Question -
Answer -
Solution:
3x тАУ y = 2 тАж(i)
2x -3y = 2 тАж(ii)
x + 2y = 8 тАж(iii)
Let the equations ofthe line (i), (ii) and (iii) represent the side of a тИЖABC.
On solving (i) and(ii),
We get,
[First,multiply Eq. (i) by 3 in Eq. (i) and then subtract]
(9x-3y)-(2x-3y) = 9-2
7x = 7
x = 1
Substituting x=1 inEq. (i), we get
3├Ч1-y = 3
y = 0
So, the coordinate ofpoint B is (1, 0)
On solving lines (ii)and (iii),
We get,
[First,multiply Eq. (iii) by 2 and then subtract]
(2x + 4y)-(2x-3y) =16-2
7y = 14
y = 2
Substituting y=2 inEq. (iii), we get
x + 2 ├Ч 2 = 8
x + 4 = 8
x = 4
Hence, the coordinateof point C is (4, 2).
On solving lines (iii)and (i),
We get,
[First,multiply in Eq. (i) by 2 and then add]
(6x-2y) + (x + 2y) = 6+ 8
7x = 14
x = 2
Substituting x=2 inEq. (i), we get
3 ├Ч 2 тАУ y = 3
y = 3
So, the coordinate ofpoint A is (2, 3).
Hence, the vertices ofthe тИЖABC formed by the given lines are as follows,
A (2, 3), B (1, 0) andC (4, 2).