Question Statement

A civil engineer is tasked with designing a roundabout where three main roads converge. The equations of the roads are:

$2x+y-11.5=0$ $x-4y+1=0$ $3x-2y-12=0$

Find the coordinates of the point where the roundabout should be designed.

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Background and Explanation

When three or more lines intersect at a single common point, they are called concurrent lines. To verify concurrency, we can check if the determinant of the coefficient matrix equals zero. Alternatively, we can find the intersection point of any two lines and verify it lies on the third.

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Solution

Let the three equations of the roads be:

$\begin{array}{rl}2x+y-11.5& =0\text{(i)}\\ x-4y+1& =0\text{(ii)}\\ 3x-2y-12& =0\text{(iii)}\end{array}$

Checking for Concurrency

First, we verify that all three roads meet at a single point by calculating the determinant of the coefficient matrix:

$\left|\begin{array}{ccc}2& 1& -11.5\\ 1& -4& 1\\ 3& -2& -12\end{array}\right|$

Expanding along the first row:

$=2\left|\begin{array}{cc}-4& 1\\ -2& -12\end{array}\right|-1\left|\begin{array}{cc}1& 1\\ 3& -12\end{array}\right|-11.5\left|\begin{array}{cc}1& -4\\ 3& -2\end{array}\right|$

$=2(48+2)-1(-12-3)-11.5(-2+12)$

$=2(50)-1(-15)-11.5(10)$

$=100+15-115=0$

Since the determinant equals zero, the three lines are concurrent (they converge at the same point).

Finding the Point of Intersection

To find the coordinates of the roundabout, we solve any two of the three equations. Let's solve equations (ii) and (iii).

Multiply equation (ii) by 3: $3(x-4y+1)=0\Rightarrow 3x-12y+3=0$

Subtract equation (iii) from this result: $\begin{array}{r}3x-12y+3=0\\ -(3x-2y-12=0)\\ -10y+15=0\end{array}$

Solving for $y$: $-10y=-15\Rightarrow y=\frac{3}{2}=1.5$

Substitute $y=\frac{3}{2}$ into equation (ii): $x-4\left(\frac{3}{2}\right)+1=0$ $x-6+1=0$ $x-5=0$ $x=5$

Therefore, the point of convergence is $\left(5,\frac{3}{2}\right)$ or $(5,1.5)$.

Verification: Substituting $(5,1.5)$ into equation (i): $2(5)+1.5-11.5=10+1.5-11.5=0$ ✓

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Key Formulas or Methods Used

• Concurrency condition for three lines: Determinant $\left|\begin{array}{ccc}{a}_1& b_1& c_1\\ a_2& b_2& c_2\\ a_3& b_3& c_3\end{array}\right|=0$ for lines $a_{i}x+b_{i}y+c_i=0$
• Solving simultaneous linear equations: Elimination method (multiply and subtract equations)
• Point of intersection: The common solution $(x,y)$ satisfying all three line equations

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Summary of Steps

1. Identify the three line equations representing the roads
2. Check concurrency by calculating the $3\times 3$ determinant of coefficients; verify it equals zero
3. Select any two equations and solve simultaneously using the elimination method
4. Solve for one variable (here, $y=\frac{3}{2}$) by eliminating the other variable
5. Back-substitute to find the remaining coordinate ($x=5$)
6. State the final coordinates $\left(5,\frac{3}{2}\right)$ where the roundabout should be constructed