The vertices of the triangle ABC are A(-4,1), B(-4,-2) and C(2,5). Find
(a) the equations of the lines AB, BC, CA
(b) coordinates of the point where AC cuts the y-axis
(c) area of triangle ABC
(d) perpendicular distance from A to BC
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Answer:
(a) Line AB is a vertical line
Eqn of line AB: x = -4
Gradient of BC = (5 - (-2)) / (2 - (-4)) = 7/6
y - 5 = 7/6 (x - 2)
y = 7x/6 + 8/3
Gradient of CA = (5 - 1)/(2 - (-4)) = 2/3
y - 5 = 2/3 (x - 2)
y = 2x/3 + 11/3
(b) Where AC cuts x-axis => x = 0
Hence, y = 11/3
Coordinates of point where AC cuts x-axis = (0, 11/3)
(c) Using shoelace formula,
area of triangle =
= ½ [ (-4)(-2) + (-4)(5) + (2)(1) - (-4)(1) - (2)(-2) - (-4)(5) ]
= ½ [ 8 - 20 + 2 + 4 + 4 + 20 ]
= 9 units2
(d)
length BC = √(72 + 62) = √(85)
0.5 * length BC * perpendicular distance from A to BC = area of triangle ABC
0.5 * √(85) * perpendicular distance from A to BC = 9 (from part c)
Perpendicular distance from A to BC = 1.95 units
