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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You're on a roll — a) Find the equation of line AB. Step 1: Find the gradient of the given line. The given line is 3y - x = 4. Rearrange it into the slope-intercept form y = mx + c: 3y = x + 4 y = (1)/(3)x + (4)/(3) The gradient of this line is m_1 = (1)/(3). Step 2: Determine the gradient of line AB. Since line AB is parallel to the given line, their gradients are equal. So, the gradient of line AB is m_AB = (1)/(3). Step 3: Find the equation of line AB. Line AB passes through point A (-3, 1) and has a gradient of (1)/(3). Using the point-slope form y - y_1 = m(x - x_1): y - 1 = (1)/(3)(x - (-3)) y - 1 = (1)/(3)(x + 3) Multiply by 3 to clear the fraction: 3(y - 1) = x + 3 3y - 3 = x + 3 Rearrange into the general form Ax + By + C = 0: x - 3y + 6 = 0 The equation of line AB is x - 3y + 6 = 0. b) Find the equation of line AD. Step 1: Determine the gradient of line AD. In a rectangle, adjacent sides are perpendicular. Therefore, line AD is perpendicular to line AB. The gradient of line AB is m_AB = (1)/(3). The gradient of a line perpendicular to AB is the negative reciprocal of m_AB: m_AD = -(1)/(m_AB) = -(1)/(1/3) = -3 Step 2: Find the equation of line AD. Line AD passes through point A (-3, 1) and has a gradient of -3. Using the point-slope form y - y_1 = m(x - x_1): y - 1 = -3(x - (-3)) y - 1 = -3(x + 3) y - 1 = -3x - 9 Rearrange into the general form Ax + By + C = 0: 3x + y + 8 = 0 The equation of line AD is 3x + y + 8 = 0. c) If C has coordinates (2, 6), find the equations of the line BC and CD in the double intercept form, (x)/(a) + (y)/(b) = 1. Equation of line BC: Step 1: Determine the gradient of line BC. In a rectangle, opposite sides are parallel. Therefore, line BC is parallel to line AD. The gradient of line AD is m_AD = -3. So, the gradient of line BC is m_BC = -3. Step 2: Find the equation of line BC. Line BC passes through point C (2, 6) and has a gradient of -3. Using the point-slope form y - y_1 = m(x - x_1): y - 6 = -3(x - 2) y - 6 = -3x + 6 Rearrange into the form Ax + By = C: 3x + y = 12 Step 3: Convert the equation to double intercept form. To get (x)/(a) + (y)/(b) = 1, divide the entire equation by 12: (3x)/(12) + (y)/(12) = (12)/(12) (x)/(4) + (y)/(12) = 1 The equation of line BC is (x)/(4) + (y)/(12) = 1. Equation of line CD: Step 1: Determine the gradient of line CD. In a rectangle, opposite sides are parallel. Therefore, line CD is parallel to line AB. The gradient of line AB is m_AB = (1)/(3). So, the gradient of line CD is m_CD = (1)/(3). Step 2: Find the equation of line CD. Line CD passes through point C (2, 6) and has a gradient of (1)/(3). Using the point-slope form y - y_1 = m(x - x_1): y - 6 = (1)/(3)(x - 2) Multiply by 3 to clear the fraction: 3(y - 6) = x - 2 3y - 18 = x - 2 Rearrange into the form Ax + By = C: x - 3y = -16 Step 3: Convert the equation to double intercept form. To get (x)/(a) + (y)/(b) = 1, divide the entire equation by -16: (x)/(-16) - (3y)/(-16) = (-16)/(-16) (x)/(-16) + (3y)/(16) = 1 This can also be written as: (x)/(-16) + (y)/(16/3) = 1 The equation of line CD is (x)/(-16) + (y)/(16/3) = 1. d) Find the coordinates of B. Step 1: Identify the lines that intersect at B. Point B is the intersection of line AB and line BC. From part (a), the equation of line AB is x - 3y + 6 = 0 x - 3y = -6 (1). From part (c), the equation of line BC is 3x + y = 12 (2). Step 2: Solve the system of equations. From equation (2), express y in terms of x: y = 12 - 3x Substitute this expression for y into equation (1): x - 3(12 - 3x) = -6 x - 36 + 9x = -6 10x - 36 = -6 10x = -6 + 36 10x = 30 x = 3 Now substitute x=3 back into the expression for y: y = 12 - 3(3) y = 12 - 9 y = 3 The coordinates of B are (3, 3). Got more? Send 'em! 📸