Three points have coordinates A(6,6), B(-3,3) and C(9,k). The foot of the perpendicular from A to BC is the midpoint of BC. Calculate the possible values of k?

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Answer 1 · 2018-02-28T03:11:56

$color (green)(k = 15, -3)$

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Given D is the mid point of BC and AD is also perpendicular to BC.

Hence, $AD$ is perpendicular bisector of side $BC$ .

Or, $AC = AB$ . I.e., ABC is an isosceles triangle.

$vec(AC)^2 = (9-6)^2 +( k-6)^2=> 9 + (k-6)^2$

$vec(AB)^2 = (6+3)^2 + (6-3)^2 = 90$

Since $vec(AB) = vec (AC)$ ,

$(9 + (k-6)^2 )= 90$

$(k-6)^2 = 90 - 9 = 81$

$k - 6 = +-9$ or k = 15, -3#

Point C (9, 15) or (9, -3)#