-3)(-3) + 4(2) + 5(1)|$$= \frac{1}{2} |9 + 8 + 5| = \frac{1}{2} |22| = 11$.Since the area of a triangle is given by $\frac{1}{2} 	imes 	ext{base} 	imes 	ext{altitude}$,we have:$11 = \frac{1}{2} 	imes \sqrt{10} 	imes 	ext{altitude}$$	ext{altitude} = \frac{22}{\sqrt{10}}$.

The length of the altitude through $A$ of the triangle $ABC$,where $A \equiv (-3, 0)$,$B \equiv (4, -1)$,and $C \equiv (5, 2)$,is

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