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Calculate the size of angle A and the coordinates of the intersection of diagonals of parallelogram ACES, given coordinates for A, C, and E.

Mathematics

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2 parts
3.2

34.70∘34.70^\circ

3.3
(5; 5)
(16; 16)

Step-by-step solution

11 steps

Here are the solutions for 3.2, 3.3, and 4:

Part 3.2· 4 steps

Size of A^\hat{A}

  1. Calculate the gradient of ACAC.

    Given A(−2;10)A(-2; 10) and C(4;−2)C(4; -2).

    mAC=−2−104−(−2)=−126=−2m_{AC} = \frac{-2 - 10}{4 - (-2)} = \frac{-12}{6} = -2

  2. Use the gradient of ABAB, mAB=7m_{AB} = 7, from 1.3.

  3. Use the formula for the angle θ\theta between two lines with gradients m1m_1 and m2m_2

    tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|.
    Let m1=mAB=7m_1 = m_{AB} = 7 and m2=mAC=−2m_2 = m_{AC} = -2.

    tan⁡A^=∣7−(−2)1+(7)(−2)∣\tan \hat{A} = \left| \frac{7 - (-2)}{1 + (7)(-2)} \right| tan⁡A^=∣91−14∣\tan \hat{A} = \left| \frac{9}{1 - 14} \right| tan⁡A^=∣9−13∣\tan \hat{A} = \left| \frac{9}{-13} \right| tan⁡A^=913\tan \hat{A} = \frac{9}{13}

  4. Calculate the angle A^\hat{A}.

    A^=arctan⁡(913)\hat{A} = \arctan\left(\frac{9}{13}\right) A^≈34.70∘\hat{A} \approx 34.70^\circ

    The size of A^\hat{A} is 34.70∘\boxed{34.70^\circ}.

Part 3.3(5; 5)

Coordinates of the point of intersection of the diagonals of parallelogram ACES, where S is a point in the first quadrant.

  1. The diagonals of a parallelogram bisect each other.

    The point of intersection is the midpoint of either diagonal.

  2. Use the coordinates of A(−2;10)A(-2; 10) and E(12;0)E(12; 0) to find the midpoint MM of diagonal AEAE.

    Mx=xA+xE2=−2+122=102=5M_x = \frac{x_A + x_E}{2} = \frac{-2 + 12}{2} = \frac{10}{2} = 5 My=yA+yE2=10+02=102=5M_y = \frac{y_A + y_E}{2} = \frac{10 + 0}{2} = \frac{10}{2} = 5

    The coordinates of the point of intersection of the diagonals are (5;5)\boxed{(5; 5)}.

    4 Calculate the coordinates of TT.

  3. Calculate the length of BEBE using B(−4;−4)B(-4; -4) and E(12;0)E(12; 0).

    BE=(12−(−4))2+(0−(−4))2BE = \sqrt{(12 - (-4))^2 + (0 - (-4))^2} BE=(16)2+(4)2BE = \sqrt{(16)^2 + (4)^2} BE=256+16BE = \sqrt{256 + 16} BE=272BE = \sqrt{272} BE=16×17=417 unitsBE = \sqrt{16 \times 17} = 4\sqrt{17} \text{ units}

    This confirms the given length.

  4. Use the distance formula for ETET with E(12;0)E(12; 0) and T(p;p)T(p; p), and set ET=417ET = 4\sqrt{17}.

    ET=(p−12)2+(p−0)2ET = \sqrt{(p - 12)^2 + (p - 0)^2} 417=(p−12)2+p24\sqrt{17} = \sqrt{(p - 12)^2 + p^2}

  5. Square both sides and solve for pp.

    (417)2=(p−12)2+p2(4\sqrt{17})^2 = (p - 12)^2 + p^2 16×17=p2−24p+144+p216 \times 17 = p^2 - 24p + 144 + p^2 272=2p2−24p+144272 = 2p^2 - 24p + 144 0=2p2−24p+144−2720 = 2p^2 - 24p + 144 - 272 0=2p2−24p−1280 = 2p^2 - 24p - 128

    Divide by 2:

    0=p2−12p−640 = p^2 - 12p - 64

  6. Factor the quadratic equation.

    (p−16)(p+4)=0(p - 16)(p + 4) = 0

    This gives two possible values for pp: p=16p = 16 or p=−4p = -4.

  7. Apply the condition p>0p > 0.

    Since p>0p > 0, we choose p=16p = 16.
    The coordinates of T(p;p)T(p; p) are (16;16)\boxed{(16; 16)}.

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