Let's calculate the slope of both lines and then compare them.

The first line passes through points

(0, - 4)

and

(2, 2) .

Let's substitute these into the Slope Formula.

m_{1} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Substitute $(0, - 4)$ & $(2, 2)$

m_{1} = \frac{2 - ( - 4 )}{2 - 0}

▼

Simplify right-hand side

$a - (- b) = a + b$

m_{1} = \frac{2 + 4}{2 - 0}

Add and subtract terms

m_{1} = \frac{6}{2}

Calculate quotient

m_{1} = 3

The slope of the first line is

3 .

Similarly, let's substitute the points that lie on this line,

(0, - 4)

and

(4, 5),

into the Slope Formula.

m_{2} = \frac{y _{2} - y _{1}}{x _{2} - x _{1}}

Substitute $(0, - 4)$ & $(4, 5)$

m_{2} = \frac{5 - ( - 4 )}{4 - 0}

▼

Simplify right-hand side

$a - (- b) = a + b$

m_{2} = \frac{5 + 4}{4 - 0}

Add and subtract terms

m_{2} = \frac{9}{4}

Write fraction as a mixed number

m_{2} = 2 \frac{1}{4}

The second line has a slope of

2 \frac{1}{4} .

To find which line is steeper, we need to compare the absolute values of their slopes.

m_{1} = 3 \Rightarrow ∣ m_{1} ∣ = 3 m_{2} = 2 \frac{1}{4} \Rightarrow ∣ m_{2} ∣ = 2 \frac{1}{4}

Whichever line's slope has a greater absolute value is the steeper line.

3 > 2 \frac{1}{4} ⇓ ∣ m_{1} ∣ > ∣ m_{2} ∣

The absolute value of the first slope is greater, so the first line has a steeper slope.

Exercise 42 Page 194

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