Introduction to Linear Polynomials | EOT

Question 7

Draw the graph of the following equations, and identify their slopes and $y$ -intercepts. Also, find the coordinates of the points where these lines cut the $y$ -axis.

(i) $y = -3x + 4$

(ii) $2y = 4x + 7$

(iii) $5y = 6x - 10$

(iv) $3y = 6x - 11$

Are any of the lines parallel?

Solution

Understand the Question

The slope-intercept form of a linear equation is: $y = mx + c$ where $m$ is the slope, $c$ is the $y$ -intercept, and the line intersects the $y$ -axis at the point $(0, c)$ .
To find the slope and $y$ -intercept of any linear equation, rearrange it into the form $y = mx + c$ .
To graph a line, find at least two points $(x, y)$ that satisfy the equation and draw a straight line through them.
Two lines are parallel if they have the same slope ( $m_1 = m_2$ ) but different $y$ -intercepts ( $c_1 \neq c_2$ ).

(i) Draw the graph of $y = -3x + 4$ , identify its slope, $y$ -intercept, and the point where it cuts the $y$ -axis.

Step 1 · Identify Slope and y-Intercept

Given equation $y = -3x + 4$

Comparing with the slope-intercept form $y = mx + c$ :

$\text{Slope } (m) = -3$
$y\text{-intercept } (c) = 4$

The line intersects the $y$ -axis at $(0, c) = (0, 4)$ .

Step 2 · Find Coordinates to Plot the Graph

Calculate coordinates for plotting:

For $x = 0$ : $y = -3(0) + 4 = 4 \implies (0, 4)$
For $x = 1$ : $y = -3(1) + 4 = 1 \implies (1, 1)$
For $x = 2$ : $y = -3(2) + 4 = -2 \implies (2, -2)$

Answer

(i) $\text{Slope} = -3$ , $y\text{-intercept} = 4$ , Point on $y\text{-axis} = (0, 4)$

(ii) Draw the graph of $2y = 4x + 7$ , identify its slope, $y$ -intercept, and the point where it cuts the $y$ -axis.

Step 1 · Convert to Slope-Intercept Form

Given equation $2y = 4x + 7$

Dividing both sides by $2$ $y = \dfrac{4}{2}x + \dfrac{7}{2} = 2x + \dfrac{7}{2}$

Comparing with $y = mx + c$ :

$\text{Slope } (m) = 2$
$y\text{-intercept } (c) = \dfrac{7}{2}$

The line intersects the $y$ -axis at $\left(0, \dfrac{7}{2}\right)$ .

Step 2 · Find Coordinates to Plot the Graph

Calculate coordinates for plotting:

For $x = 0$ : $y = 2(0) + \dfrac{7}{2} = \dfrac{7}{2} \implies \left(0, \dfrac{7}{2}\right)$
For $x = 1$ : $y = 2(1) + 3.5 = 5.5 \implies (1, 5.5)$
For $x = -1$ : $y = 2(-1) + 3.5 = 1.5 \implies (-1, 1.5)$

Answer

(ii) $\text{Slope} = 2$ , $y\text{-intercept} = \dfrac{7}{2}$ , Point on $y\text{-axis} = \left(0, \dfrac{7}{2}\right)$

(iii) Draw the graph of $5y = 6x - 10$ , identify its slope, $y$ -intercept, and the point where it cuts the $y$ -axis.

Step 1 · Convert to Slope-Intercept Form

Given equation $5y = 6x - 10$

Dividing both sides by $5$ $y = \dfrac{6}{5}x - \dfrac{10}{5} = \dfrac{6}{5}x - 2$

Comparing with $y = mx + c$ :

$\text{Slope } (m) = \dfrac{6}{5}$
$y\text{-intercept } (c) = -2$

The line intersects the $y$ -axis at $(0, -2)$ .

Step 2 · Find Coordinates to Plot the Graph

Calculate coordinates for plotting:

For $x = 0$ : $y = \dfrac{6}{5}(0) - 2 = -2 \implies (0, -2)$
For $x = 5$ : $y = \dfrac{6}{5}(5) - 2 = 4 \implies (5, 4)$
For $x = -5$ : $y = \dfrac{6}{5}(-5) - 2 = -8 \implies (-5, -8)$

Answer

(iii) $\text{Slope} = \dfrac{6}{5}$ , $y\text{-intercept} = -2$ , Point on $y\text{-axis} = (0, -2)$

(iv) Draw the graph of $3y = 6x - 11$ , identify its slope, $y$ -intercept, and the point where it cuts the $y$ -axis.

Step 1 · Convert to Slope-Intercept Form

Given equation $3y = 6x - 11$

Dividing both sides by $3$ $y = \dfrac{6}{3}x - \dfrac{11}{3} = 2x - \dfrac{11}{3}$

Comparing with $y = mx + c$ :

$\text{Slope } (m) = 2$
$y\text{-intercept } (c) = -\dfrac{11}{3}$

The line intersects the $y$ -axis at $\left(0, -\dfrac{11}{3}\right)$ .

Step 2 · Find Coordinates to Plot the Graph

Calculate coordinates for plotting:

For $x = 0$ : $y = 2(0) - \dfrac{11}{3} = -\dfrac{11}{3} \implies \left(0, -\dfrac{11}{3}\right)$
For $x = 1$ : $y = 2(1) - \dfrac{11}{3} = -\dfrac{5}{3} \implies \left(1, -\dfrac{5}{3}\right)$
For $x = 4$ : $y = 2(4) - \dfrac{11}{3} = \dfrac{13}{3} \implies \left(4, \dfrac{13}{3}\right)$

Answer

(iv) $\text{Slope} = 2$ , $y\text{-intercept} = -\dfrac{11}{3}$ , Point on $y\text{-axis} = \left(0, -\dfrac{11}{3}\right)$

Are any of the lines parallel?

Step 1 · Compare Slopes of the Given Lines

Listing the slopes of all four lines:

Line (i): $m_1 = -3$
Line (ii): $m_2 = 2$
Line (iii): $m_3 = \dfrac{6}{5}$
Line (iv): $m_4 = 2$

Since $m_2 = m_4 = 2$ and their $y$ -intercepts are different $\left(\dfrac{7}{2} \neq -\dfrac{11}{3}\right)$ , lines (ii) and (iv) are parallel to each other.

Answer

Yes, lines (ii) and (iv) are parallel because both have a slope of $2$ .

Common Mistakes

Forgetting to Isolate $y$ : Reading the slope directly from the coefficient of $x$ before making the coefficient of $y$ equal to $1$ (e.g., mistaking the slope of $2y = 4x + 7$ as $4$ instead of $\dfrac{4}{2} = 2$ ).
Negative Signs: Forgetting to include the negative sign in the $y$ -intercept when the equation has a minus sign, such as $c = -2$ in $y = \dfrac{6}{5}x - 2$ .
Coordinate Order: Writing the point on the $y$ -axis as $(c, 0)$ instead of $(0, c)$ .

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