The co-ordinates of A, B and C are (3, 1), (1, 5) and (4, 2) respectively. P is the mid-pt. of BC and Q lies on AC and is such that `CQ : QA=3 : 1`, R lies on AB and is such that `AR : RB=1 : 3`. Find the equation of the lines AP, BQ and CR.

Answer

To solve the problem step by step, we will find the coordinates of points P, Q, and R first, and then derive the equations of lines AP, BQ, and CR.

### Step 1: Find the coordinates of point P (midpoint of BC)
Given the coordinates of B (1, 5) and C (4, 2), we can find the midpoint P using the midpoint formula:

\[ P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) \]

Substituting the coordinates of B and C:

\[ P = \left( \frac{1 + 4}{2}, \frac{5 + 2}{2} \right) = \left( \frac{5}{2}, \frac{7}{2} \right) \]

### Step 2: Find the coordinates of point Q (on AC such that CQ:QA = 3:1)
Using the section formula, the coordinates of Q can be calculated as follows. Let A = (3, 1) and C = (4, 2). The ratio CQ:QA = 3:1 means that Q divides AC in the ratio 3:1.

Using the section formula:

\[ Q = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]

where \( m = 3 \) and \( n = 1 \):

\[ Q = \left( \frac{3 \cdot 4 + 1 \cdot 3}{3 + 1}, \frac{3 \cdot 2 + 1 \cdot 1}{3 + 1} \right) = \left( \frac{12 + 3}{4}, \frac{6 + 1}{4} \right) = \left( \frac{15}{4}, \frac{7}{4} \right) \]

### Step 3: Find the coordinates of point R (on AB such that AR:RB = 1:3)
Using the section formula again, we find R which divides AB in the ratio 1:3. The coordinates of A are (3, 1) and B are (1, 5).

Using the section formula:

\[ R = \left( \frac{m x_2 + n x_1}{m+n}, \frac{m y_2 + n y_1}{m+n} \right) \]

where \( m = 1 \) and \( n = 3 \):

\[ R = \left( \frac{1 \cdot 1 + 3 \cdot 3}{1 + 3}, \frac{1 \cdot 5 + 3 \cdot 1}{1 + 3} \right) = \left( \frac{1 + 9}{4}, \frac{5 + 3}{4} \right) = \left( \frac{10}{4}, \frac{8}{4} \right) = \left( \frac{5}{2}, 2 \right) \]

### Step 4: Find the equation of line AP
Using the coordinates of A (3, 1) and P (5/2, 7/2), we can find the slope \( m \):

\[ m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{\frac{7}{2} - 1}{\frac{5}{2} - 3} = \frac{\frac{7}{2} - \frac{2}{2}}{\frac{5}{2} - \frac{6}{2}} = \frac{\frac{5}{2}}{-\frac{1}{2}} = -5 \]

Using the point-slope form \( y - y_1 = m(x - x_1) \):

\[ y - 1 = -5\left(x - 3\right) \]

Expanding this:

\[ y - 1 = -5x + 15 \implies y + 5x = 16 \]

### Step 5: Find the equation of line BQ
Using the coordinates of B (1, 5) and Q (15/4, 7/4), we find the slope:

\[ m = \frac{\frac{7}{4} - 5}{\frac{15}{4} - 1} = \frac{\frac{7}{4} - \frac{20}{4}}{\frac{15}{4} - \frac{4}{4}} = \frac{-\frac{13}{4}}{\frac{11}{4}} = -\frac{13}{11} \]

Using the point-slope form:

\[ y - 5 = -\frac{13}{11}(x - 1) \]

Expanding this:

\[ y - 5 = -\frac{13}{11}x + \frac{13}{11} \implies 11y - 55 = -13x + 13 \implies 13x + 11y = 68 \]

### Step 6: Find the equation of line CR
Using the coordinates of C (4, 2) and R (5/2, 2):

The slope \( m \):

\[ m = \frac{2 - 2}{\frac{5}{2} - 4} = \frac{0}{\frac{5}{2} - \frac{8}{2}} = 0 \]

Since the slope is 0, the line CR is horizontal and passes through y = 2:

\[ y = 2 \]

### Final Equations
1. Equation of line AP: \( y + 5x = 16 \)
2. Equation of line BQ: \( 13x + 11y = 68 \)
3. Equation of line CR: \( y = 2 \)