Question

2. Vertices of the triangles taken in order and their areas are given below. In each of the
following find the value of a.
Vertices Area (in sq. units)
(i) ( , 0 0), (4, a), (6, 4) 17
(ii) (a, a), (4, 5), (6,-1) 9
(iii) (a, -3), (3, a), (-1,5) 12

Answer 1

Hi ,

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The area of the triangle formed by the points

A( x1,y1) , B( x2,y2) , C(x3 , y3) is

1/2|x1 ( y2 - y1) + x2 (y3-y1) + x3 ( y1 - y2 )|

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Now ,

1) Let A ( 0,0), B( 4 , a ) , C( 6 , 4 ) are three
vertices of the triangle ,

area of ∆ = 17 sq units

1/2| 0( a - 4 ) + 4( 4 - 0 ) + 6( 0 - a ) | = 17

16 - 6a = 17 × 2

-6a = 34 - 16

-6a = 18
a = 18/(-6)
a = -3

2 ) Let A( a , a ) , B( 4 , 5 ) , C( 6 , -1 ) are three

vertices of the triangle ,

area of the triangle = 9
1/2 | a [ 5 -(-1 )] + 4 ( -1 - a ) + 6 ( a - 5 ) | = 9

6a - 4 - 4a + 6a - 30 = 18

8a - 34 = 18

8a = 42
a = 42/8
a = 21/4
3 ) Let A ( a , -3 ) , B( 3 , a ), C( -1 , 5 ) are three

vertices of the triangle ,

area of the triangle = 12 sq units

1/2| a ( a - 5 ) + 3[ 5 - ( -3 ) ] + ( -1 )( -3 - a ) | = 12

a² - 5a + 24 + 3 + a = 24
a²-4a + 27 - 24 = 0
a² - 4a + 3 = 0

a² - 3a - a + 3 = 0
a( a - 3 ) - ( a - 3 ) = 0
( a - 3 )( a - 1 ) = 0

Therefore ,

a - 3 = 0 or a - 1 = 0

a = 3 or a = 1
I hope this helps you.

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