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In the diagram below, trapezium ABCD is drawn with vertices A(-6; 7), B(9; 4), C(0; -4) and D(t; 3). Calculate the distance of AB.

Mathematics

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6 parts
1.1

234 or approximately 15.30 units\text{}\sqrt{234} \text{ or approximately 15.30 units}

1.2

(−3,1.5)\text{}(-3, 1.5)

1.3

89\text{}\frac{8}{9}

1.4

y=89x−4\text{}y = \frac{8}{9}x - 4

1.5

approximately 41.63∘\text{approximately } 41.63^\circ

1.6

−10.5\text{} -10.5

Step-by-step solution

14 steps

Here are the solutions to the questions:

Part 1.1· 3 steps

Calculate the distance of AB.

  1. Use the distance formula d=(x2−x1)2+(y2−y1)2d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2} with points A(−6,7)(-6, 7) and B(9,4)(9, 4).

    dAB=(9−(−6))2+(4−7)2d_{AB} = \sqrt{(9 - (-6))^2 + (4 - 7)^2}

  2. Simplify the expression.

    dAB=(9+6)2+(−3)2d_{AB} = \sqrt{(9 + 6)^2 + (-3)^2} dAB=(15)2+9d_{AB} = \sqrt{(15)^2 + 9} dAB=225+9d_{AB} = \sqrt{225 + 9} dAB=234d_{AB} = \sqrt{234}

  3. Calculate the numerical value.

    dAB≈15.297d_{AB} \approx 15.297

    The distance of AB is 234 or approximately 15.30 units\boxed{\text{}\sqrt{234} \text{ or approximately 15.30 units}}.

Part 1.2

(−3,1.5)\text{}(-3, 1.5)

Determine the midpoint of AC.

  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) with points A(−6,7)(-6, 7) and C(0,−4)(0, -4).

    MAC=(−6+02,7+(−4)2)M_{AC} = \left(\frac{-6 + 0}{2}, \frac{7 + (-4)}{2}\right)

  2. Simplify the coordinates.

    MAC=(−62,32)M_{AC} = \left(\frac{-6}{2}, \frac{3}{2}\right) MAC=(−3,1.5)M_{AC} = \left(-3, 1.5\right)

    The midpoint of AC is \boxed{\text{}(-3, 1.5)}}.

Part 1.3

89\text{}\frac{8}{9}

Determine the gradient of BC.

  1. Use the gradient formula m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 - x_1} with points B(9,4)(9, 4) and C(0,−4)(0, -4).

    mBC=−4−40−9m_{BC} = \frac{-4 - 4}{0 - 9}

  2. Simplify the expression.

    mBC=−8−9m_{BC} = \frac{-8}{-9} mBC=89m_{BC} = \frac{8}{9}

    The gradient of BC is \boxed{\text{}\frac{8}{9}}}.

Part 1.4

y=89x−4\text{}y = \frac{8}{9}x - 4

Find the equation of the line BC.

  1. Use the point-slope form y−y1=m(x−x1)y - y_1 = m(x - x_1) with the gradient mBC=89m_{BC} = \frac{8}{9} and point C(0,−4)(0, -4).

    y−(−4)=89(x−0)y - (-4) = \frac{8}{9}(x - 0)

  2. Simplify the equation to the form y=mx+cy = mx + c.

    y+4=89xy + 4 = \frac{8}{9}x y=89x−4y = \frac{8}{9}x - 4

    The equation of the line BC is \boxed{\text{}y = \frac{8}{9}x - 4}}.

Part 1.5

approximately 41.63∘\text{approximately } 41.63^\circ

Determine the inclination angle, α\alpha, of the line BC.

  1. Use the relationship tan⁡α=m\tan \alpha = m, where mm is the gradient of BC.

    tan⁡α=89\tan \alpha = \frac{8}{9}

  2. Calculate the angle α\alpha.

    α=arctan⁡(89)\alpha = \arctan\left(\frac{8}{9}\right) α≈41.6339∘\alpha \approx 41.6339^\circ

    The inclination angle α\alpha is \boxed{\text{approximately } 41.63^\circ}}.

Part 1.6

−10.5\text{} -10.5

Determine the value of tt.

  1. For ABCD to be a trapezium, at least one pair of opposite sides must be parallel.

    From the diagram, AD appears parallel to BC. If AD is parallel to BC, their gradients must be equal: mAD=mBCm_{AD} = m_{BC}.
    We know mBC=89m_{BC} = \frac{8}{9} from question 1.3.

  2. Calculate the gradient of AD using points A(−6,7)(-6, 7) and D(t,3)(t, 3).

    mAD=3−7t−(−6)m_{AD} = \frac{3 - 7}{t - (-6)} mAD=−4t+6m_{AD} = \frac{-4}{t + 6}

  3. Set mAD=mBCm_{AD} = m_{BC} and solve for tt.

    −4t+6=89\frac{-4}{t + 6} = \frac{8}{9} −4×9=8×(t+6)-4 \times 9 = 8 \times (t + 6) −36=8t+48-36 = 8t + 48 −36−48=8t-36 - 48 = 8t −84=8t-84 = 8t t=−848t = \frac{-84}{8} t=−212t = -\frac{21}{2} t=−10.5t = -10.5

    The value of tt is \boxed{\text{} -10.5}}.

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