题目
Problem
Figure 2 shows the points P, Q and R.
Figure 2
Points P and Q have coordinates (−1,4) and (4,7) respectively.
(a) Find an equation for the straight line passing through points P and Q.
Give your answer in the form ax+by+c=0 where a, b and c are integers.
(3)
The point R has coordinates (p,−3), where p is a positive constant.
Given that angle QPR=90∘,
(b) find the value of p.
(Solutions relying on calculator technology are not acceptable.)
(3)
解答
(a)
解法一
思路
展开
先用两点求斜率,再代入点斜式,最后整理到 ax+by+c=0。
答题过程
展开
The gradient of PQ is
4−(−1)7−4=53.
Using point P(−1,4),
y−4=5y−20=3x−5y+23=53(x+1)3x+30.
(b)
解法一
思路
展开
因为 ∠QPR=90∘,所以 PR 垂直于 PQ。先求出 PR 的斜率应为 −35,再用 P(−1,4) 和 R(p,−3) 写斜率方程。
答题过程
展开
The gradient of PQ is 53, so the gradient of PR is
−35.
Using P(−1,4) and R(p,−3),
p−(−1)−3−4=p+1−7=−35−35.
Hence
21=21=5p=p=5(p+1)5p+516516.