Question Statement

Three jet fighters are flying straight in different directions along the lines: $2x-y-14=0$ $x+y-1=0$ $3x+2y-7=0$

Check whether the jet fighters will pass through a single point. If yes, find the point.

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Background and Explanation

Three straight lines are concurrent if they all pass through a single common point. For lines in the form $ax+by+c=0$, we can check concurrency using the determinant condition: the determinant of the coefficient matrix must equal zero. Alternatively, we can find the intersection point of two lines and verify it lies on the third.

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Solution

The three given lines representing the flight paths are:

$\begin{array}{rl}2x-y-14& =0\text{(i)}\\ x+y-1& =0\text{(ii)}\\ 3x+2y-7& =0\text{(iii)}\end{array}$

Step 1: Check for concurrency using the determinant condition

The jet fighters will pass through the same point if the three lines are concurrent. For three lines $a_{1}x+b_{1}y+c_1=0$, $a_{2}x+b_{2}y+c_2=0$, and $a_{3}x+b_{3}y+c_3=0$, the condition for concurrency is:

$\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$

Setting up the determinant with our coefficients:

$\left|\begin{array}{ccc}2& -1& -14\\ 1& 1& -1\\ 3& 2& -7\end{array}\right|$

Step 2: Evaluate the determinant

Expanding along the first row:

$=2\left|\begin{array}{cc}1& -1\\ 2& -7\end{array}\right|-(-1)\left|\begin{array}{cc}1& -1\\ 3& -7\end{array}\right|+(-14)\left|\begin{array}{cc}1& 1\\ 3& 2\end{array}\right|$

$=2(1\cdot (-7)-(-1)\cdot 2)+1(1\cdot (-7)-(-1)\cdot 3)-14(1\cdot 2-1\cdot 3)$

$=2(-7+2)+1(-7+3)-14(2-3)$

$=2(-5)+1(-4)-14(-1)$

$=-10-4+14$

$=0$

Since the determinant equals 0, the three lines are concurrent. Thus, the jet fighters will pass through the same point.

Step 3: Find the point of concurrency

To find the specific point, we solve equations (i) and (ii) simultaneously:

Adding equations (i) and (ii): $\begin{array}{rl}(2x-y-14)+(x+y-1)& =0\\ 3x-15& =0\\ 3x& =15\\ x& =5\end{array}$

Substitute $x=5$ into equation (ii): $\begin{array}{rl}5+y-1& =0\\ 4+y& =0\\ y& =-4\end{array}$

Therefore, the point of concurrency is $(5,-4)$.

(Verification: Substituting $(5,-4)$ into equation (iii): $3(5)+2(-4)-7=15-8-7=0$ ✓)

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Key Formulas or Methods Used

• Concurrency condition for three lines: $\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$
• Determinant expansion (Laplace expansion along first row)
• Solving simultaneous linear equations by elimination method

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Summary of Steps

1. Identify the line equations in standard form $ax+by+c=0$
2. Set up the $3\times 3$ determinant using coefficients $(a,b,c)$ from each line
3. Expand and evaluate the determinant:
   • If value $=0$: lines are concurrent (intersect at one point)
   • If value $\ne 0$: lines are not concurrent
4. Solve any two equations simultaneously to find the intersection coordinates $(x,y)$
5. State the final answer with the point of concurrency (if it exists)