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Given the vertices of a triangle P(1, -3), Q(7, 5), and R(-3, -5).

Mathematics

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[68]\begin{bmatrix} 6 \\ 8 \end{bmatrix}

[−4−2]\begin{bmatrix} -4 \\ -2 \end{bmatrix}

Step-by-step solution

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Here's the solution to the problem:

Given the vertices of a triangle P(1, -3), Q(7, 5), and R(-3, -5).

Part i· 2 steps

Express PQ⃗\vec{PQ} and PR⃗\vec{PR} as column vectors.

  1. Find vector PQ⃗\vec{PQ}.

    To find the vector from point P to point Q, subtract the coordinates of P from the coordinates of Q.

    PQ⃗=Q−P=[75]−[1−3]=[7−15−(−3)]=[65+3]\vec{PQ} = Q - P = \begin{bmatrix} 7 \\ 5 \end{bmatrix} - \begin{bmatrix} 1 \\ -3 \end{bmatrix} = \begin{bmatrix} 7-1 \\ 5-(-3) \end{bmatrix} = \begin{bmatrix} 6 \\ 5+3 \end{bmatrix} PQ⃗=[68]\vec{PQ} = \boxed{\begin{bmatrix} 6 \\ 8 \end{bmatrix}}

  2. Find vector PR⃗\vec{PR}.

    To find the vector from point P to point R, subtract the coordinates of P from the coordinates of R.

    PR⃗=R−P=[−3−5]−[1−3]=[−3−1−5−(−3)]=[−4−5+3]\vec{PR} = R - P = \begin{bmatrix} -3 \\ -5 \end{bmatrix} - \begin{bmatrix} 1 \\ -3 \end{bmatrix} = \begin{bmatrix} -3-1 \\ -5-(-3) \end{bmatrix} = \begin{bmatrix} -4 \\ -5+3 \end{bmatrix} PR⃗=[−4−2]\vec{PR} = \boxed{\begin{bmatrix} -4 \\ -2 \end{bmatrix}}

Part ii· 4 steps

Show that triangle PQR is isosceles.

To show that a triangle is isosceles, we need to calculate the lengths of its three sides. If at least two sides have equal length, the triangle is isosceles. The length (magnitude) of a vector [xy]\begin{bmatrix} x \\ y \end{bmatrix} is given by x2+y2\sqrt{x^2 + y^2}.

  1. Calculate the length of side PQ.

    Using PQ⃗=[68]\vec{PQ} = \begin{bmatrix} 6 \\ 8 \end{bmatrix}:

    ∣PQ⃗∣=62+82=36+64=100=10|\vec{PQ}| = \sqrt{6^2 + 8^2} = \sqrt{36 + 64} = \sqrt{100} = 10

    The length of PQ is 10 units.

  2. Calculate the length of side PR.

    Using PR⃗=[−4−2]\vec{PR} = \begin{bmatrix} -4 \\ -2 \end{bmatrix}:

    ∣PR⃗∣=(−4)2+(−2)2=16+4=20|\vec{PR}| = \sqrt{(-4)^2 + (-2)^2} = \sqrt{16 + 4} = \sqrt{20}

    The length of PR is 20\sqrt{20} units, which can be simplified to 252\sqrt{5} units.

  3. Calculate the length of side QR.

    First, find the vector QR⃗\vec{QR}. Subtract the coordinates of Q from the coordinates of R.

    QR⃗=R−Q=[−3−5]−[75]=[−3−7−5−5]=[−10−10]\vec{QR} = R - Q = \begin{bmatrix} -3 \\ -5 \end{bmatrix} - \begin{bmatrix} 7 \\ 5 \end{bmatrix} = \begin{bmatrix} -3-7 \\ -5-5 \end{bmatrix} = \begin{bmatrix} -10 \\ -10 \end{bmatrix}

    Now, calculate the length of QR⃗\vec{QR}:

    ∣QR⃗∣=(−10)2+(−10)2=100+100=200|\vec{QR}| = \sqrt{(-10)^2 + (-10)^2} = \sqrt{100 + 100} = \sqrt{200}

    The length of QR is 200\sqrt{200} units, which can be simplified to 10210\sqrt{2} units.

  4. Compare the lengths of the

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