Determine graphically, the coordinates of vertices of a triangle whose equations are $2x - 3y + 6 = 0$ ; $2x + 3y - 18 = 0$ and $x = 0$ . Also, find the area of this triangle.

Q: 32. Determine graphically, the coordinates of vertices of a triangle whose equations are $2x - 3y + 6 = 0$; $2x + 3y - 18 = 0$ and $x = 0$. Also, find the area of this triangle.

The three lines $2x - 3y + 6 = 0$, $2x + 3y - 18 = 0$ and $x = 0$ form a triangle with vertices $A(0, 2)$, $B(0, 6)$ and $C(3, 4)$, and its area is 6 square units. To plot the lines, points are found for each equation: for $2x - 3y + 6 = 0$, the points $(0, 2), (3, 4) \text{ and } (-3, 0)$ lie on the line; for $2x + 3y - 18 = 0$, the points $(0, 6), (3, 4) \text{ and } (9, 0)$ lie on the line. The third line $x = 0$ is simply the y-axis. Plotting all three lines on graph paper shows that line 1 meets the y-axis at $A(0, 2)$, line 2 meets the y-axis at $B(0, 6)$, and the two slanting lines intersect each other at $C(3, 4)$ — this can be verified algebraically by adding the two equations to get $4x - 12 = 0$, so $x = 3$, and substituting back gives $y = 4$. Since A and B both lie on the y-axis, side AB is vertical with length $|6-2| = 4\text{ units}$, taken as the base. The perpendicular (horizontal) distance from $C(3, 4)$ to the y-axis is simply its x-coordinate, 3 units, taken as the height. Hence $\text{Area} = \frac{1}{2}\times \text{base} \times \text{height} = \frac{1}{2}\times 4 \times 3 = 6\text{ square units}$.

Pair of Linear Equations in Two Variables

Graphical solution and area of triangle formed

Official Answer

The three lines $2x - 3y + 6 = 0$ , $2x + 3y - 18 = 0$ and $x = 0$ form a triangle with vertices $A(0, 2)$ , $B(0, 6)$ and $C(3, 4)$ , and its area is 6 square units.

To plot the lines, points are found for each equation: for $2x - 3y + 6 = 0$ , the points $(0, 2), (3, 4) \text{ and } (-3, 0)$ lie on the line; for $2x + 3y - 18 = 0$ , the points $(0, 6), (3, 4) \text{ and } (9, 0)$ lie on the line. The third line $x = 0$ is simply the y-axis. Plotting all three lines on graph paper shows that line 1 meets the y-axis at $A(0, 2)$ , line 2 meets the y-axis at $B(0, 6)$ , and the two slanting lines intersect each other at $C(3, 4)$ — this can be verified algebraically by adding the two equations to get $4x - 12 = 0$ , so $x = 3$ , and substituting back gives $y = 4$ .

Since A and B both lie on the y-axis, side AB is vertical with length $|6-2| = 4\text{ units}$ , taken as the base. The perpendicular (horizontal) distance from $C(3, 4)$ to the y-axis is simply its x-coordinate, 3 units, taken as the height. Hence $\text{Area} = \frac{1}{2}\times \text{base} \times \text{height} = \frac{1}{2}\times 4 \times 3 = 6\text{ square units}$ .

graphical methodlinear equationsvertices(0,2)(0,6)(3,4)area of triangle6 square unitsy-axis x=0base × height

Marking Scheme

11 mark: Correct table of at least two/three points for line $2x - 3y + 6 = 0$ .
21 mark: Correct table of at least two/three points for line $2x + 3y - 18 = 0$ .
31 mark: Drawing all three lines (including $x = 0$ , the y-axis) accurately to scale on graph paper.
41 mark: Reading off the correct vertices $A(0, 2)$ , $B(0, 6)$ , $C(3, 4)$ .
51 mark: Computing the $\text{area} = \frac{1}{2} \times \text{base}(4) \times \text{height}(3) = 6\text{ square units}$ with correct units. (Full marks for the equivalent determinant/coordinate-geometry area method.)

Hint

$x = 0$ is the y-axis. Find where each slanting line cuts the y-axis $(0,2)$ and $(0,6)$ , and where the two slanting lines meet $(3,4)$ . Base is along the y-axis (length 4), height is the x-coordinate of the third vertex (3).

Quick Oral Answer

The line $x = 0$ is the y-axis; the two slanting lines cut it at $(0,2)$ and $(0,6)$ and meet each other at $(3,4)$ . Base along the y-axis is 4 and the height is 3, so the area is half of 4 times 3, that is 6 square units.

Analysis & Explanation

A 5-mark question integrating graphical solving of linear equations with computing the area of the enclosed triangle.

Concept & strategy

Treat $x = 0$ (the y-axis) as one full side of the triangle — many students freeze because this equation 'looks different' from the other two.
With two vertices lying on the y-axis, the side between them is vertical; the third vertex's x-coordinate directly gives the height, avoiding the distance formula entirely.
Cross-check the area using the coordinate/determinant formula: $\frac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$ .

Common mistakes / exam caution

Forgetting that $x = 0$ is a valid line (the y-axis) and not plotting it.
Misreading the intersection point C from the graph, or giving only algebraic working without the required graph — since the question explicitly says 'determine graphically', graph marks are lost for algebra-only answers.

Real-world relevance

Graphical solving of simultaneous linear equations models real break-even and equilibrium analysis in economics, such as the intersection of supply-demand or cost-revenue lines.

Common Mistakes

1Solving the equations only algebraically and not drawing the graph — the question says 'Determine graphically', so the plotted graph is required for full marks.
2Treating $x = 0$ as a puzzling condition instead of recognising it as the y-axis, one of the three sides of the triangle.
3Using the wrong base or height for the area — the vertical side on the y-axis is the base (4 units) and the height is the x-coordinate of the opposite vertex (3 units), not a slant length.

Interesting Facts

The intersection point of two lines on a graph is exactly the algebraic solution of the pair of equations — here $(3, 4)$ satisfies both $2x - 3y + 6 = 0$ and $2x + 3y - 18 = 0$ .

When one side of a triangle lies along a coordinate axis, its area needs no distance formula: the height is simply the perpendicular coordinate of the opposite vertex.

Graphical solution of two straight lines is the mathematical heart of 'break-even analysis' in business and 'market equilibrium' in economics, where the crossing point of cost/revenue or supply/demand lines gives the answer.

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Frequently Asked Questions

What does the line $x = 0$ represent?

$x = 0$ is the equation of the y-axis. Every point on the y-axis has x-coordinate 0. So one full side of the triangle lies along the y-axis, which makes finding the base and area very easy.

How do I find the third vertex without a graph?

Solve the two slanting equations together. Adding $2x - 3y + 6 = 0$ and $2x + 3y - 18 = 0$ eliminates y: $4x - 12 = 0$ , so $x = 3$ ; substituting back gives $y = 4$ . Thus $C = (3, 4)$ . But remember the question demands the graph too, so you must also plot it.

Why is the area 6 square units?

Vertices $A(0,2)$ and $B(0,6)$ lie on the y-axis, so the base $AB = 6-2 = 4\text{ units}$ . The height is the horizontal distance of $C(3,4)$ from the y-axis, which is its x-coordinate, 3 units. $\text{Area} = \frac{1}{2}\times 4 \times 3 = 6\text{ square units}$ .

Determine graphically, the coordinates of vertices of a triangle whose equations are 2x−3y+6=02x - 3y + 6 = 02x−3y+6=0; 2x+3y−18=02x + 3y - 18 = 02x+3y−18=0 and x=0x = 0x=0. Also, find the area of this triangle.