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1 Determine the equation of line AD in the form y = mx + c. 3.2 Determine the coordinates of D.

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5 parts
3.1

y=−12x+92y = -\frac{1}{2}x + \frac{9}{2}

3.2
(9; 0)
3.4

55 units5\sqrt{5} \text{ units}

3.5

25 square units25 \text{ square units}

3.6

45∘45^\circ

Step-by-step solution

15 steps

Here's the solution to the analytical geometry problem.

Given:
Parallelogram ABCD
A (−1;5)(-1; 5)
B (2;6)(2; 6)
Vertex D lies on the x-axis.
Equation of BC is x+2y=14x + 2y = 14.

Part 3.1· 3 steps

Determine the equation of line AD in the form y=mx+cy = mx + c.

  1. Find the gradient of BC.

    The equation of BC is x+2y=14x + 2y = 14.
    Rearrange to y=mx+cy = mx + c form:

    2y=−x+142y = -x + 14 y=−12x+7y = -\frac{1}{2}x + 7

    The gradient of BC is mBC=−12m_{BC} = -\frac{1}{2}.

  2. Determine the gradient of AD.

    Since ABCD is a parallelogram, AD is parallel to BC. Therefore, their gradients are equal.

    mAD=mBC=−12m_{AD} = m_{BC} = -\frac{1}{2}

  3. Use point A (−1;5)(-1; 5) and the gradient mADm_{AD} to find the equation of AD.

    Using the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1):

    y−5=−12(x−(−1))y - 5 = -\frac{1}{2}(x - (-1)) y−5=−12(x+1)y - 5 = -\frac{1}{2}(x + 1) y−5=−12x−12y - 5 = -\frac{1}{2}x - \frac{1}{2} y=−12x−12+5y = -\frac{1}{2}x - \frac{1}{2} + 5 y=−12x+92y = -\frac{1}{2}x + \frac{9}{2}

    The equation of line AD is y=−12x+92\boxed{y = -\frac{1}{2}x + \frac{9}{2}}.

Part 3.2(9; 0)

Determine the coordinates of D.

  1. Use the fact that D lies on the x-axis.

    If D lies on the x-axis, its y-coordinate is 0. So, D (xD;0)(x_D; 0).

  2. Substitute y=0y = 0 into the equation of line AD.

    From 3.1, the equation of AD is y=−12x+92y = -\frac{1}{2}x + \frac{9}{2}.

    0=−12xD+920 = -\frac{1}{2}x_D + \frac{9}{2}

    Multiply by 2 to clear the denominators:

    0=−xD+90 = -x_D + 9 xD=9x_D = 9

    The coordinates of D are (9;0)\boxed{(9; 0)}.

Part 3.3· 2 steps

If the coordinates of F are $(10; 2)$, show that DF is perpendicular to BC.

  1. Calculate the gradient of DF.

    Given F (10;2)(10; 2) and D (9;0)(9; 0).

    mDF=yF−yDxF−xD=2−010−9=21=2m_{DF} = \frac{y_F - y_D}{x_F - x_D} = \frac{2 - 0}{10 - 9} = \frac{2}{1} = 2

  2. Compare mDFm_{DF} with mBCm_{BC}.

    From 3.1, mBC=−12m_{BC} = -\frac{1}{2}.
    For two lines to be perpendicular, the product of their gradients must be −1-1.

    mDF×mBC=2×(−12)=−1m_{DF} \times m_{BC} = 2 \times \left(-\frac{1}{2}\right) = -1

    Since the product of their gradients is −1-1, DF is perpendicular to BC.

Part 3.4

55 units5\sqrt{5} \text{ units}

Calculate the length of AD. (Leave your answer in surd form.)

  1. Use the distance formula.

    A (−1;5)(-1; 5) and D (9;0)(9; 0).

    AD=(xD−xA)2+(yD−yA)2AD = \sqrt{(x_D - x_A)^2 + (y_D - y_A)^2} AD=(9−(−1))2+(0−5)2AD = \sqrt{(9 - (-1))^2 + (0 - 5)^2} AD=(9+1)2+(−5)2AD = \sqrt{(9 + 1)^2 + (-5)^2} AD=(10)2+25AD = \sqrt{(10)^2 + 25} AD=100+25AD = \sqrt{100 + 25} AD=125AD = \sqrt{125}

  2. Simplify the surd.

    AD=25×5=25×5=55AD = \sqrt{25 \times 5} = \sqrt{25} \times \sqrt{5} = 5\sqrt{5}

    The length of AD is 55 units\boxed{5\sqrt{5} \text{ units}}.

Part 3.5

25 square units25 \text{ square units}

Hence, or otherwise, calculate the area of parallelogram ABCD.

  1. Find the length of BC.

    Since ABCD is a parallelogram, AD=BCAD = BC.
    From 3.4, AD=55AD = 5\sqrt{5}.
    So, BC=55BC = 5\sqrt{5}.

  2. Find the perpendicular height of the parallelogram.

    The perpendicular height from D to BC is the length of DF, since DF is perpendicular to BC (shown in 3.3).
    D (9;0)(9; 0) and F (10;2)(10; 2).

    DF=(xF−xD)2+(yF−yD)2DF = \sqrt{(x_F - x_D)^2 + (y_F - y_D)^2} DF=(10−9)2+(2−0)2DF = \sqrt{(10 - 9)^2 + (2 - 0)^2} DF=(1)2+(2)2DF = \sqrt{(1)^2 + (2)^2} DF=1+4DF = \sqrt{1 + 4} DF=5DF = \sqrt{5}

  3. Calculate the area of the parallelogram.

    Area of parallelogram = base ×\times perpendicular height.
    Using BC as the base and DF as the perpendicular height:

    Area=BC×DF\text{Area} = BC \times DF Area=(55)×(5)\text{Area} = (5\sqrt{5}) \times (\sqrt{5}) Area=5×(5×5)\text{Area} = 5 \times (\sqrt{5} \times \sqrt{5}) Area=5×5\text{Area} = 5 \times 5 Area=25\text{Area} = 25

    The area of parallelogram ABCD is 25 square units\boxed{25 \text{ square units}}.

Part 3.6

45∘45^\circ

Calculate the size of ∠ABC\angle ABC.

  1. Find the gradient of AB.

    A (−1;5)(-1; 5) and B (2;6)(2; 6).

    mAB=yB−yAxB−xA=6−52−(−1)=13m_{AB} = \frac{y_B - y_A}{x_B - x_A} = \frac{6 - 5}{2 - (-1)} = \frac{1}{3}

  2. Find the angle of inclination for AB and BC.

    Let α\alpha be the angle of inclination for AB and β\beta be the angle of inclination for BC.

    mAB=tan⁡α=13  ⟹  α=arctan⁡(13)≈18.43∘m_{AB} = \tan \alpha = \frac{1}{3} \implies \alpha = \arctan\left(\frac{1}{3}\right) \approx 18.43^\circ mBC=tan⁡β=−12m_{BC} = \tan \beta = -\frac{1}{2}

    Since mBCm_{BC} is negative, β\beta is an obtuse angle.
    The reference angle is arctan⁡(∣−12∣)=arctan⁡(12)≈26.57∘\arctan\left(\left|-\frac{1}{2}\right|\right) = \arctan\left(\frac{1}{2}\right) \approx 26.57^\circ.
    So, β=180∘−26.57∘=153.43∘\beta = 180^\circ - 26.57^\circ = 153.43^\circ.

  3. Calculate ∠ABC\angle ABC.

    ∠ABC\angle ABC is the angle between line AB and line BC.
    Using the formula for the angle between two lines: tan⁡θ=∣m1−m21+m1m2∣\tan \theta = \left|\frac{m_1 - m_2}{1 + m_1 m_2}\right|.
    Let m1=mBC=−12m_1 = m_{BC} = -\frac{1}{2} and m2=mAB=13m_2 = m_{AB} = \frac{1}{3}.

    tan⁡(∠ABC)=∣−12−131+(−12)(13)∣\tan(\angle ABC) = \left|\frac{-\frac{1}{2} - \frac{1}{3}}{1 + \left(-\frac{1}{2}\right)\left(\frac{1}{3}\right)}\right| tan⁡(∠ABC)=∣−36−261−16∣\tan(\angle ABC) = \left|\frac{-\frac{3}{6} - \frac{2}{6}}{1 - \frac{1}{6}}\right| tan⁡(∠ABC)=∣−5656∣\tan(\angle ABC) = \left|\frac{-\frac{5}{6}}{\frac{5}{6}}\right| tan⁡(∠ABC)=∣−1∣\tan(\angle ABC) = |-1| tan⁡(∠ABC)=1\tan(\angle ABC) = 1 ∠ABC=arctan⁡(1)\angle ABC = \arctan(1) ∠ABC=45∘\angle ABC = 45^\circ

    The size of ∠ABC\angle ABC is 45∘\boxed{45^\circ}.

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