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    for each triangle ABC, show that its right angled, state the hypotenuse, find the area and find tan A.
    THE POINTS ARE A(13, -4) B(7,4) C(3,1)
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    (Original post by Anum Liaquet)
    for each triangle ABC, show that its right angled, state the hypotenuse, find the area and find tan A
    THE POINTS ARE A(13, -4) B(7,4) C(3,1)
    First get the vectors \vec{AB}, \vec{AC}, \vec{BC}
    Two vector is perpendicular to each other if their scalar product is zero.
    The area is the cross product of the two suitable vectors
    You vill get cosA from the scalar product, then tanA with help of
    an trigonometric identity.
    This only a method for solution certainly there are other methods, too.
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    If you have done vectors using them makes this easy to do.

    If not, start by finding the gradients of the lines passing through the 3 pairs of points (AB, AC, BC). Then consider what the condition is for 2 gradients to be perpendicular. Sketch a right-angled triangle. Where is the hypotenuse with respect to the right angle? Then calculate the distance between the two points defining this hypotenuse. Finally, tanA can be found by considering the relationship between the gradient and tan(theta) where theta is the angle between the line and the x-axis.
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    (Original post by ztibor)
    First get the vectors \vec{AB}, \vec{AC}, \vec{BC}
    Two vector is perpendicular to each other if their scalar product is zero.
    The area is the cross product of the two suitable vectors
    You vill get cosA from the scalar product, then tanA with help of
    an trigonometric identity.
    This only a method for solution certainly there are other methods, too.
    The area is actually half the modulus of the cross product of two of the vectors.
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    Thanks it really helped me ...
 
 
 
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