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Question 1180577: Find the constant m for which all three lines 2x - y - 4 = 0 , x + 2y + 3 = 0 and mx + y - 1 = 0 intersect at one point.
Note: Can you please show your full solution? Thank you!
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52782)   (Show Source):
You can put this solution on YOUR website! .
Find the constant m for which all three lines 2x - y - 4 = 0 , x + 2y + 3 = 0 and mx + y - 1 = 0 intersect at one point.
Note: Can you please show your full solution? Thank you!
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First, solve the system to find the coordinates of the intersection point.
2x - y = 4 (1)
x + 2y = -3 (2)
I will solve by the Substitution method.
From equation (1), express y = 2x - 4 and substitute this expression into equation (2)
x + 2*(2x-4) = -3
x + 4x - 8 = -3
5x = -3 + 8 = 5
x = 5/5 = 1
Then y = 2x-4 = 2*1 - 4 = 2 - 4 = -2.
Thus the solution to the system is (x,y) = (1,-2); so, the intersection point is (1,-2).
Ok, we just solved half of the problem.
Now our goal is to find "m" in the last equation.
For it, substitute the found coordinates of the intersection point x= 1, y= -2 into the equation
m*1 + (-2) -1 = 0,
which gives the ANSWER m = 3.
Solved, explained, and full solution is shown, step by step in all details.
Have a nice day (!)
Answer by MathLover1(20850)  (Show Source):
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