Let's work through each part of the question step-by-step.

Given Points:

The points provided are:

• $D(-1, 2)$
• $E(4, -2)$
• $F(-5, -3)$

Part 1.1.1: Gradient of DF

To find the gradient of line $DF$, we use the formula: $\text{Gradient} = \frac{y_2 - y_1}{x_2 - x_1}$ where $D(x_1, y_1) = (-1, 2)$ and $F(x_2, y_2) = (-5, -3)$.

Plugging in these values: $\text{Gradient of DF} = \frac{-3 - 2}{-5 + 1} = \frac{-5}{-4} = \frac{5}{4}$

So, the gradient of $DF$ is $\frac{5}{4}$.

Part 1.1.2: Gradient of DE if it is Perpendicular to DF

If line $DE$ is perpendicular to line $DF$, then the gradient of $DE$ would be the negative reciprocal of the gradient of $DF$.

Since the gradient of $DF$ is $\frac{5}{4}$, the gradient of $DE$ would be: $\text{Gradient of DE} = -\frac{4}{5}$

Part 1.1.3: Equation of Line DE in the Form $y = mx + c$

Now, we need to find the equation of line $DE$ with a gradient of $-\frac{4}{5}$ and passing through point $D(-1, 2)$.

Using the point-slope form of a line: $y - y_1 = m(x - x_1)$ Substitute $m = -\frac{4}{5}$, $x_1 = -1$, and $y_1 = 2$: $y - 2 = -\frac{4}{5}(x + 1)$ Expanding and simplifying: $y - 2 = -\frac{4}{5}x - \frac{4}{5}$ $y = -\frac{4}{5}x + 2 - \frac{4}{5}$ $y = -\frac{4}{5}x + \frac{10}{5} - \frac{4}{5}$ $y = -\frac{4}{5}x + \frac{6}{5}$

So, the equation of line $DE$ in the form $y = mx + c$ is: $y = -\frac{4}{5}x + \frac{6}{5}$

Let me know if you'd like further details on any part of the solution!

Relative Questions to Deepen Understanding:

1. How do you determine if two lines are perpendicular based on their gradients?
2. How can the distance formula be applied to find the length of line $DF$?
3. What would the equation of line $DF$ look like?
4. How can you find the midpoint of $DF$?
5. How would the equation change if point $E$ had different coordinates?

Quick Tip:

When working with perpendicular lines, remember that their gradients are negative reciprocals (e.g., if one gradient is $\frac{a}{b}$, the other will be $-\frac{b}{a}$).