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talitha
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Tyro
Asked: 2021-03-31T13:32:21+00:00 In: ,

Geometry: Triangle

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Determine the kind of triangle formed with vertices at A(2,-1), B(3,4), C(-7, 6).

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1 Answer

  2. acedstud

    We approach this question by finding the interior angles of the triangle. First, we find the slopes of the lines that make up the sides of the triangle. The slope of a line passing through points (x_1, y_1) and (x_2, y_2) is given by m=\dfrac{y_2-y_1}{x_2-x_1}.

    Slope of side AB with coordinates A(2, -1) and B(3, 4) is given by m_1=\dfrac{4-(-1)}{3-2}=\dfrac{5}{1}=5.

    Slope of side AC with coordinates A(2, -1) and C(-7, 6) is given by m_2=\dfrac{6-(-1)}{-7-2}=\dfrac{7}{-9}=-\dfrac{7}{9}.

    Slope of side BC with coordinates B(3, 4) and C(-7, 6) is given by m_3=\dfrac{6-4}{-7-3}=\dfrac{2}{-10}=-\dfrac{1}{5}.

    The angle (\theta) between two lines with slopes m_1 and m_2 is given by \tan\theta=\left|\dfrac{m_2-m_1}{1-m_1m_2}\right|.

    Thus, the angle between AB and AC is given by \tan\theta=\left|\dfrac{-\frac{7}{9}-5}{1-(-\frac{7}{9})(5)}\right|=\left|\dfrac{-\frac{52}{9}}{1+\frac{35}{9}}\right|

    =\left|\dfrac{-\frac{52}{9}}{\frac{44}{9}}\right|=\dfrac{13}{11}

    \Rightarrow\theta=\tan^{-1}(\frac{13}{11})\approx49.8^o.

    The angle between AB and BC is given by \tan\theta=\left|\dfrac{-\frac{1}{5}-5}{1-(-\frac{1}{5})(5)}\right|=\left|\dfrac{-\frac{26}{5}}{1+1}\right|

    =\left|\dfrac{-\frac{26}{5}}{2}\right|=\dfrac{13}{5}

    \Rightarrow\theta=\tan^{-1}(\frac{13}{5})\approx69.0^o.

    Thus, the third angle is 180 - 49.8 - 69.0 = 61.2^o.

    Since the three angles are less than 90^o then the triangle formed by ABC is an acute triangle.

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