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Given points P(0, 3), Q(4, 1), R(2, -3), and S(-2, -1), show that PR = QS, determine the coordinates of N, the midpoint of QS, prove that PN ⊥ QS, and prove that P, N and R are collinear.

Mathematics

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2 parts
1.1
Shown:

PR = QS

1.2
(1, 0)

Step-by-step solution

10 steps

Part 1.1· 2 steps

Show that PR = QS

Given points P(0,3)P(0, 3), Q(4,1)Q(4, 1), R(2,−3)R(2, -3), and S(−2,−1)S(-2, -1).

  1. Calculate the length of PR using the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}.

    PR=(2−0)2+(−3−3)2PR = \sqrt{(2 - 0)^2 + (-3 - 3)^2} PR=(2)2+(−6)2PR = \sqrt{(2)^2 + (-6)^2} PR=4+36PR = \sqrt{4 + 36} PR=40PR = \sqrt{40}

  2. Calculate the length of QS using the distance formula.

    QS=(−2−4)2+(−1−1)2QS = \sqrt{(-2 - 4)^2 + (-1 - 1)^2} QS=(−6)2+(−2)2QS = \sqrt{(-6)^2 + (-2)^2} QS=36+4QS = \sqrt{36 + 4} QS=40QS = \sqrt{40}

    Since PR=40PR = \sqrt{40} and QS=40QS = \sqrt{40}, then PR=QSPR = QS.

Part 1.2(1, 0)

Determine the coordinates of N, the midpoint of QS.

  1. Use the midpoint formula M=(x1+x22,y1+y22)M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right) for points Q(4,1)Q(4, 1) and S(−2,−1)S(-2, -1).

    N=(4+(−2)2,1+(−1)2)N = \left(\frac{4 + (-2)}{2}, \frac{1 + (-1)}{2}\right) N=(22,02)N = \left(\frac{2}{2}, \frac{0}{2}\right) N=(1,0)N = (1, 0)

    The coordinates of N are (1,0)\boxed{(1, 0)}.

Part 1.3· 3 steps

Prove that PN ⊥\perp QS.

  1. Calculate the gradient of PN using points $P(0, 3)$ and $N(1, 0)$.

    mPN=0−31−0=−31=−3m_{PN} = \frac{0 - 3}{1 - 0} = \frac{-3}{1} = -3

  2. Calculate the gradient of QS using points Q(4,1)Q(4, 1) and S(−2,−1)S(-2, -1).

    mQS=−1−1−2−4=−2−6=13m_{QS} = \frac{-1 - 1}{-2 - 4} = \frac{-2}{-6} = \frac{1}{3}

  3. Multiply the gradients.

    mPN×mQS=−3×13=−1m_{PN} \times m_{QS} = -3 \times \frac{1}{3} = -1

    Since the product of their gradients is −1-1, PN⊥QSPN \perp QS.

Part 1.4· 2 steps

Prove that P, N and R are collinear.

  1. Calculate the gradient of PN using points $P(0, 3)$ and $N(1, 0)$.

    mPN=0−31−0=−3m_{PN} = \frac{0 - 3}{1 - 0} = -3

  2. Calculate the gradient of NR using points N(1,0)N(1, 0) and R(2,−3)R(2, -3).

    mNR=−3−02−1=−31=−3m_{NR} = \frac{-3 - 0}{2 - 1} = \frac{-3}{1} = -3

    Since mPN=mNRm_{PN} = m_{NR} and they share a common point N, P, N, and R are collinear.

Part 1.5· 2 steps

What kind of quadrilateral is PQRS? Give a reason.

  1. Determine properties of the diagonals.

    From 1.1, PR=QS=40PR = QS = \sqrt{40}, so the diagonals are equal in length.
    From 1.2, N is the midpoint of QS.
    From 1.4, N is also the midpoint of PR (as MPR=(0+22,3+(−3)2)=(1,0)M_{PR} = \left(\frac{0+2}{2}, \frac{3+(-3)}{2}\right) = (1,0), which is N).
    Since the diagonals bisect each other, PQRS is a parallelogram.
    From 1.3, PN⊥QSPN \perp QS. Since N lies on PR, this means the diagonals PR and QS are perpendicular.

  2. Classify the quadrilateral based on these properties.

    A parallelogram with equal diagonals is a rectangle.
    A parallelogram with perpendicular diagonals is a rhombus.
    A quadrilateral that is both a rectangle and a rhombus is a square.
    Therefore, PQRS is a square.

    Reason: The diagonals PR and QS are equal in length, bisect each other, and are perpendicular.

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