I want to print the lines that with two columns equal to a variable, for example, input:
2607s1  NC_000067.6 95.92   49  1   1   3   50  1e-14   84.2
2607s1  NC_000067.6 97.73   44  1   0   7   50  4e-14   84.2
2607s1  NC_000067.6 97.67   43  1   0   8   50  1e-13   75.0
and variables for first and last column:
a="2607s1"; b="84.2"
and using awk command, output:
2607s1  NC_000067.6 95.92   49  1   1   3   50  1e-14   84.2
2607s1  NC_000067.6 97.73   44  1   0   7   50  4e-14   84.2
I have tried the following but not work:
awk -v '$1==$a' && '$10==$b' test_file
cat test_file|awk '$1=="($a)" && $10=="($b)"'
cat test_file|awk '$1==($a) && $10==($b)'
cat test_file|awk '$1=="$a" && $10=="$b"'
Moreover, I am running it in a while loop, so the $a and $b keep changing Please help..
You are passing the shell variables to the awk command using a wrong method. It should be like
awk -v a="$a" -v b="$b" '$1==a && $10 == b' 
What it does
-v a="$a" creates an awk variable a and assigns the value of shell variable $a to it.
-v b="$b" Creates awk variable b.
OR
awk '$1==a && $10 == b' a="$a" b="$b" file
When you write a statement like this
awk -v '$1==$a' && '$10==$b' test_file
awk doesn't know what $a $b is because both are shell variables. 
And the correct method of using -v for passing shell variables is as in demonstrated in the examples.
From awk manuals
-v var=val --assign var=val Assign the value val to the variable var, before execution of the program begins. Such variable values are available to the BEGIN block of an AWK program.
$ cat file 
2607s1  NC_000067.6 95.92   49  1   1   3   50  1e-14   84.2
2607s1  NC_000067.6 97.73   44  1   0   7   50  4e-14   84.2
2607s1  NC_000067.6 97.67   43  1   0   8   50  1e-13   75.0
$ a="2607s1"; b="84.2"
$ awk -v a="$a" -v b="$b" '$1==a && $10 == b' file 
2607s1  NC_000067.6 95.92   49  1   1   3   50  1e-14   84.2
2607s1  NC_000067.6 97.73   44  1   0   7   50  4e-14   84.2
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