Answer:
After solve the equations we get value of k=3
Step-by-step explanation:
We need to find value of k for which the lines x+2y=0, 3x-4y-10=0 and 5x+ky-7=0 are concurrent.
If the lines are concurrent, they pass through same point.
Let:
[tex]x+2y=0--eq(1)\\ 3x-4y-10=0--eq(2)\\ 5x+ky-7=0--eq(3)[/tex]
First solving equation 1 and 2 to find values of x and y
From eq(1) we find value of x and put it in eq(2)
[tex]From \ eq(1) x+2y=0\\x=-2y\\Put x=-2y \ in \ eq(2)\\3x-4y-10=0\\3(-2y)-4y-10=0 \\-6y-4y=10\\-10y=10\\y=\frac{10}{-10}\\y=-1[/tex]
After solving we get value of y=-1
Now putting in eq(1) to get value of x
[tex]x+2y=0\\x+2(-1)=0\\x-2=0\\x=2[/tex]
So, Value of x= 2
Now put value of x=2 and y=-1 into eq(3) to find value of k
[tex]5x+ky-7=0\\5(2)+k(-1)-7=0\\10-k-7=0\\-k+3=0\\-k=-3\\k=3[/tex]
So, After solve the equations we get value of k=3
