The vertices of triangle ABC are A(-1,2), B (1,5) and C(4,3).
(a) Find the lengths of the sides AB, BC and CA
(b) What type of triangle is ABC.
(c) Find the perpendicular distance form B to AC.
(d) Find the coordinates of the point at which the line AC cuts the x-axis
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Answer:
(a) Length of side AB = √[(1 - (-1))2 + (5 - 2)2] = √13
Length of side BC = √[(4 - 1)2 + (3 - 5)2] = √13
Length of side CA = √[(4 - (-1))2 + (3 - 2)2] = √26
(b) 2 sides are equal
Therefore, triangle ABC is an isosceles triangle.
(c) Let the perpendicular distance from B to AC be h
h2 + (0.5 √26)2 = (√13)2
h2 = 13 - 6.5
h2 = 6.5
h = 2.55 units
(d) Gradient of AC = (3 - 2) / (4 - (-1)) = 1/5
Eqn of AC: y - 3 = 1/5 (x - 4)
y = x/5 + 11/5
Cut the x-axis:
When y = 0, x = -11
Hence, the coordinates of the point at which the line AC cuts the x-axis is (-11, 0)
