(5) If the three points (h, 0)
(a, b) and (0, k) lie on a straight line, then using the area of the triangle
formula show that (a/h) + (b/k) = 1,where h, k ≠ 0
Solution :
Let A(h, 0) B (a, b) and C (0, k) are the three points
Since the three points A (h, 0) B (a, b) and C (0, k) lie on a straight line we can say that the three points are collinear
So , area of triangle ABC = 0
(1/2) [ (h b + a k + 0) – (0 + 0 + kh) ] = 0
[ h b + a k – k h ] = 0 x 2
h b + a k - k h = 0
h b + a k = k h
Divided (k h) on both sides,
(h b)/(k h) + (a k)/(k h) = (k h)/(k h)
(b/k) + (a/h) = 1
(a/h) + (b/k) = 1
(6) Find the area of the triangle formed by joining the midpoints of the sides of a triangle whose vertices are (0, -1) (2, 1) and (0, 3). Find the ratio of this area to the area of the given triangle.
Solution :
Let A (0,-1) B (2,1) and C (0,3) are the vertices of the triangle. D,E and F are the midpoints of the side AB,BC and CA respectively.
Midpoint of AB = (x_{1 }+ x_{2})/2 , (y_{1 }+ y_{2})/2
= (0 + 2)/2 , (-1 + 1)/2
= 2/2 , 0/2
= D (1,0)
Midpoint of BC = (x_{1 }+ x_{2})/2 , (y_{1 }+ y_{2})/2
= (2 + 0)/2 , (1 + 3)/2
= 2/2 , 4/2
= E (1,2)
Midpoint of CA = (x1+x2)/2 , (y1+y2)/2
= (0 + 0)/2 , (3 + (-1))/2
= 0/2 , 2/2
= F (0,1)
= (1/2) [(0 + 6 + 0) – (-2 + 0 + 0)]
= (1/2) [6+2]
= (1/2) [8]
= 4 square units
= (1/2) [(2 + 1 + 0) – (0 + 0 + 1)]
= (1/2) [3-1]
= (1/2) [2]
= 1 square units
Area of triangle ABC: Area of triangle DEF
4 : 1
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