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प्रश्न
Let P(11, 7), Q(13.5, 4) and R(9.5, 4) be the midpoints of the sides AB, BC and AC respectively of ∆ABC. Find the coordinates of the vertices A, B and C. Hence find the area of ∆ABC and compare this with area of ∆PQR.
उत्तर
Let the vertices of the ∆ABC be A(x1, y1), B(x2, y2), C(x3, y3)
Mid point of AB =
(11, 7) =
x1 + x2 = 22 ...(1) |
y1 + y2 = 14 ...(2) |
Mid point of BC =
⇒ (13.5, 4) =
x2 + x3 = 27 ...(3) |
y2 + y3 = 8 ...(4) |
Mid point of AC =
(9.5, 4) =
x1 + x3 = 19 ...(5) |
y1 + y3 = 8 ...(6) |
Add (1), (3) and (5)
2x1 + 2x2 + 2x3 = 22 + 27 + 19
2(x1 + x2 + x3) = 68
x1 + x2 + x3 = 34
From (1) ⇒ x1 + x2 = 22
x3 = 34 – 22 = 12
From (3) ⇒ x2 + x3 = 27
x1 = 34 – 27 = 7
From (5) ⇒ x1 + x3 = 19
x2 = 34 – 19 = 15
Add (2), (4) and (6)
2y1 + 2y2 + 2y3 = 14 + 8 + 8
2(y1 + y2 + y3) = 30
y1 + y2 + y3 = 15
From (2) ⇒ y1 + y2 = 14
y3 = 15 – 14 = 1
From (4) ⇒ y2 + y3 = 18
y1 = 15 – 8 = 7
From (6) ⇒ y1 + y3 = 8
y2 = 15 – 8 = 7
The vertices of a ΔABC are A(7, 7), B(15, 7) and C(12, 1)
Area of ΔABC =
=
=
=
= 24 sq. units
Area of ΔPRQ =
=
=
= 6 sq. units
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