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Find the area of a triangles whose vertices are
(i) (6, 3), (-3, 5) and (4, - 2)
(ii) [(at12, at1),( at22, 2at2)( at32, 2at3)]
(iii) (a, c + a), (a, c) and (-a, c - a)
(i) Area of a triangle is given by
1/2[x1(y2 - y3) + x2(y3 - y1) + x3(y1 + y2)]
Here, x1 = 6,y1 = 3, x2 = -3, y2 = 5, x3 = 4,y3 = -2]
Let A(6, 3), B(-3, 5) and C(4,-2) be the given points
Area of ∆ABC = 1/2 [6(5+2)+(-3)(- 2 -3)+ 4(3 - 5)]
=1/2 [6 × 7- 3 × ( – 5) + 4( – 2)]
= 1/2[42 +15 - 8]
= 49/2 sq. units
(ii) Let A = (x1, y1) = (at12, 2at1)
B = (x2,y2) = (at22, 2at2)
= (x3, y3) = (at32, 2at2) be the given points.
The area of ∆ABC
(iii) Let A = (x1,y1) = (a, c + a)
B = (x2, y2) = (a, c)
C = (x3, y3) = (- a, c - a) be the given points
= 1/2[a ( – {c - a}) + a(c – a - (c + a)) +( - a)(c + a - a)]
= 1/2 [a(c – c + a) + a(c – a – c - a) - a(c + a - c)]
= 1/2[a × a + ax( - 2a) – a × a]
= 1/2[a2 - 2a2 - a2]
= 1/2×(-2a)2
= - a2
Find the area of the quadrilaterals, the coordinates of whose vertices are
(i) (-3, 2), (5, 4), (7, -6) and (-5,- 4)
(ii) (1, 2), (6, 2), (5, 3) and (3, 4)
(iii) (-4, -2), (-3, -5), (3, -2), (2, 3)
(i) Let A(-3, 2), B(5, 4), C(7,- 6) and D ( -5, – 4) be the given points.
Area of ∆ABC
= 1/2[-3(4 + 6) + 5(- 6 - 2) + 7(2 - 4)]
= 1/2[-3×1 + 5×(-8) + 7(-2)]
= 1/2[- 30 – 40 -14]
= - 42
But area cannot be negative
∴ Area of ∆ADC = 42 square units
Area of ∆ADC
= 1/2[-3( – 6 + 4) + 7(- 4 - 2) + (- 5)(2 + 6)]
= 1/2[- 3( – 2) + 7(- 6) – 5 × 8]
= 1/2[6 – 42 - 40]
= 1/2 × – 76
= - 38
∴ Area of ∆ADC = 38 square units
Now, area of quadrilateral ABCD
= Ar. of ABC+ Ar of ADC
= (42 + 38)
= 80 square. Units
(ii) Let A(1, 2) , B (6, 2) , C (5, 3) and (3, 4) be the given points
= 1/2[1(2 - 3) + 6(3 - 2) + 5(2 - 2)]
= 1/2[ -1 + 6 × (1) + 0]
= 1/2[ – 1 + 6]
= 5/2
= 1/2[1(3 - 4) + 5(4 - 2) + 3(2 - 3)]
= 1/2[-1 × 5 × 2 + 3(-1)]
= 1/2[-1 + 10 - 3]
= 1/2[6]
= 3
Now, Area of quadrilateral ABCD
= Area of ABC + Area of ADC
(iii) Let A (- 4, 2), B( – 3, – 5), C (3,- 2) and D(2, 3) be the given points
Area of ∆ABC = 1/2|(- 4)(- 5 + 2) - 3(-2 + 2) + 3(- 2 + 5)|
= 1/2|(-4)(-3) – 3(0) + 3(3)|
= 21/2
Area of ∆ACD = 1/2|( – 4)(3 + 2) + 2( – 2 + 2) + 3( – 2 – 3)|
= 1/2|- 4(5) + 2(0) + 3(- 5)|= (- 35)/2
But area can’t neative, hence area of ∆ADC = 35/2
Now, of quadrilateral (ABCD) = ar(∆ABC) + ar(∆ADC)
Area (quadrilateral ABCD) = 21/2 + 35/2
Area (quadrilateral ABCD) = 56/2
Area (quadrilateral ABCD) = 28 square. Units
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Chapter 14: Coordinate Geometry Exercise –...