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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.
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उत्तर
Let the vertices of the ∆ABC be A(x1, y1), B(x2, y2), C(x3, y3)
Mid point of AB = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
(11, 7) = `((x_1 + x_2)/2, (y_1 + y_2)/2)`
|
`(x_1 + x_2)/2` = 11 x1 + x2 = 22 ...(1) |
`(y_1 + y_2)/2` = 7 y1 + y2 = 14 ...(2) |
Mid point of BC = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
⇒ (13.5, 4) = `((x_2 + x_3)/2, (y_2 + y_3)/2)`
|
`(x_2 + x_3)/2` = 13.5 x2 + x3 = 27 ...(3) |
`(y_2 + y_3)/2` = 4 y2 + y3 = 8 ...(4) |
Mid point of AC = `((x_1 + x_3)/2, (y_1 + y_3)/2)`
(9.5, 4) = `((x_1 + x_2)/2, (y_1 + y_3)/2)`
|
`(x_1 + x_3)/2` = 9.5 x1 + x3 = 19 ...(5) |
`(y_1 + y_3)/2` = 4 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 = `1/2[(x_1y_2 + x_2y_3 + x_3y_1) - (x_2y_1 + x_3y_2 + x_1y_3)]`
= `1/2[(7 + 84 + 105) - (84 + 15 + 49)]`
= `1/2[196 - 148]`
= `1/2 xx 48`
= 24 sq. units
Area of ΔPRQ = `1/2[(44 + 8 + 94.5) - (66.5 + 54 + 44)]`
= `1/2[176.5 - 164.5]`
= `1/2 xx 12`
= 6 sq. units
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