上图显示了一个三角形 ABCABC,其顶点为 B(7,2)B(7,2) 和 C(−6,−2)C(-6,-2)。 ABAB 垂直于 ACAC,且平行于直线 y=−2xy = -2x。 MM 和 NN 分别是 ABAB 和 BCBC 的中点。

  1. (a)

    求 AA 的坐标。

    求 AA 的坐标。

  2. (b)

    解释为什么四边形 ACNM 是一个梯形。

  3. (c)

    求四边形 ACNM 的面积。

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完整解法

7(a)

1. Determine the gradient of line $AB$

Since ABAB is parallel to y=−2xy = -2x, the gradient of ABAB, mAB=−2m_{AB} = -2.

2. Find the equation of line $AB$

Using point B(7,2)B(7, 2) and mAB=−2m_{AB} = -2:
y−2=−2(x−7)y - 2 = -2(x - 7)
y=−2x+14+2y = -2x + 14 + 2
y=−2x+16y = -2x + 16

3. Determine the gradient of line $AC$

Since ABAB is perpendicular to ACAC, mAB×mAC=−1m_{AB} \times m_{AC} = -1.
−2×mAC=−1  ⟹  mAC=12-2 \times m_{AC} = -1 \implies m_{AC} = \frac{1}{2}.

4. Find the equation of line $AC$

Using point C(−6,−2)C(-6, -2) and mAC=12m_{AC} = \frac{1}{2}:
y−(−2)=12(x−(−6))y - (-2) = \frac{1}{2}(x - (-6))
y+2=12x+3y + 2 = \frac{1}{2}x + 3
y=12x+1y = \frac{1}{2}x + 1

5. Solve the equations simultaneously to find the coordinates of $A$

Equating the two expressions for yy:
−2x+16=12x+1-2x + 16 = \frac{1}{2}x + 1
15=52x15 = \frac{5}{2}x
x=15×25=6x = 15 \times \frac{2}{5} = 6
Substitute x=6x = 6 into y=−2x+16y = -2x + 16:
y=−2(6)+16=4y = -2(6) + 16 = 4
The coordinates of AA are (6,4)(6, 4).

答案

(6,4)(6, 4)

7(b)

1. Use the Midpoint Theorem to establish parallel lines

In △ABC\triangle ABC, MM and NN are given as the mid-points of sides ABAB and BCBC respectively. By the Midpoint Theorem, the line segment connecting these mid-points is parallel to the third side, so MN∥ACMN \parallel AC.

2. Conclude the shape of the quadrilateral

A quadrilateral with at least one pair of parallel opposite sides is a trapezium. Since MN∥ACMN \parallel AC, quadrilateral ACNMACNM is a trapezium.

答案

By the Midpoint Theorem, MNMN is parallel to ACAC. Since quadrilateral ACNMACNM has a pair of parallel sides, it is a trapezium.

7(c)

1. State the coordinates of $A$ (from part a) and find the coordinates of $M$ and $N$.

From part (a), AA is (6,4)(6, 4).
MM is the mid-point of ABAB:
M=(6+72,4+22)=(6.5,3)M = \left(\frac{6+7}{2}, \frac{4+2}{2}\right) = (6.5, 3)
NN is the mid-point of BCBC:
N=(7+(−6)2,2+(−2)2)=(0.5,0)N = \left(\frac{7+(-6)}{2}, \frac{2+(-2)}{2}\right) = (0.5, 0)

2. Apply the shoelace formula to find the area of quadrilateral $ACNM$.

Using the vertices A(6,4)A(6, 4), C(−6,−2)C(-6, -2), N(0.5,0)N(0.5, 0), and M(6.5,3)M(6.5, 3) in counter-clockwise order:
Area=12∣6−60.56.564−2034∣\text{Area} = \frac{1}{2} \left| \begin{matrix} 6 & -6 & 0.5 & 6.5 & 6 \\ 4 & -2 & 0 & 3 & 4 \end{matrix} \right|
Area=12∣(6)(−2)+(−6)(0)+(0.5)(3)+(6.5)(4)−[(4)(−6)+(−2)(0.5)+(0)(6.5)+(3)(6)]∣\text{Area} = \frac{1}{2} \left| (6)(-2) + (-6)(0) + (0.5)(3) + (6.5)(4) - [ (4)(-6) + (-2)(0.5) + (0)(6.5) + (3)(6) ] \right|
Area=12∣(−12+0+1.5+26)−(−24−1+0+18)∣\text{Area} = \frac{1}{2} \left| (-12 + 0 + 1.5 + 26) - (-24 - 1 + 0 + 18) \right|
Area=12∣15.5−(−7)∣\text{Area} = \frac{1}{2} \left| 15.5 - (-7) \right|
Area=12(22.5)=11.25\text{Area} = \frac{1}{2} (22.5) = 11.25

答案

11.2511.25 units2^2

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