the vertice of the triangleABC are A(-4,1), B(-4,-2) and C(2,5). find
just answer question d. the earlier questions may well serve you as a reference .
(a) the equations of the lines AB, BC, CA
(b) coordinates of the point where AC cuts tge y0axis
(c) area of triangle ABC
(d) perpendicular distance from A to BC
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pls give the answers for the first few parts so that there's no necessity to do extra work. If possible, provide the answer for part d too. Thanks.
(a) AB : x = -4
BC : y = 7/6 x + 8/3
CA : y = 2/3x +11/3
b. (0, 11/3)
(c) 9 units 2
(d) 1.95 units
length BC = sqrt (7^2 + 6^2) = sqrt(85)
0.5 * length BC * perpendicular distance from A to BC = area of triangle ABC
0.5 * sqrt(85) * perpendicular distance from A to BC = 9 (from part c)
perpendicular distance from A to BC = 1.95 units