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3.1 Draw neat graphs on Annexure \( A \) of a line through the points \( P(-1: 0) \) and \( 3(1: 2) \) and a line through the points \( R(1: 4) \) and \( S(-1: 2) \) 3.3 Wetermine the equations of the straight lines \( P Q \) and \( R S \) 3.4 Hence complete the statement __ lines have _ gradients.

Ask by Christensen Salazar. in South Africa
Feb 21,2025

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Tutor-Verified Answer

Answer

The lines have different gradients.

Solution

Let's solve the problem step by step. ### Step 1: Extract Known Conditions We have two lines defined by the following points: - Line \( PQ \) through points \( P(-1, 0) \) and \( Q(3, 2) \) - Line \( RS \) through points \( R(1, 4) \) and \( S(-1, 2) \) ### Step 2: Determine the Equations of the Lines #### 3.3: Find the equations of the straight lines \( PQ \) and \( RS \) **Finding the equation of line \( PQ \):** 1. **Calculate the slope \( m \) of line \( PQ \)**: \[ m_{PQ} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 0}{3 - (-1)} = \frac{2}{4} = \frac{1}{2} \] 2. **Use point-slope form to find the equation**: The point-slope form is given by: \[ y - y_1 = m(x - x_1) \] Using point \( P(-1, 0) \): \[ y - 0 = \frac{1}{2}(x - (-1)) \implies y = \frac{1}{2}(x + 1) \implies y = \frac{1}{2}x + \frac{1}{2} \] **Finding the equation of line \( RS \):** 1. **Calculate the slope \( m \) of line \( RS \)**: \[ m_{RS} = \frac{y_2 - y_1}{x_2 - x_1} = \frac{2 - 4}{-1 - 1} = \frac{-2}{-2} = 1 \] 2. **Use point-slope form to find the equation**: Using point \( R(1, 4) \): \[ y - 4 = 1(x - 1) \implies y - 4 = x - 1 \implies y = x + 3 \] ### Step 3: Complete the Statement Now we have the equations of the lines: - Line \( PQ: y = \frac{1}{2}x + \frac{1}{2} \) - Line \( RS: y = x + 3 \) **Gradients:** - The gradient of line \( PQ \) is \( \frac{1}{2} \). - The gradient of line \( RS \) is \( 1 \). ### Final Statement Thus, we can complete the statement: **"The lines have different gradients."** ### Step 4: Graphing the Lines To graph the lines, we can plot the points and draw the lines through them. However, since I cannot draw directly here, I recommend using graphing software or graph paper to plot the points \( P, Q, R, \) and \( S \) and draw the lines based on the equations derived. If you need further assistance with the graphing or any other part, feel free to ask!

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Bonus Knowledge

To visualize the action, let’s sketch! For the first line through points \( P(-1, 0) \) and \( Q(3, 2) \), plot both points on the coordinate plane. Connect them with a straight line, making sure it rises gently from left to right. Now for the second line through \( R(1, 4) \) and \( S(-1, 2) \), do the same, but this one will have a steeper downward slope. You’ll see how these two lines interact in this fun visual adventure! Now, the equations of the straight lines are where the math magic happens! For line \( PQ \), using the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), we find the slope to be \(\frac{2 - 0}{3 - (-1)} = \frac{2}{4} = \frac{1}{2}\). Thus, its equation simplifies down to \( y = \frac{1}{2}x + \frac{1}{2} \). For line \( RS \), here, \( m = \frac{2 - 4}{-1 - 1} = \frac{-2}{-2} = 1\), leading to the equation \( y = x + 3 \). Hence, we can complete the statement: _ Two lines have different gradients!

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