A line segment joining P (2, 3) and Q (0, -1) is cut by the x-axis at the point R. A line AB cuts the y axis at T(0,6) and is perpendicular to PQ at S. Find the:

(a) equation of line PQ

(b) equation of line AB

(c) coordinates of points R and S. Draw a diagram for the same 

Let's go step by step to solve the problem.

### (a) Equation of Line PQ
The first step is to find the equation of the line segment \( PQ \) joining points \( P(2, 3) \) and \( Q(0, -1) \).

1. **Find the slope of \( PQ \)**:
   \[
   \text{Slope of } PQ = m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-1 - 3}{0 - 2} = \frac{-4}{-2} = 2
   \]

2. **Use the point-slope form to find the equation**:
   The equation of a line in point-slope form is given by:
   \[
   y - y_1 = m(x - x_1)
   \]
   Substituting the values of \( m = 2 \) and point \( P(2, 3) \) into the equation:
   \[
   y - 3 = 2(x - 2)
   \]
   Simplifying:
   \[
   y - 3 = 2x - 4
   \]
   \[
   y = 2x - 1
   \]
   So, the equation of line \( PQ \) is:
   \[
   y = 2x - 1
   \]

### (b) Equation of Line AB
The line \( AB \) is perpendicular to \( PQ \) and cuts the y-axis at \( T(0, 6) \).

1. **Find the slope of \( AB \)**:
   Since \( AB \) is perpendicular to \( PQ \), the slope of \( AB \) will be the negative reciprocal of the slope of \( PQ \).
   \[
   \text{Slope of } AB = -\frac{1}{2}
   \]

2. **Use the slope-intercept form to find the equation**:
   The equation of a line in slope-intercept form is given by:
   \[
   y = mx + c
   \]
   Substituting the slope \( m = -\frac{1}{2} \) and using the point \( T(0, 6) \) (where \( c = 6 \)):
   \[
   y = -\frac{1}{2}x + 6
   \]
   So, the equation of line \( AB \) is:
   \[
   y = -\frac{1}{2}x + 6
   \]

### (c) Coordinates of Points R and S

1. **Point R (intersection of line PQ with the x-axis)**:
   To find the x-coordinate where the line \( PQ \) intersects the x-axis, set \( y = 0 \) in the equation of line \( PQ \):
   \[
   0 = 2x - 1
   \]
   Solving for \( x \):
   \[
   x = \frac{1}{2}
   \]
   So, the coordinates of \( R \) are \( \left(\frac{1}{2}, 0\right) \).

2. **Point S (intersection of lines PQ and AB)**:
   To find the coordinates of \( S \), we need to solve the system of equations given by the equations of lines \( PQ \) and \( AB \):
   \[
   y = 2x - 1
   \]
   \[
   y = -\frac{1}{2}x + 6
   \]
   Setting these equal to each other to find the x-coordinate of \( S \):
   \[
   2x - 1 = -\frac{1}{2}x + 6
   \]
   Adding \( \frac{1}{2}x \) to both sides:
   \[
   2.5x - 1 = 6
   \]
   Adding 1 to both sides:
   \[
   2.5x = 7
   \]
   Solving for \( x \):
   \[
   x = \frac{7}{2.5} = \frac{14}{5} = 2.8
   \]
   Substituting \( x = 2.8 \) back into the equation of \( PQ \) to find \( y \):
   \[
   y = 2(2.8) - 1 = 5.6 - 1 = 4.6
   \]
   So, the coordinates of \( S \) are \( (2.8, 4.6) \).

### Diagram
Let's create a diagram to visualize the situation.{"size":"1024x1024","prompt":"A geometric diagram showing the line segment PQ joining points P(2, 3) and Q(0, -1) on a Cartesian plane. The line PQ intersects the x-axis at point R (1/2, 0). Another line AB is shown, which is perpendicular to PQ and cuts the y-axis at point T(0, 6). The lines AB and PQ intersect at point S (2.8, 4.6). The axes are labeled, and the key points P, Q, R, T, and S are clearly marked with their respective coordinates."}

A line segment joining P (2, 3) and Q (0, -1) is cut by the x-axis at the point R. A line AB cuts the y axis at T(0,6) and is perpendicular to PQ at S. Find the: (a) equation of line PQ (b) equation of line AB (c) coordinates of points R and S. Draw a diagram for the same.

Solve a coordinate geometry problem involving the line segment joining points P (2, 3) and Q (0, -1), where it intersects the x-axis and another line AB perpendicular to PQ. Detailed step-by-step solutions are provided for finding the equations of the lines and the coordinates of the points of intersection. Suitable for students in Grades 10-12.

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