Show that BT is an altitude of triangle ABC and determine the coordinates of C, given A(p; 8), T(1; q), B(-2; -2), and the equations of lines AC and BC.

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Asked on April 19, 2026Mathematics

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.

Accepted Answer · 2026-04-19T18:49:47.182Z

Welcome back — been a while! Let's pick up where you left off. Step 1: Show that p = -7 and q = 2. Point A(p; 8) lies on the line AC with equation 4y + 3x = 11. Substitute x=p and y=8 into the equation: 4(8) + 3p = 11 32 + 3p = 11 3p = 11 - 32 3p = -21 p = -7 Point T(1; q) lies on the line AC with equation 4y + 3x = 11. Substitute x=1 and y=q into the equation: 4q + 3(1) = 11 4q + 3 = 11 4q = 11 - 3 4q = 8 q = 2 Thus, p = -7 and q = 2. Step 2: Show that BT is an altitude of ABC. An altitude is perpendicular to the opposite side. We need to show that BT AC. First, find the gradient of BT using points B(-2; -2) and T(1; 2): m_BT = (y_2 - y_1)/(x_2 - x_1) = (2 - (-2))/(1 - (-2)) = (4)/(3) Next, find the gradient of AC from its equation 4y + 3x = 11. Rearrange to y = mx + c form: 4y = -3x + 11 y = -(3)/(4)x + (11)/(4) So, the gradient of AC is m_AC = -(3)/(4). Now, check if BT and AC are perpendicular by multiplying their gradients: m_BT × m_AC = (4)/(3) × (-(3)/(4)) = -1 Since the product of their gradients is -1, BT AC. Therefore, BT is an altitude of ABC. Step 3: Determine the coordinates of C. Point C is the intersection of line AC and line BC. Equation of AC: 4y + 3x = 11 (1) Equation of BC: 7y - x = -12 (2) From equation (2), express x in terms of y: x = 7y + 12 Substitute this expression for x into equation (1): 4y + 3(7y + 12) = 11 4y + 21y + 36 = 11 25y + 36 = 11 25y = 11 - 36 25y = -25 y = -1 Substitute y = -1 back into the expression for x: x = 7(-1) + 12 x = -7 + 12 x = 5 The coordinates of C are (5; -1). Step 4: Determine the area of ABC. We can use the formula for the area of a triangle: Area = (1)/(2) × base × height. Since BT is an altitude to AC, we can use AC as the base and BT as the height. First, calculate the length of AC using points A(-7; 8) and C(5; -1): AC = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) AC = sqrt((5 - (-7))^2 + (-1 - 8)^2) AC = sqrt((12)^2 + (-9)^2) AC = sqrt(144 + 81) AC = sqrt(225) AC = 15 units Next, calculate the length of BT using points B(-2; -2) and T(1; 2): BT = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) BT = sqrt((1 - (-2))^2 + (2 - (-2))^2) BT = sqrt((3)^2 + (4)^2) BT = sqrt(9 + 16) BT = sqrt(25) BT = 5 units Now, calculate the area of ABC: Area = (1)/(2) × AC × BT Area = (1)/(2) × 15 × 5 Area = (1)/(2) × 75 Area = 37.5 square units The area of ABC is 37.5 square units. Send me the next one 📸

Accepted Answer · 2026-04-19T18:49:47.182Z

Welcome back — been a while! Let's pick up where you left off. Step 1: Show that p = -7 and q = 2. Point A(p; 8) lies on the line AC with equation 4y + 3x = 11. Substitute x=p and y=8 into the equation: 4(8) + 3p = 11 32 + 3p = 11 3p = 11 - 32 3p = -21 p = -7 Point T(1; q) lies on the line AC with equation 4y + 3x = 11. Substitute x=1 and y=q into the equation: 4q + 3(1) = 11 4q + 3 = 11 4q = 11 - 3 4q = 8 q = 2 Thus, p = -7 and q = 2. Step 2: Show that BT is an altitude of ABC. An altitude is perpendicular to the opposite side. We need to show that BT AC. First, find the gradient of BT using points B(-2; -2) and T(1; 2): m_BT = (y_2 - y_1)/(x_2 - x_1) = (2 - (-2))/(1 - (-2)) = (4)/(3) Next, find the gradient of AC from its equation 4y + 3x = 11. Rearrange to y = mx + c form: 4y = -3x + 11 y = -(3)/(4)x + (11)/(4) So, the gradient of AC is m_AC = -(3)/(4). Now, check if BT and AC are perpendicular by multiplying their gradients: m_BT × m_AC = (4)/(3) × (-(3)/(4)) = -1 Since the product of their gradients is -1, BT AC. Therefore, BT is an altitude of ABC. Step 3: Determine the coordinates of C. Point C is the intersection of line AC and line BC. Equation of AC: 4y + 3x = 11 (1) Equation of BC: 7y - x = -12 (2) From equation (2), express x in terms of y: x = 7y + 12 Substitute this expression for x into equation (1): 4y + 3(7y + 12) = 11 4y + 21y + 36 = 11 25y + 36 = 11 25y = 11 - 36 25y = -25 y = -1 Substitute y = -1 back into the expression for x: x = 7(-1) + 12 x = -7 + 12 x = 5 The coordinates of C are (5; -1). Step 4: Determine the area of ABC. We can use the formula for the area of a triangle: Area = (1)/(2) × base × height. Since BT is an altitude to AC, we can use AC as the base and BT as the height. First, calculate the length of AC using points A(-7; 8) and C(5; -1): AC = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) AC = sqrt((5 - (-7))^2 + (-1 - 8)^2) AC = sqrt((12)^2 + (-9)^2) AC = sqrt(144 + 81) AC = sqrt(225) AC = 15 units Next, calculate the length of BT using points B(-2; -2) and T(1; 2): BT = sqrt((x_2 - x_1)^2 + (y_2 - y_1)^2) BT = sqrt((1 - (-2))^2 + (2 - (-2))^2) BT = sqrt((3)^2 + (4)^2) BT = sqrt(9 + 16) BT = sqrt(25) BT = 5 units Now, calculate the area of ABC: Area = (1)/(2) × AC × BT Area = (1)/(2) × 15 × 5 Area = (1)/(2) × 75 Area = 37.5 square units The area of ABC is 37.5 square units. Send me the next one 📸

Show that BT is an altitude of triangle ABC and determine the coordinates of C, given A(p; 8), T(1; q), B(-2; -2), and the equations of lines AC and BC.

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Show that BT is an altitude of triangle ABC and determine the coordinates of C, given A(p; 8), T(1; q), B(-2; -2), and the equations of lines AC and BC.

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