Obtain the equation of the diagonal PR in the form y = mx + c.

Question 20: a) Obtain the equation of the diagonal PR in the form y = mx + c. Step 1: Find the midpoint of the line segment AB. The points are A(-10, -1) and B(2, 7). The midpoint M is given by the formula M = ((x_1+x_2)/(2), (y_1+y_2)/(2)). M = ((-10+2)/(2), (-1+7)/(2)) = ((-8)/(2), (6)/(2)) = (-4, 3) Step 2: Find the gradient of the line segment AB. The gradient m_AB is given by the formula m = (y_2-y_1)/(x_2-x_1). m_AB = (7 - (-1))/(2 - (-10)) = (7+1)/(2+10) = (8)/(12) = (2)/(3) Step 3: Find the gradient of the perpendicular bisector (diagonal PR). Since PR is the perpendicular bisector of AB, its gradient m_PR is the negative reciprocal of m_AB. m_PR = -(1)/(m_AB) = -(1)/(2)3 = -(3)/(2) Step 4: Obtain the equation of the line PR. The line PR passes through the midpoint M(-4, 3) and has a gradient of -(3)/(2). Using the point-slope form y - y_1 = m(x - x_1): y - 3 = -(3)/(2)(x - (-4)) y - 3 = -(3)/(2)(x + 4) y - 3 = -(3)/(2)x - (3)/(2) × 4 y - 3 = -(3)/(2)x - 6 y = -(3)/(2)x - 6 + 3 y = -(3)/(2)x - 3 The equation of the diagonal PR is y = -(3)/(2)x - 3. b) Given that the x and y-intercepts of the diagonal PR are the centre of the rhombus and the vertex R respectively. Find the coordinates of point R. Step 1: Find the y-intercept of the diagonal PR. The equation of PR is y = -(3)/(2)x - 3. The y-intercept occurs when x=0. y = -(3)/(2)(0) - 3 y = -3 The y-intercept is (0, -3). According to the question, the y-intercept is vertex R. The coordinates of point R are (0, -3). c) Point Q(-8, -4) is one of the vertices of the rhombus in (a) above. Obtain the equation of QS leaving it in the form ax + by = c, where a, b and c are integers. Hence state the coordinate of point S. Step 1: Find the gradient of the diagonal QS. In a rhombus, the diagonals are perpendicular. The gradient of PR is m_PR = -(3)/(2). The gradient of QS, m_QS, is the negative reciprocal of m_PR. m_QS = -(1)/(m_PR) = -(1)/(-3)2 = (2)/(3) Step 2: Obtain the equation of the line QS. The line QS passes through Q(-8, -4) and has a gradient of (2)/(3). Using the point-slope form y - y_1 = m(x - x_1): y - (-4) = (2)/(3)(x - (-8)) y + 4 = (2)/(3)(x + 8) Multiply by 3 to clear the fraction: 3(y + 4) = 2(x + 8) 3y + 12 = 2x + 16 Rearrange into the form ax + by = c: 2x - 3y = 12 - 16 2x - 3y = -4 The equation of QS is 2x - 3y = -4. Step 3: Find the coordinates of the center of the rhombus. The center of the rhombus is the intersection of the diagonals PR and QS. Equation of PR: y = -(3)/(2)x - 3 Equation of QS: 2x - 3y = -4 Substitute the expression for y from the PR equation into the QS equation: 2x - 3(-(3)/(2)x - 3) = -4 2x + (9)/(2)x + 9 = -4 Multiply by 2 to eliminate the fraction: 4x + 9x + 18 = -8 13x = -8 - 18 13x = -26 x = -2 Substitute x = -2 back into the equation for PR to find y: y = -(3)/(2)(-2) - 3 y = 3 - 3 y = 0 The center of the rhombus, let's call it C, is (-2, 0). Step 4: Find the coordinates of point S. The center C(-2, 0) is the midpoint of the diagonal QS. Let S = (x_S, y_S). Using the midpoint formula C = ((x_Q+x_S)/(2), (y_Q+y_S)/(2)) with Q(-8, -4): For the x-coordinate: -2 = (-8 + x_S)/(2) -4 = -8 + x_S x_S = -4 + 8 x_S = 4 For the y-coordinate: 0 = (-4 + y_S)/(2) 0 = -4 + y_S y_S = 4 The coordinates of point S are (4, 4). d) Find the gradient of the line RS. Step 1: Use the coordinates of R and S to find the gradient. From part (b), R = (0, -3). From part (c), S = (4, 4). The gradient m_RS is given by m = (y_2-y_1)/(x_2-x_1). m_RS = (4 - (-3))/(4 - 0) m_RS = (4 + 3)/(4) m_RS = (7)/(4) The gradient of the line RS is (7)/(4). That's 2 down. 3 left today — send the next one.