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Use the distance formula d = sqrt((x2-x1)2 + (y2-y1)2) with points C(2, 1) and G(1, -4).

Mathematics

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3 parts
13.2

26\sqrt{26}

13.3
-3
13.4

108.4∘108.4^\circ

Step-by-step solution

9 steps

are of G are (1,−4)\boxed{(1, -4)}.

Part 13.2· 2 steps

Calculate the length of CG. Leave your answer in surd form.

  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 C(2,1)C(2, 1) and G(1,−4)G(1, -4).

    CG=(1−2)2+(−4−1)2CG = \sqrt{(1-2)^2 + (-4-1)^2}

  2. Simplify the expression.

    CG=(−1)2+(−5)2CG = \sqrt{(-1)^2 + (-5)^2} CG=1+25CG = \sqrt{1 + 25} CG=26CG = \sqrt{26}

    The length of CG is 26\boxed{\sqrt{26}}.

Part 13.3-3

Calculate 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(4,−5)B(4, -5) and C(2,1)C(2, 1).

    mBC=1−(−5)2−4m_{BC} = \frac{1 - (-5)}{2 - 4}

  2. Simplify the expression.

    mBC=1+5−2m_{BC} = \frac{1 + 5}{-2} mBC=6−2m_{BC} = \frac{6}{-2} mBC=−3m_{BC} = -3

    The gradient of BC is −3\boxed{-3}.

Part 13.4

108.4∘108.4^\circ

Calculate the value of θ\theta, the angle of inclination of BC, rounded off to ONE decimal digit.

  1. Use the relationship tan⁡θ=mBC\tan \theta = m_{BC}.

    From 13.3, mBC=−3m_{BC} = -3.

    tan⁡θ=−3\tan \theta = -3

  2. Calculate θ\theta using the arctangent function.

    θ=arctan⁡(−3)\theta = \arctan(-3) θ≈−71.565∘\theta \approx -71.565^\circ

  3. Adjust the angle to be between 0∘0^\circ and 180∘180^\circ for the angle of inclination.

    θ=−71.565∘+180∘\theta = -71.565^\circ + 180^\circ θ=108.435∘\theta = 108.435^\circ

  4. Round the angle to one decimal digit.

    θ≈108.4∘\theta \approx 108.4^\circ

    The value of θ\theta is 108.4∘\boxed{108.4^\circ}.

Part 13.5· 1 step

Show that △ABC\triangle ABC is an isosceles triangle.

  1. Calculate the length of side AB using points A(−2,−3)A(-2, -3) and $B(4, -

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