7. Draw the graphs of the equations and . Determine the coordinates of the vertices of the triangle formed by these lines and the -axis, and shade the triangular region.

Understand the Question To solve graphically, express in terms of for both linear equations and find corresponding pairs of coordinates. Plot the lines on a Cartesian plane. The triangle is bounded by the line , the line , and the -axis (). Determine the three vertices of this triangle and shade the enclosed triangular region. Step 1 · Find Points for the First Line Given equation: Calculate for different values of : For : For : For : Step 2 · Find Points for the Second Line Given equation: Calculate for different values of : For : For : For : Step 3 · Plot the Lines and Determine the Vertices of the Triangle 1. Intersection of with the -axis (): 2. Intersection of with the -axis (): 3. Intersection of the two lines: Substitute into : Substitute into : Answer > The coordinates of the vertices of the triangle are , , and . Common Mistakes Wrong Axis Boundary: Confusing the -axis () with the -axis (), leading to incorrect base vertices such as or instead of and . Coordinate Order: Writing instead of , e.g., writing instead of .

Pair of Linear Equations in Two Variables | Exercise 3.1

Question 7

Draw the graphs of the equations $x - y + 1 = 0$ and $3x + 2y - 12 = 0$ . Determine the coordinates of the vertices of the triangle formed by these lines and the $x$ -axis, and shade the triangular region.

Solution

Understand the Question

To solve graphically, express $y$ in terms of $x$ for both linear equations and find corresponding pairs of $(x, y)$ coordinates.
Plot the lines on a Cartesian plane.
The triangle is bounded by the line $x - y + 1 = 0$ , the line $3x + 2y - 12 = 0$ , and the $x$ -axis ( $y = 0$ ).
Determine the three vertices of this triangle and shade the enclosed triangular region.

Step 1 · Find Points for the First Line

Given equation: $x - y + 1 = 0 \implies y = x + 1$

Calculate $y$ for different values of $x$ :

For $x = 0$ : $y = 0 + 1 = 1 \implies (0, 1)$
For $x = 1$ : $y = 1 + 1 = 2 \implies (1, 2)$
For $x = -1$ : $y = -1 + 1 = 0 \implies (-1, 0)$

Step 2 · Find Points for the Second Line

Given equation:

\begin{aligned} 3x + 2y - 12 &= 0 \\[0.6em] 2y &= 12 - 3x \\[0.6em] y &= \dfrac{12 - 3x}{2} \end{aligned}

Calculate $y$ for different values of $x$ :

For $x = 0$ : $y = \dfrac{12 - 3(0)}{2} = \dfrac{12}{2} = 6 \implies (0, 6)$
For $x = 2$ : $y = \dfrac{12 - 3(2)}{2} = \dfrac{12 - 6}{2} = 3 \implies (2, 3)$
For $x = 4$ : $y = \dfrac{12 - 3(4)}{2} = \dfrac{0}{2} = 0 \implies (4, 0)$

Step 3 · Plot the Lines and Determine the Vertices of the Triangle

Diagram 1

1. Intersection of $x - y + 1 = 0$ with the $x$ -axis ( $y = 0$ ):

\begin{aligned} x - 0 + 1 &= 0 \\ x &= -1 \implies (-1, 0) \end{aligned}

2. Intersection of $3x + 2y - 12 = 0$ with the $x$ -axis ( $y = 0$ ):

\begin{aligned} 3x + 2(0) - 12 &= 0 \\ 3x - 12 &= 0 \\ 3x &= 12 \\ x &= 4 \implies (4, 0) \end{aligned}

3. Intersection of the two lines: Substitute $y = x + 1$ into $3x + 2y - 12 = 0$ :

\begin{aligned} 3x + 2(x + 1) - 12 &= 0 \\ 3x + 2x + 2 - 12 &= 0 \\ 5x - 10 &= 0 \\ 5x &= 10 \\ x &= 2 \end{aligned}

Substitute $x = 2$ into $y = x + 1$ :

\begin{aligned} y &= 2 + 1 \\ y &= 3 \implies (2, 3) \end{aligned}

Answer

The coordinates of the vertices of the triangle are $(-1, 0)$ , $(4, 0)$ , and $(2, 3)$ .

Common Mistakes

Wrong Axis Boundary: Confusing the $x$ -axis ( $y = 0$ ) with the $y$ -axis ( $x = 0$ ), leading to incorrect base vertices such as $(0, 1)$ or $(0, 6)$ instead of $(-1, 0)$ and $(4, 0)$ .
Coordinate Order: Writing $(y, x)$ instead of $(x, y)$ , e.g., writing $(0, -1)$ instead of $(-1, 0)$ .

Question 7

More questions in Exercise 3.1