图中显示了三个点 P(−6,0)P(-6, 0)、Q(4k,k−2)Q(4k, k-2) 和 R(8,5)R(8, 5),其中 PQPQ 垂直于 QRQR 且 k<0.k < 0.

  1. (i)

    证明 k=−1k = -1。

  2. (ii)

    PRPR 的垂直平分线交 y 轴于 SS。求 SS 的坐标。

  3. (iii)

    求四边形 PQRSPQRS 的面积。

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完整解法

13(i)

1. Find the gradients of the line segments $PQ$ and $QR$ in terms of $k$.

Gradient of PQPQ, mPQ=(k−2)−04k−(−6)=k−24k+6m_{PQ} = \frac{(k-2) - 0}{4k - (-6)} = \frac{k-2}{4k+6}

Gradient of QRQR, mQR=5−(k−2)8−4k=7−k8−4km_{QR} = \frac{5 - (k-2)}{8 - 4k} = \frac{7-k}{8-4k}

2. Apply the perpendicularity condition to form an equation in $k$.

Since PQ⊥QRPQ \perp QR, mPQ×mQR=−1m_{PQ} \times m_{QR} = -1.

(k−24k+6)(7−k8−4k)=−1\left(\frac{k-2}{4k+6}\right) \left(\frac{7-k}{8-4k}\right) = -1

(k−2)(7−k)=−(4k+6)(8−4k)(k-2)(7-k) = -(4k+6)(8-4k)

7k−k2−14+2k=−(32k−16k2+48−24k)7k - k^2 - 14 + 2k = -(32k - 16k^2 + 48 - 24k)

−k2+9k−14=16k2−8k−48-k^2 + 9k - 14 = 16k^2 - 8k - 48

3. Solve the resulting quadratic equation for $k$ and reject the invalid root.

17k2−17k−34=017k^2 - 17k - 34 = 0

k2−k−2=0k^2 - k - 2 = 0

(k−2)(k+1)=0(k-2)(k+1) = 0

k=2k = 2 or k=−1k = -1

Since it is given that k<0k < 0, we reject k=2k = 2.

Therefore, k=−1k = -1 (shown).

答案

k=−1k = -1 (shown)

13(ii)

1. Find the coordinates of the midpoint of $PR$.

Midpoint of PR=(−6+82,0+52)=(1,2.5)PR = \left( \frac{-6 + 8}{2}, \frac{0 + 5}{2} \right) = (1, 2.5)

2. Find the gradient of the line segment $PR$.

Gradient of PR=5−08−(−6)=514PR = \frac{5 - 0}{8 - (-6)} = \frac{5}{14}

3. Determine the gradient of the perpendicular bisector.

Gradient of perpendicular bisector =−1514=−145=−2.8= -\frac{1}{\frac{5}{14}} = -\frac{14}{5} = -2.8

4. Find the equation of the perpendicular bisector and determine its $y$-intercept.

Equation of the perpendicular bisector:
y−2.5=−2.8(x−1)y - 2.5 = -2.8(x - 1)
Since SS lies on the yy-axis, x=0x = 0:
y−2.5=−2.8(0−1)y - 2.5 = -2.8(0 - 1)
y=2.5+2.8=5.3y = 2.5 + 2.8 = 5.3
Coordinates of SS are (0,5.3)(0, 5.3).

答案

(0,5.3)(0, 5.3)

13(iii)

1. State the coordinates of the vertices and set up the shoelace formula

From earlier parts, the coordinates are P(−6,0)P(-6, 0), Q(−4,−3)Q(-4, -3), R(8,5)R(8, 5), and S(0,5.3)S(0, 5.3).
Arranging the vertices in counter-clockwise order, the area of quadrilateral PQRSPQRS is given by:
Area=12∣−6−480−60−355.30∣\text{Area} = \frac{1}{2} \left| \begin{matrix} -6 & -4 & 8 & 0 & -6 \\ 0 & -3 & 5 & 5.3 & 0 \end{matrix} \right|

2. Evaluate the shoelace formula to find the area

Area=12∣[(−6)(−3)+(−4)(5)+(8)(5.3)+(0)(0)]−[(0)(−4)+(−3)(8)+(5)(0)+(5.3)(−6)]∣\text{Area} = \frac{1}{2} | [(-6)(-3) + (-4)(5) + (8)(5.3) + (0)(0)] - [(0)(-4) + (-3)(8) + (5)(0) + (5.3)(-6)] |
=12∣(18−20+42.4+0)−(0−24+0−31.8)∣= \frac{1}{2} | (18 - 20 + 42.4 + 0) - (0 - 24 + 0 - 31.8) |
=12∣40.4−(−55.8)∣= \frac{1}{2} | 40.4 - (-55.8) |
=12(96.2)=48.1= \frac{1}{2} (96.2) = 48.1

答案

48.148.1 units2^2

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