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Calculating the area of a triangle from points on straight lines watch

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    I have worked out three points on two different straight lines:

     Point A: (\frac{5}{2}, 0) 



This point is on line l1.



Point B: (-5, 0) This point is on line l2.



Point C: (-10, -10) This point is a point of intersection between lines l1 and l2.

    How do I calculate the area of triangle ABC from this? I understand that the area of a triangle is defined as follows:

     \frac{1}{2} \times base \times height

    What I do not fully understand is how the base and height are derived from these points.
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    if you have the coordinates of the 3 vertics you can find the 3 lengths of the sides.

    once you have the 3 lengths you can find an angle using the cosine rule.

    then you can use 0.5abSInC for the area.

    if you are feeling adventurous you can use Heron's Rule for the area which does not involve angles.
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    (Original post by Illidan2)
    I have worked out three points on two different straight lines:

     Point A: (\frac{5}{2}, 0) 



This point is on line l1.



Point B: (-5, 0) This point is on line l2.



Point C: (-10, -10) This point is a point of intersection between lines l1 and l2.

    How do I calculate the area of triangle ABC from this? I understand that the area of a triangle is defined as follows:

     \frac{1}{2} \times base \times height

    What I do not fully understand is how the base and height are derived from these points.
    You may find a diagram useful.

    Notice that two of the points, A and B, have the same y-coordinate. So, make that the base of your triangle.

    The length of the base is simply the difference between their two x-coordinates.

    And the perpendicular height of the triangle is the difference between their common y-coordinate and the y-coordinate of the third point.
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    Got it! Thank you
 
 
 
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