In Delta ABC,if A(1, 7),B(2, 4),and C(k, 5) a | vedclass

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(D) Given that $\angle A = 90^{\circ}$,the sides $AB$ and $AC$ are perpendicular to each other.Therefore,the product of their slopes must be $-1$,i.e.

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$-5$

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(D) Given that $\angle A = 90^{\circ}$,the sides $AB$ and $AC$ are perpendicular to each other.Therefore,the product of their slopes must be $-1$,i.e.,$m_{AB} 	imes m_{AC} = -1$.The slope of a line passing through $(x_1, y_1)$ and $(x_2, y_2)$ is given by $m = \frac{y_2 - y_1}{x_2 - x_1}$.Slope of $AB$ $(m_{AB})$ = $\frac{4 - 7}{2 - 1} = \frac{-3}{1} = -3$.Slope of $AC$ $(m_{AC})$ = $\frac{5 - 7}{k - 1} = \frac{-2}{k - 1}$.Since $m_{AB} 	imes m_{AC} = -1$,we have $(-3) 	imes \left( \frac{-2}{k - 1} ight) = -1$.$\frac{6}{k - 1} = -1$.$6 = -(k - 1)$.$6 = -k + 1$.$k = 1 - 6 = -5$.

In $\Delta ABC$,if $A(1, 7)$,$B(2, 4)$,and $C(k, 5)$ and $\angle A$ is a right angle,then find the value of $k$.

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