without drawing the graphs show that the following equation are of concurrent lines y =5x-3 ; y= 4 - 2x and 2x + 3y =8
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Lines are concurrent only if all meet at a point...i.e all have one common point...
On solving first and second equation we get
5x - 3 = 4 - 2x
7x = 7
x = 1
Putting x= 1 in first equation of line we get
y = 5 X 1 - 3 = 5 - 3 = 2
Here we get a point of intersection of first two lines as ( 1, 2 )...If there lines are concurrent then x=1 and y=2 must satisfy the third equation of line...
putting x = 1 and y = 2 on LHS side of third line we get
2 X 1 + 3 X 2 = 8 = RHS
Thus lines are concurrent..
On solving first and second equation we get
5x - 3 = 4 - 2x
7x = 7
x = 1
Putting x= 1 in first equation of line we get
y = 5 X 1 - 3 = 5 - 3 = 2
Here we get a point of intersection of first two lines as ( 1, 2 )...If there lines are concurrent then x=1 and y=2 must satisfy the third equation of line...
putting x = 1 and y = 2 on LHS side of third line we get
2 X 1 + 3 X 2 = 8 = RHS
Thus lines are concurrent..
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Answer:
here it is likewise eee
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