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Let's go through the problem step by step, starting with part **(a)**.

### a) Show that angle \( \angle ABC \) is a right angle.

To show that angle \( \angle ABC \) is a right angle, we need to prove that the vectors (or lines) representing the sides \( AB \) and \( BC \) are perpendicular. For this, we can use the slopes of the lines:

- The slope of a line through two points \( (x_1, y_1) \) and \( (x_2, y_2) \) is given by:
\[
m = \frac{y_2 - y_1}{x_2 - x_1}
\]

#### Slope of \( AB \):
- Point \( A = (-3, 6) \)
- Point \( B = (-5, -2) \)
\[
m_{AB} = \frac{6 - (-2)}{-3 - (-5)} = \frac{6 + 2}{-3 + 5} = \frac{8}{2} = 4
\]

#### Slope of \( BC \):
- Point \( B = (-5, -2) \)
- Point \( C = (7, 5) \)
\[
m_{BC} = \frac{5 - (-2)}{7 - (-5)} = \frac{5 + 2}{7 + 5} = \frac{7}{12}
\]

To prove that \( \angle ABC \) is a right angle, the slopes of \( AB \) and \( BC \) should satisfy:
\[
m_{AB} \times m_{BC} = -1
\]

Substituting the slopes:
\[
4 \times \frac{7}{12} = \frac{28}{12} = \frac{7}{3}
\]

Since \( 4 \times \frac{7}{12} \neq -1 \), this means \( \angle ABC \) is **not** a right angle based on this calculation. However, based on the visual geometry of the problem, this needs closer inspection, possibly due to assumptions about the axes or scale.

### Moving on to part **(b)**:

### b) i. What is the slope of \( DE \)?

Points \( D \) and \( E \) are the midpoints of sides \( AB \) and \( AC \), respectively.

#### Coordinates of \( D \) (midpoint of \( AB \)):
The midpoint formula is given by:
\[
D = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)
\]

For \( AB \) where \( A = (-3, 6) \) and \( B = (-5, -2) \):
\[
D = \left( \frac{-3 + (-5)}{2}, \frac{6 + (-2)}{2} \right) = \left( \frac{-8}{2}, \frac{4}{2} \right) = (-4, 2)
\]

#### Coordinates of \( E \) (midpoint of \( AC \)):
For \( AC \) where \( A = (-3, 6) \) and \( C = (7, 5) \):
\[
E = \left( \frac{-3 + 7}{2}, \frac{6 + 5}{2} \right) = \left( \frac{4}{2}, \frac{11}{2} \right) = (2, 5.5)
\]

Now, we can find the slope of \( DE \):
\[
m_{DE} = \frac{5.5 - 2}{2 - (-4)} = \frac{3.5}{6} = \frac{7}{12}
\]

### b) ii. What are the coordinates of the point of intersection of \( DE \) with the x-axis?

The x-axis has the equation \( y = 0 \). To find the intersection of line \( DE \) with the x-axis, we need to use the point-slope form of the equation of a line.

The slope of \( DE \) is \( \frac{7}{12} \), and we can use point \( D(-4, 2) \) to write the equation of the line:
\[
y - 2 = \frac{7}{12}(x + 4)
\]

Setting \( y = 0 \) to find the x-intercept:
\[
0 - 2 = \frac{7}{12}(x + 4)
\]
\[
-2 = \frac{7}{12}(x + 4)
\]
Multiplying both sides by 12:
\[
-24 = 7(x + 4)
\]
Dividing by 7:
\[
x + 4 = \frac{-24}{7}
\]
\[
x = \frac{-24}{7} - 4 = \frac{-24}{7} - \frac{28}{7} = \frac{-52}{7}
\]

So, the x-coordinate of the point where \( DE \) intersects the x-axis is \( \frac{-52}{7} \), and the point is \( \left( \frac{-52}{7}, 0 \right) \).

---

Would you like more details on any part of the solution? Here are 5 related questions to explore:

1. How do you calculate the distance between two points on a Cartesian plane?
2. Can you determine if a triangle is right-angled using the Pythagorean Theorem in this scenario?
3. How do you derive the equation of a line using the general form \( y = mx + c \)?
4. What is the significance of finding midpoints in geometry problems involving triangles?
5. How do you verify the collinearity of points in a Cartesian plane?

**Tip**: Always double-check slope calculations and midpoints to avoid errors in complex geometrical proofs!

Triangle ABC is drawn on a Cartesian plane. Show that angle ABC is a right angle. Points D and E are the midpoints of sides AB and AC. What is the slope of DE and where does it intersect the x-axis?

Explore how to prove that a triangle has a right angle using slopes and how to find the slope of a line connecting midpoints of two sides. This example involves triangle ABC with vertices at A(-3, 6), B(-5, -2), and C(7, 5) on a Cartesian plane. Ideal for students in Grade 10-12.

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