Question Statement

A surveyor is mapping out a triangular park where three straight walkways meet. The walkways are represented by the equations of lines: $x+y-4=0$ $2x-y-2=0$ $x-2y+2=0$

Find the coordinates of the points of intersection of the walkways.

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Background and Explanation

To find the intersection point of two lines, we solve their equations simultaneously using elimination or substitution methods. For three lines, we solve each pair of equations to determine if they form a triangle (three distinct vertices) or if they are concurrent (all meeting at a single point).

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Solution

The three walkways are represented by:

• Line (i): $x+y-4=0$
• Line (ii): $2x-y-2=0$
• Line (iii): $x-2y+2=0$

Finding Intersection of Lines (i) and (ii) — Point $A$

To find where the first two walkways meet, solve equations (i) and (ii) simultaneously:

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

Adding equations (i) and (ii) to eliminate $y$: $\begin{array}{rl}(x+y-4)+(2x-y-2)& =0\\ 3x-6& =0\\ 3x& =6\\ x& =2\end{array}$

Substitute $x=2$ into equation (i): $\begin{array}{rl}2+y-4& =0\\ y-2& =0\\ y& =2\end{array}$

Point $A$ is $(2,2)$.

Finding Intersection of Lines (i) and (iii) — Point $B$

To find where the first and third walkways meet, solve equations (i) and (iii):

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

Subtract equation (iii) from equation (i): $\begin{array}{rl}(x+y-4)-(x-2y+2)& =0\\ x+y-4-x+2y-2& =0\\ 3y-6& =0\\ 3y& =6\\ y& =2\end{array}$

Substitute $y=2$ into equation (i): $\begin{array}{rl}x+2-4& =0\\ x-2& =0\\ x& =2\end{array}$

Point $B$ is $(2,2)$.

Finding Intersection of Lines (ii) and (iii) — Point $C$

To find where the second and third walkways meet, solve equations (ii) and (iii):

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

Multiply equation (iii) by 2: $2x-4y+4=0$

Subtract this from equation (ii): $\begin{array}{rl}(2x-y-2)-(2x-4y+4)& =0\\ 2x-y-2-2x+4y-4& =0\\ 3y-6& =0\\ 3y& =6\\ y& =2\end{array}$

Substitute $y=2$ into equation (ii): $\begin{array}{rl}2x-2-2& =0\\ 2x-4& =0\\ 2x& =4\\ x& =2\end{array}$

Point $C$ is $(2,2)$.

Conclusion

All three pairs of lines intersect at the same point $(2,2)$. Therefore, the three walkways are concurrent (they all meet at a single point) rather than forming a triangular park with three distinct vertices.

The coordinates of the point of intersection are $(2,2)$.

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Key Formulas or Methods Used

• Solving simultaneous linear equations: Finding values of $x$ and $y$ that satisfy both equations simultaneously
• Elimination method: Adding or subtracting equations to eliminate one variable
• Substitution method: Solving for one variable and substituting back into the other equation
• Concurrency of lines: Three lines are concurrent if all pairs intersect at the same point

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Summary of Steps

1. Label the equations: Identify the three line equations as (i), (ii), and (iii)
2. Solve pair (i) and (ii): Use elimination by adding the equations to eliminate $y$, finding $x=2$, then substitute to find $y=2$
3. Solve pair (i) and (iii): Use elimination by subtracting (iii) from (i) to eliminate $x$, finding $y=2$, then substitute to find $x=2$
4. Solve pair (ii) and (iii): Multiply (iii) by 2, then subtract from (ii) to eliminate $x$, finding $y=2$, then substitute to find $x=2$
5. Verify concurrency: Confirm that all three intersection points are identical at $(2,2)$, indicating the lines meet at a single point