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Find the equation of the altitude from C to AB and the coordinates of the foot of the altitude (D).

Mathematics

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Answer

3.1

(4,5)\mathbf{(4, 5)}

Step-by-step solution

3 steps

Step 1: Calculate the gradient of AB.
The coordinates of A are (−10,−2)(-10, -2) and B are (10,8)(10, 8).

mAB=yB−yAxB−xA=8−(−2)10−(−10)=1020=12m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{8 - (-2)}{10 - (-10)} = \frac{10}{20} = \frac{1}{2}

Step 2: Determine the gradient of the altitude from C to AB.
The altitude is perpendicular to AB. If two lines are perpendicular, the product of their gradients is −1-1.

mCD=−1mAB=−11/2=−2m_{CD} = -\frac{1}{m_{AB}} = -\frac{1}{1/2} = -2

Step 3: Find the equation of the altitude using point C (11,−9)(11, -9) and mCD=−2m_{CD} = -2.
Using the point-gradient form y−y1=m(x−x1)y - y_1 = m(x - x_1):

y−(−9)=−2(x−11)y - (-9) = -2(x - 11) y+9=−2x+22y + 9 = -2x + 22 y=−2x+22−9y = -2x + 22 - 9 y=−2x+13y = -2x + 13

This matches the given equation.

Part 3.1· 2 steps

2

  1. Find the equation of line AB.

    Using the gradient mAB=12m_{AB} = \frac{1}{2} and point A (−10,−2)(-10, -2):

    y−yA=mAB(x−xA)y - y_A = m_{AB}(x - x_A) y−(−2)=12(x−(−10))y - (-2) = \frac{1}{2}(x - (-10)) y+2=12(x+10)y + 2 = \frac{1}{2}(x + 10) y+2=12x+5y + 2 = \frac{1}{2}x + 5 y=12x+3y = \frac{1}{2}x + 3

  2. Solve the system of equations for line AB and the altitude CD to find point D.

    Equation of AB: y=12x+3y = \frac{1}{2}x + 3
    Equation of CD: y=−2x+13y = -2x + 13
    Set the y-values equal:

    12x+3=−2x+13\frac{1}{2}x + 3 = -2x + 13

    Multiply by 2 to eliminate the fraction:

    x+6=−4x+26x + 6 = -4x + 26 x+4x=26−6x + 4x = 26 - 6 5x=205x = 20 x=4x = 4

    Substitute x=4x = 4 into the equation for the altitude:

    y=−2(4)+13y = -2(4) + 13 y=−8+13y = -8 + 13 y=5y = 5

    The coordinates of point D are (4,5)\boxed{\mathbf{(4, 5)}}.

Part 3.2· 1 step

To calculate the area of △ABC\triangle ABC, we can use the formula: Area =12×base×height= \frac{1}{2} \times \text{base} \times \text{height}.
The base is AB and the height is CD (the altitude from C to AB).

  1. Calculate the length of AB.

    Using the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} for A (−10,−2)(-10, -2) and B (10,8)(10, 8):

    AB=(10−(−10))2+(8−(−2))2AB = \sqrt{(10 - (-10))^2 + (8 - (-2))^2}
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