Math, asked by Anonymous, 1 year ago

without drawing the graph, show that lines y=5x-3,y=4-2x and 2x-3y=8 are concurrent.

Answers

Answered by perfect2003
3
hey mate

QUESTION : without drawing the graph, show that lines y=5x-3,y=4-2x and 2x-3y=8 are concurrent.

ANSWER : 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..

HOPE IT HELPS YOU

PLZ MARK AS BRAINLIST
Similar questions