Question

两条相互垂直的直线交于点 $A(6,8)$ ，它们的 $y$ 轴截距为 $P$ 和 $Q$ ，且两个截距之和为零。求 $\triangle APQ$ 的面积。

The $y$ -intercepts, $P$ and $Q,$ of two perpendicular lines intersecting at the point $A(6,8)$ have a sum of zero. What is the area of $\triangle APQ?$

$45$

$48$

$54$

$60$

$72$

答案：D

知识点：坐标几何中线（几何）三角形面积

难度评级：1660

解答：

两个 $y$ 轴截距到原点的距离相等，因为它们的数值之和为 $0.$

设这个距离为 $z.$ 因为两条给定直线互相垂直， $\triangle APQ$ 在 $A$ 处为直角。原点是斜边 $PQ$ 的中点，所以它到 $P$ 、 $Q$ 和 $A$ 的距离相等。因此 $A$ 到原点的距离也是 $z$ 。

由距离公式， $z = \sqrt{6^2 + 8^2} = 10$ 。从 $A$ 到 $\overline{PQ}$ 的高为 $6$ （也就是 $A$ 的 $x$ 坐标）。

又有 $PQ = 2 \cdot 10 = 20,$ 所以面积为 $[APQ] = \dfrac{1}{2} \cdot 6 \cdot 20 = 60.$

所以正确答案是 D。

We have that the $y$ -intercepts are an equal distance from the origin since their values sum to $0.$

Let this distance be $z.$ Because the two given lines are perpendicular, $\triangle APQ$ is right at $A$ . The origin is the midpoint of its hypotenuse $PQ$ , so it is equidistant from $P$ , $Q$ , and $A$ . Hence the distance from $A$ to the origin is also $z$ .

We then know that $z = \sqrt{6^2 + 8^2} = 10$ by the distance formula. We know the altitude from $A$ to $\overline{PQ}$ is $6$ (it is just the $x$ -value of $A$ ).

We also know that $PQ = 2 \cdot 10 = 20,$ which tells us that the area $[APQ] = \dfrac{1}{2} \cdot 6 \cdot 20 = 60.$

Thus, D is the correct answer.

2014 AMC 10A 第 14 题

答案：D

难度评级：1660

解答：

其他年份的第 14 题