For points A(1, 7),B(2, 4),and C(k, 5),if ang | vedclass

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Question

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

Vedclass · Accepted Answer

$-5$

Vedclass · Answer

(D) Given that $\angle A = 90^\circ$,the line segments $AB$ 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 $AB$ is $m_{AB} = \frac{4 - 7}{2 - 1} = \frac{-3}{1} = -3$.The slope of $AC$ is $m_{AC} = \frac{5 - 7}{k - 1} = \frac{-2}{k - 1}$.Substituting these into the condition: $(-3) 	imes \left( \frac{-2}{k - 1} ight) = -1$.$\frac{6}{k - 1} = -1$.$6 = -(k - 1)$.$6 = -k + 1$.$k = 1 - 6 = -5$.

For points $A(1, 7)$,$B(2, 4)$,and $C(k, 5)$,if $\angle A = 90^\circ$,then find the value of $k$.

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