The points A, B and Care (4, 0), (2, 2) and (0, 6) respectively. Find the equations of AB and BC.
If AB cuts the y-axis at P and BC cuts the x-axis at Q, find the co-ordinates of P and Q. 

Answer

To solve the problem step by step, we will find the equations of lines AB and BC, and then determine the coordinates of points P and Q where these lines intersect the y-axis and x-axis, respectively.

### Step 1: Find the equation of line AB

**Given points:**
- A(4, 0)
- B(2, 2)

**Calculate the slope (m) of line AB:**
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{2 - 4} = \frac{2}{-2} = -1 \]

**Using the point-slope form of the line equation:**
\[ y - y_1 = m(x - x_1) \]
Using point A(4, 0):
\[ y - 0 = -1(x - 4) \]
\[ y = -x + 4 \]

**Rearranging to standard form:**
\[ x + y = 4 \]

### Step 2: Find the equation of line BC

**Given points:**
- B(2, 2)
- C(0, 6)

**Calculate the slope (m) of line BC:**
\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{6 - 2}{0 - 2} = \frac{4}{-2} = -2 \]

**Using the point-slope form of the line equation:**
Using point C(0, 6):
\[ y - 6 = -2(x - 0) \]
\[ y - 6 = -2x \]
\[ y + 2x = 6 \]

### Step 3: Find the coordinates of point P where line AB cuts the y-axis

To find the y-intercept (P), set \(x = 0\) in the equation of line AB:
\[ 0 + y = 4 \implies y = 4 \]
Thus, the coordinates of point P are:
\[ P(0, 4) \]

### Step 4: Find the coordinates of point Q where line BC cuts the x-axis

To find the x-intercept (Q), set \(y = 0\) in the equation of line BC:
\[ 0 + 2x = 6 \implies 2x = 6 \implies x = 3 \]
Thus, the coordinates of point Q are:
\[ Q(3, 0) \]

### Final Answer:
- The equation of line AB is \(x + y = 4\).
- The equation of line BC is \(y + 2x = 6\).
- The coordinates of point P are \((0, 4)\).
- The coordinates of point Q are \((3, 0)\).