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The image contains a problem asking to determine the gradient and the equation of the straight line that passes through two given points. Here is a translation of the question:

**3-3. Determine the gradient and the equation of the straight line passing through:**
(a) Point A (2, 5) and point B (-1, 4).  
(b) Point A (-2, 3) and point B (6, -3).  
(c) Point A (2, 3) and point B (5, 7).

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### General Steps to Solve Each Part:

1. **Calculate the Gradient (Slope)** using the formula:
   \[
   m = \frac{y_2 - y_1}{x_2 - x_1}
   \]
   Where \( (x_1, y_1) \) and \( (x_2, y_2) \) are the coordinates of the two points.

2. **Equation of the Line** in slope-intercept form:
   \[
   y - y_1 = m(x - x_1)
   \]
   This is the point-slope form of the line equation, where \( m \) is the gradient and \( (x_1, y_1) \) is one of the points.

Let’s go through each part:

### (a) Points A(2, 5) and B(-1, 4)
1. **Gradient:**
   \[
   m = \frac{4 - 5}{-1 - 2} = \frac{-1}{-3} = \frac{1}{3}
   \]

2. **Equation:**
   Using point \( A(2, 5) \) and the gradient \( m = \frac{1}{3} \):
   \[
   y - 5 = \frac{1}{3}(x - 2)
   \]
   Simplifying this:
   \[
   y - 5 = \frac{1}{3}x - \frac{2}{3}
   \]
   \[
   y = \frac{1}{3}x + \frac{13}{3}
   \]

### (b) Points A(-2, 3) and B(6, -3)
1. **Gradient:**
   \[
   m = \frac{-3 - 3}{6 - (-2)} = \frac{-6}{6 + 2} = \frac{-6}{8} = \frac{-3}{4}
   \]

2. **Equation:**
   Using point \( A(-2, 3) \) and the gradient \( m = \frac{-3}{4} \):
   \[
   y - 3 = \frac{-3}{4}(x - (-2)) = \frac{-3}{4}(x + 2)
   \]
   Simplifying:
   \[
   y - 3 = \frac{-3}{4}x - \frac{6}{4} = \frac{-3}{4}x - \frac{3}{2}
   \]
   \[
   y = \frac{-3}{4}x + \frac{3}{2}
   \]

### (c) Points A(2, 3) and B(5, 7)
1. **Gradient:**
   \[
   m = \frac{7 - 3}{5 - 2} = \frac{4}{3}
   \]

2. **Equation:**
   Using point \( A(2, 3) \) and the gradient \( m = \frac{4}{3} \):
   \[
   y - 3 = \frac{4}{3}(x - 2)
   \]
   Simplifying:
   \[
   y - 3 = \frac{4}{3}x - \frac{8}{3}
   \]
   \[
   y = \frac{4}{3}x - \frac{8}{3} + 3 = \frac{4}{3}x + \frac{1}{3}
   \]

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Would you like further details on any part of this explanation, or help with a different topic? 

Here are some related questions to expand your understanding:
1. How do you find the intercepts of a line?
2. What is the general form of a linear equation, and how does it relate to slope-intercept form?
3. How can you verify if two lines are parallel or perpendicular?
4. How do you convert from point-slope form to standard form of a line?
5. Can a vertical line have a gradient, and if not, why?

**Tip:** Always double-check your calculations, especially the signs when subtracting coordinates to find the gradient.

Tentukanlah gradien dan persamaan garis lurus yang melalui: (a) Titik A (2, 5) dan titik B (-1, 4). (b) Titik A (-2, 3) dan titik B (6, -3). (c) Titik A (2, 3) dan titik B (5, 7).

Learn how to find the gradient and equation of a straight line passing through two points. This problem involves calculating the slope using the formula m = (y2 - y1) / (x2 - x1) and deriving the equation in slope-intercept form. Suitable for students in Grades 8-10.

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