CBSE Class 10 Answered
determine algebrically the vertices of the triangle formed by the lines 3x-y=3,2x-3y=2,x+2y=8
Asked by sanya wason | 04 Jul, 2011, 12:26: AM
Expert Answer
3x-y=3 -----(i)
2x-3y=2 -----(ii)
x+2y=8 -----(iii)
The vertices will be the points of intersection of the three equations, taking two at a time. Corresponding to the three pairs of equations, we get the three vertices.
Consider (i) and (ii) :
Multiplying (i) by (-3) and adding to (ii), we get
- 9x + 3y + 2x - 3y = -9 + 2
=> x = 1
(i) => y = 3x - 3 = 3.1 - 3 = 0
1st vertex is (1, 0)
Consider (ii) and (iii) :
Multiplying (iii) by (-2) and adding to (ii), we get,
2x - 3y - 2x - 4y = 2 - 16
=> y = 2
(iii) => x = 8 - 2y = 8 - 2.2 = 4
2nd vertex is ( 4, 2)
Consider (i) and (iii) :
Multiplying (i) by 2 and adding to (iii), we get
6x - 2y + x + 2y = 6 + 8
=> x = 2
(i) => y = 3x - 3 = 3.2 - 3 = 3
3rd vertex is ( 2,3)
Hence the three vertices are (1,0) , (4,2) & ( 2,3)
Answered by | 04 Jul, 2011, 11:07: AM
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