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Problem of the Week
Problem D and Solution
Slope on a Plane

Problem

Percy drew \(x\) and \(y\) axes on grid paper and then plotted the point \(P(8,5).\) Quinlan then chose a different point, \(Q\), and said its coordinates were each positive integers less than or equal to \(20.\) Determine the probability that the slope of \(PQ\) is \(0\), \(1\), or \(2.\)

Solution

First, suppose the slope of \(PQ\) is \(0.\) Then \(Q\) must have a \(y\)-coordinate of \(5.\) Since the coordinates of \(Q\) are each positive integers less than or equal to \(20\), the smallest possible \(x\)-coordinate is \(1\) and the largest is \(20.\) Thus, there are \(20-1+1=20\) possible points. However, this includes the point \(P\), so there are \(20-1=19\) possibilities for \(Q\) such that \(PQ\) has slope \(0.\)

Next, suppose the slope of \(PQ\) is \(1.\) Let the coordinates of \(Q\) be \((a,b).\) Then \[\begin{aligned} \frac{b-5}{a-8} &= 1\\ b -5 &= a-8\\ b &= a-3 \end{aligned}\] Since \(a\) and \(b\) are each positive integers less than or equal to \(20\), the point with the smallest possible value for \(a\) is \((4,1).\) Similarly, the point with the largest possible value for \(a\) is \((20,17).\) Since \(b =a-3\), there is a possible value for \(b\) for each value of \(a\) between \(4\) and \(20.\) Thus, there are \(20-4+1=17\) possible points. However, this includes the point \(P\), so there are \(17-1=16\) possibilities for \(Q\) such that \(PQ\) has slope \(1.\)

Finally, suppose the slope of \(PQ\) is \(2.\) Let the coordinates of \(Q\) be \((a,b).\) Then \[\begin{aligned} \frac{b-5}{a-8} &= 2\\ b -5 &= 2(a-8)\\ b-5 &= 2a -16\\ b &= 2a-11 \end{aligned}\] Since \(a\) and \(b\) are each positive integers less than or equal to \(20\), the point with the smallest possible value for \(a\) is \((6,1).\) To determine the point with the largest possible value for \(a\), we first notice that if \(a=20\), then \(b>20.\) Then we can set \(b=2a-11<20.\) Thus \(2a<31\), or \(a<15.5.\) It follows that the point with the largest possible value for \(a\) is \((15,19).\) Since \(b =2a-11\), there is a possible value for \(b\) for each value of \(a\) between \(6\) and \(15.\) Thus, there are \(15-6+1=10\) possible points. However, this includes the point \(P\), so there are \(10-1=9\) possibilities for \(Q\) such that \(PQ\) has slope \(2.\)

Thus, the total number of possibilities for \(Q\) such that \(PQ\) has slope \(0\), \(1\), or \(2\) is \(19+16+9=44.\) The total number of possibilities for \(Q\) is \(20 \times 20 -1 =399.\) Thus, the probability that the slope of \(PQ\) is \(0\), \(1\), or \(2\) is equal to \(\frac{44}{399}\), or approximately \(11\%.\)