Question

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.

Hint:

To double-check your graphical intersection points, solve the equations algebraically.
Add the two equations:
\[ (2x - 3y + 6) + (2x + 3y - 18) = 0 \implies 4x - 12 = 0 \implies x = 3 \]
Substitute $x = 3$ in the first equation:
\[ 2(3) - 3y + 6 = 0 \implies 12 = 3y \implies y = 4 \]
This confirms that the intersection vertex is exactly $(3, 4)$.

Answer 1

Step 1: Understanding the Question:
The topic is the Graphical Method of solving a pair of linear equations in two variables.
We are given three linear equations which represent three straight lines.
When plotted on a Cartesian coordinate plane, these three lines intersect to form a triangle.
We need to determine the coordinates of the vertices of this triangle and calculate its area.

Step 2: Key Formula or Approach:

• Find at least two or three points for each line to plot them accurately.

• Identify the intersection points of the lines, which form the vertices $A$, $B$, and $C$.

• Use the standard area formula for a triangle:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} \]
by identifying a convenient base along one of the coordinate axes.


Step 3: Detailed Explanation:

• Find points for the first line: $2x - 3y + 6 = 0 \implies x = \frac{3y - 6}{2}$
If $y = 0$, then $x = -3 \implies (-3, 0)$
If $y = 2$, then $x = 0 \implies (0, 2)$
If $y = 4$, then $x = 3 \implies (3, 4)$

• Find points for the second line: $2x + 3y - 18 = 0 \implies x = \frac{18 - 3y}{2}$
If $y = 0$, then $x = 9 \implies (9, 0)$
If $y = 6$, then $x = 0 \implies (0, 6)$
If $y = 4$, then $x = 3 \implies (3, 4)$

• The third equation is $x = 0$, which is the $y$-axis.

• Plotting these lines on a graph sheet:
The first line crosses the $y$-axis at $B(0, 2)$.
The second line crosses the $y$-axis at $C(0, 6)$.
The two lines intersect each other at $A(3, 4)$.
Thus, the vertices of the triangle formed by these three lines are:
\[ A(3, 4), \quad B(0, 2), \quad C(0, 6) \]

• Calculate the Area of the triangle $ABC$:
Let the base of the triangle be segment $BC$ on the $y$-axis.
The length of base $BC$ is:
\[ \text{Base} = |y_C - y_B| = |6 - 2| = 4\text{ units} \]
The height ($h$) of the triangle is the perpendicular distance from vertex $A(3, 4)$ to the $y$-axis, which is equal to the absolute value of the $x$-coordinate of point $A$:
\[ \text{Height} = 3\text{ units} \]
Now, substitute these into the triangle area formula:
\[ \text{Area} = \frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 4 \times 3 = 6\text{ square units} \]


Step 4: Final Answer:
The coordinates of the vertices of the triangle are $A(3, 4)$, $B(0, 2)$, and $C(0, 6)$, and the area of the triangle is $6\text{ square units}$.