This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
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Welcome back 💸🚸Kayy — missed you this week. Here's the solution to your question. Question 3 Given points: P, R(3; 5), S(-3; -7), T(-5; k). Q(0; 5) is a point on the y-axis. PT RS. RS cuts the y-axis at D. PR cuts the x-axis at F. PT cuts the x-axis at E. PEF = . 3.1 Write down the equation of PR. Step 1: Identify the coordinates of P and R. From the diagram, P is on the y-axis and has the same y-coordinate as Q(0;5), so P(0;5). The coordinates of R are given as (3;5). Step 2: Determine the equation of the line passing through P(0;5) and R(3;5). Since both points have the same y-coordinate (y=5), the line PR is a horizontal line. The equation of PR is y=5. 3.2 Calculate the: 3.2.1 Gradient of RS Step 1: Identify the coordinates of R and S. R(3;5) and S(-3;-7). Step 2: Use the gradient formula m = (y_2 - y_1)/(x_2 - x_1). m_RS = (-7 - 5)/(-3 - 3) m_RS = (-12)/(-6) m_RS = 2 The gradient of RS is 2. 3.2.2 Size of Step 1: Identify the line associated with . is the angle PEF. This is the angle of inclination of the line PT with the positive x-axis. Step 2: Determine the gradient of PT. It is given that PT RS. Therefore, their gradients are equal. From 3.2.1, m_RS = 2. So, m_PT = 2. Step 3: Calculate the angle of inclination . The gradient m is equal to (). () = m_PT () = 2 = (2) ≈ 63.43^ The size of is 63.43^. 3.2.3 Coordinates of D Step 1: Identify the line and point D. D is the y-intercept of the line RS. Step 2: Find the equation of the line RS. We have the gradient m_RS = 2 and a point R(3;5). Using the point-slope form y - y_1 = m(x - x_1): y - 5 = 2(x - 3) y - 5 = 2x - 6 y = 2x - 1 Step 3: Find the y-intercept (where x=0). Substitute x=0 into the equation: y = 2(0) - 1 y = -1 The coordinates of D are (0;-1). 3.3 If it is given that TS = 2sqrt(5), calculate the value of k. Step 1: Identify the coordinates of T and S. T(-5; k) and S(-3; -7). Step 2: Use the distance formula d = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2). Given TS = 2sqrt(5). (2sqrt(5))^2 = (-3 - (-5))^2 + (-7 - k)^2 4 × 5 = (-3 + 5)^2 + (-7 - k)^2 20 = (2)^2 + (-(7+k))^2 20 = 4 + (7+k)^2 Step 3: Solve for k. 16 = (7+k)^2 Take the square root of both sides: ±sqrt(16) = 7+k ± 4 = 7+k Case 1: 4 = 7+k k = 4 - 7 k = -3 Case 2: -4 = 7+k k = -4 - 7 k = -11 From the diagram, point T is in the third quadrant, meaning its y-coordinate (k) must be negative. Both values are negative. However, looking at the diagram, T is below the x-axis and its y-coordinate appears to be greater than -7 (S's y-coordinate). If k=-11, T would be much lower than S. If k=-3, T is higher than S, which aligns with the visual representation. The value of k is -3. 3.4 Parallelogram TDNS, with N in the 4th quadrant, is drawn. Calculate the coordinates of N. Step 1: Identify the vertices of the parallelogram TDNS. T(-5;-3) (using k=-3 from 3.3) D(0;-1) S(-3;-7) Let N(x_N; y_N). Step 2: Use the property that the diagonals of a parallelogram bisect each other. The midpoint of TS must be the same as the midpoint of DN. Midpoint of TS: M_TS = ((-5 + (-3))/(2); (-3 + (-7))/(2)) M_TS = ((-8)/(2); (-10)/(2)) M_TS = (-4; -5) Midpoint of DN: M_DN = ((0 + x_N)/(2); (-1 + y_N)/(2)) Step 3: Equate the midpoints and solve for x_N and y_N. (x_N)/(2) = -4 x_N = -8 (-1 + y_N)/(2) = -5 -1 + y_N = -10 y_N = -9 The coordinates of N are (-8;-9). Self-correction: The question states N is in the 4th quadrant. A point in the 4th quadrant has a positive x-coordinate and a negative y-coordinate. My calculated N(-8;-9) is in the 3rd quadrant. This means the order of vertices for the parallelogram is important. Let's re-evaluate using vector addition or the property that opposite sides are parallel and equal in length. For parallelogram TDNS, TD = SN. D - T = N - S (0 - (-5); -1 - (-3)) = (x_N - (-3); y_N - (-7)) (5; 2) = (x_N + 3; y_N + 7) Equating components: 5 = x_N + 3 x_N = 2 2 = y_N + 7 y_N = -5 So, N(2;-5). This point is in the 4th quadrant (positive x, negative y), which matches the condition. The coordinates of N are (2;-5). 3.5 APRD is reflected about the y-axis to form APRD'. Calculate the size of RDR'. Step 1: Identify the coordinates of A, P, R, D. The problem statement mentions "trapezium PRST" and "parallelogram TDNS". It does not explicitly define point A. However, the question asks about APRD. Let's assume A is a point that forms a quadrilateral APRD. Given the context of reflections, we need the coordinates of R and D. R(3;5) D(0;-1) P(0;5) Step 2: Find the coordinates of R' and D' after reflection about the y-axis. When a point (x,y) is reflected about the y-axis, its image is (-x,y). R(3;5) R'(-3;5) D(0;-1) D'(0;-1) (D is on the y-axis, so it's invariant under y-axis reflection) Step 3: Calculate the size of RDR'. We need to find the angle formed by the lines DR and DR'. The coordinates are D(0;-1), R(3;5), R'(-3;5). Calculate the gradient of DR (m_DR): m_DR = (5 - (-1))/(3 - 0) = (6)/(3) = 2 Calculate the gradient of DR' (m_DR'): m_DR' = (5 - (-1))/(-3 - 0) = (6)/(-3) = -2 Step 4: Use the formula for the angle between two lines: () = |(m_1 - m_2)/(1 + m_1 m_2)|. Let m_1 = m_DR = 2 and m_2 = m_DR' = -2. ( RDR') = |(2 - (-2))/(1 + (2)(-2))| ( RDR') = |(4)/(1 - 4)| ( RDR') = |(4)/(-3)| ( RDR') = (4)/(3) RDR' = ((4)/(3)) RDR' ≈ 53.13^ The size of RDR' is 53.13^. Drop the next question.