Question

determine algebrically the vertices of triqngle formed by lines 3x-y=3,2x-3y=2and x+2y=8
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Answer 1

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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)

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