https://file.aiquickdraw.com/mathbot/file/1730360177290o25urd2h.jpg

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!

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### 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 \( -\frac{b}{a} \)).

1.1.2 If line DE is perpendicular to DF, write down the gradient of DE

Learn how to find the gradient of line DE when it is perpendicular to DF, given points D(-1, 2), E(4, -2), and F(-5, -3). This problem demonstrates key concepts in coordinate geometry, including the use of gradients and properties of perpendicular lines. Suitable for students in Grades 10-12.

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