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.

Pair of Linear Equations in Two Variables

Graphical Solution of Linear Equations — Area of Triangle

Official Answer

Rewrite each line for plotting: Line 1, $2x-3y+6=0 \Rightarrow y=\frac{2x+6}{3}$ , giving points $(0,2)$ , $(3,4)$ , $(-3,0)$ . Line 2, $2x+3y-18=0 \Rightarrow y=\frac{18-2x}{3}$ , giving points $(0,6)$ , $(3,4)$ , $(9,0)$ . Line 3 is simply $x=0$ , the y-axis.

Plotting all three lines on the same graph with a suitable scale (e.g. 1 unit = 1 cm on each axis) shows that Line 1 and Line 2 both pass through $(3,4)$ , so they intersect there. Line 1 meets the y-axis ( $x=0$ ) at $(0,2)$ , and Line 2 meets the y-axis at $(0,6)$ .

Thus the triangle formed has vertices $A(0,2)$ , $B(0,6)$ and $C(3,4)$ . These can be verified algebraically: substituting $x=0$ into Line 1 gives $y=2$ (point $A$ ); substituting $x=0$ into Line 2 gives $y=6$ (point $B$ ); adding the two equations $(2x-3y+6)+(2x+3y-18)=0$ gives $4x-12=0$ , so $x=3$ , and back-substitution gives $y=4$ (point $C$ ).

Since $A(0,2)$ and $B(0,6)$ both lie on the y-axis, side AB is vertical with length $6-2 = 4$ units, which can be taken as the base. The height is the perpendicular distance from $C(3,4)$ to the y-axis, which is simply its x-coordinate, 3 units.

$\text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 3 = 6$ square units.

Hence, the vertices of the triangle are $A(0,2)$ , $B(0,6)$ , $C(3,4)$ , and its area is 6 square units.

graphical methodlinear equationspoint of intersectionvertices of trianglearea of triangley-axiselimination method

Marking Scheme

11 mark: correct table of at least 2-3 points for each of the two given lines ( $2x-3y+6=0$ and $2x+3y-18=0$ ).
21 mark: accurate graph/plot of both lines along with the line $x=0$ (the y-axis) on the same axes with proper scale.
31 mark: correctly reading/computing the three vertices of the triangle: $(0,2)$ , $(0,6)$ , and $(3,4)$ .
41 mark: correct method for area — either using base (on y-axis) = 4 units and height = 3 units, or the coordinate area formula.
51 mark: correct final area = 6 square units, with proper unit stated.

Hint

Plot each line using at least 3 points; the vertices are the pairwise intersections. Since $x=0$ is the y-axis, use the segment between the two y-axis vertices as the base and the horizontal distance to the third vertex as the height.

Quick Oral Answer

Plotting $2x-3y+6=0$ , $2x+3y-18=0$ , and $x=0$ gives vertices $(0,2)$ , $(0,6)$ and $(3,4)$ ; since one side lies on the y-axis with length 4, and the height to the opposite vertex is 3, the area is $\frac{1}{2}\times 4\times 3=6$ square units.

Analysis & Explanation

A graphical solution of three linear equations, combining plotting skills, simultaneous equations, and area calculation.

Concept

Each line is plotted using 2-3 convenient integer points; the third equation $x=0$ is simply the y-axis.
Vertices of the triangle are the three pairwise intersection points of the lines.

Key steps

Line1 $\cap (x=0)$ and Line2 $\cap (x=0)$ are read directly by substituting $x=0$ .
Line1 ∩ Line2 is found algebraically (adding the equations to eliminate y) and used to verify the graph reading.
Since two vertices lie on the y-axis, that side is vertical, letting $\text{Area} = \frac{1}{2}\times \text{base} \times \text{height}$ without the full coordinate formula.

Common mistakes

Choosing points too close together, making the plotted line inaccurate for reading intersections.
Missing that the vertical side lies on the y-axis and instead applying the full coordinate area formula unnecessarily.

Real-world

Graphical solution of simultaneous equations models finding break-even points or the intersection of cost/revenue lines in economics and business planning.

Common Mistakes

1Plotting the lines inaccurately or using too few points, leading to a wrong read-off of the intersection coordinates from the graph.
2Applying the general coordinate area formula incorrectly (sign errors in the determinant/shoelace expansion) instead of using the simpler base-height method available here since one side lies on the y-axis.
3Confusing which two lines form which vertex — e.g., mistakenly finding the intersection of Line 1 and Line 2 as a y-axis point instead of correctly solving $2x-3y+6=0$ and $2x+3y-18=0$ simultaneously.

Interesting Facts

Solving pairs of linear equations graphically was one of the earliest applications of coordinate geometry, formalised by René Descartes in the 17th century, merging algebra and geometry into a single framework.

This particular triangle is isosceles-like in structure since it has one vertical side along the y-axis of length 4 units, illustrating how choosing axis-aligned sides can simplify area calculations without needing the general Heron's or coordinate formula.

Graphical methods for solving equations, while less commonly used in modern computational tools, remain foundational for understanding concepts like consistency, inconsistency and dependency of linear systems — visually showing whether lines intersect, are parallel, or coincide.

Spotted a mistake or something unclear?

Tell us — we fix reported answers fast.

Frequently Asked Questions

Why is x=0 used as one of the three 'lines' forming the triangle?

$x=0$ is the equation of the y-axis itself. Using it as a boundary line means one side of the triangle lies exactly on the y-axis, which is a common CBSE technique to simplify area computation using the vertical side as a base.

Is there an algebraic (non-graphical) way to verify the vertices found from the graph?

Yes — solve each pair of equations simultaneously: e.g., adding $2x-3y+6=0$ and $2x+3y-18=0$ eliminates y directly, giving $4x-12=0$ , $x=3$ , $y=4$ , confirming vertex $(3,4)$ algebraically, matching the graphical reading.

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.