Three lines are said to be concurrent if they all pass through a single common point (i.e., they intersect at one point). Rather than finding the intersection point of two lines and substituting into the third, we can use the determinant condition for concurrency, which provides a direct algebraic test.
Solution
For three lines given in the general form a1x+b1y+c1=0, a2x+b2y+c2=0, and a3x+b3y+c3=0, the condition for concurrency is:
a1a2a3b1b2b3c1c2c3=0
Step 1: Identify the coefficients from each line:
Line 1 (2x−y+3=0): a1=2,b1=−1,c1=3
Line 2 (x−y=0): a2=1,b2=−1,c2=0
Line 3 (3x+ay+1=0): a3=3,b3=a,c3=1
Step 2: Set up the determinant equation:
213−1−1a301=0
Step 3: Expand the determinant along the first row. Recall that expanding a 3×3 determinant involves multiplying each element by its corresponding minor (the 2×2 determinant formed by deleting the row and column of that element) with alternating signs:
2−1a01−(−1)1301+313−1a=0
Note: The signs follow the pattern +,−,+ for the first row, so the middle term becomes −(−1)=+1.
Step 4: Calculate each 2×2 determinant:
First minor: (−1)(1)−(0)(a)=−1−0=−1
Second minor: (1)(1)−(0)(3)=1−0=1
Third minor: (1)(a)−(−1)(3)=a+3
Step 5: Substitute these values back into the expansion:
2(−1)+1(1)+3(a+3)=0
Step 6: Simplify and solve for a:
−2+1+3a+9=0
3a+8=0
3a=−8
a=−38
Therefore, the value of a is −38.
Key Formulas or Methods Used
Condition for concurrency of three lines: Three lines a1x+b1y+c1=0, a2x+b2y+c2=0, and a3x+b3y+c3=0 are concurrent if and only if:
a1a2a3b1b2b3c1c2c3=0
Determinant expansion along first row: For a 3×3 matrix, det(A)=a11M11−a12M12+a13M13, where Mij is the minor of element aij
Determinant of 2×2 matrix:prqs=ps−qr
Summary of Steps
Identify coefficients from the three given lines: (2,−1,3), (1,−1,0), and (3,a,1)
Set up the determinant condition for concurrency equal to zero
Expand the determinant along the first row, calculating each 2×2 minor