Here we will learn how to solve the slope of the graph of y = mx + c.

The graph of y = mx + c is a straight line joining the points (0, c) and (\(\frac{-c}{m}\), 0).

Let M = (\(\frac{-c}{m}\), 0) and N = (0, c) and ∠NMX = θ.

Then, tan θ is called the slope of the line which is the graph of y = mx + c.

Now, ON = c and OM = \(\frac{c}{m}\).

Therefore, in the right-angled ∆MON, tan θ = \(\frac{ON}{OM}\) = \(\frac{c}{\frac{c}{m} }\) = m.

Thus, the slope of the line which is the graph of y = mx + c is m

And m is equal to the tangent of the angle that the line makes with the positive direction of the x-axis.

Solved examples on slope of the graph of y = mx + c:

**1.** What is the slope of the line which makes 60° with the
positive direction of the x-axis?

**Solution:**

The slope = tan 60° = √3

**2.** What is the slope of the line which is the graph of 2x –
3y + 5 = 0?

**Solution:**

Here, 2x – 3y + 5 = 0

⟹ 3y = 2x + 5

⟹ y = \(\frac{2}{3}\)x + \(\frac{5}{3}\).

Comparing with y = mx + c, we have m = \(\frac{2}{3}\).

Therefore, the slope of the line is \(\frac{2}{3}\).

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