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Figure 1 below shows a figure drawn on a cartesian plain, the vertices of which are A(2;6); B(-6; -4) and C(10; - 6): 4.1 Determine the length of AC 4.2 The midpoint of BC. 4.3 The angle of inclination of AB.

Mathematics

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The vertices of the figure are A(2; 6), B(-6; -4), and C(10; -6).

1.4.1 Determine the length of AC

  1. Use the distance formula D=(x2−x1)2+(y2−y1)2D = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}.

    For points A(2, 6) and C(10, -6):

    AC=(10−2)2+(−6−6)2AC = \sqrt{(10-2)^2 + (-6-6)^2}

  2. Calculate the differences in coordinates and square them.

    AC=(8)2+(−12)2AC = \sqrt{(8)^2 + (-12)^2} AC=64+144AC = \sqrt{64 + 144}

  3. Add the squared values and find the square root.

    AC=208AC = \sqrt{208} AC=16×13AC = \sqrt{16 \times 13} AC=413AC = 4\sqrt{13}

    The length of AC is \boxed{\text{4\sqrt{13} units}}.

    1.4.2 The midpoint of BC

  4. 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 B(-6, -4) and C(10, -6):

    MBC=(−6+102,−4+(−6)2)M_{BC} = \left(\frac{-6+10}{2}, \frac{-4+(-6)}{2}\right)

  5. Calculate the sums and divide by 2.

    MBC=(42,−102)M_{BC} = \left(\frac{4}{2}, \frac{-10}{2}\right) MBC=(2,−5)M_{BC} = (2, -5)

    The midpoint of BC is (2, -5)\boxed{\text{(2, -5)}}.

    1.4.3 The angle of inclination of AB

  6. Calculate the gradient (slope) of AB using the formula m=y2−y1x2−x1m = \frac{y_2-y_1}{x_2-x_1}.

    For points A(2, 6) and B(-6, -4):

    mAB=−4−6−6−2m_{AB} = \frac{-4-6}{-6-2} mAB=−10−8m_{AB} = \frac{-10}{-8} mAB=54m_{AB} = \frac{5}{4}

  7. Use the formula for the angle of inclination θ=arctan⁡(m)\theta = \arctan(m).

    tan⁡θ=54\tan \theta = \frac{5}{4} θ=arctan⁡(54)\theta = \arctan\left(\frac{5}{4}\right)

  8. Calculate the angle.

    θ≈51.34∘\theta \approx 51.34^\circ

    The angle of inclination of AB is approximately \boxed{\text{51.34^\circ}}.

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