a Recall that when the line is horizontal its slope is 0, and when the line is vertical its slope is undefined.
B
b What happens if two lines do not intersect? Recall that parallel lines have the same slope.
C
c You can consider y= bax and y=- abx. Consider a point on each line and the triangle formed with these points and the origin. Then, apply the Converse of the Pythagorean Theorem.
A
a See solution.
B
b See solution.
C
c See solution.
Practice makes perfect
a Let's begin by writing the given information and what we want to prove.
Given: & l_1 and l_2 such that m_1 * m_2 = -1
Prove: & l_1 ⊥ l_2
The first thing we will prove is that neither line can be horizontal nor vertical. We will do it by indirect reasoning.
Let's assume temporarily that l_1 is horizontal. This implies that m_1=0. Therefore, m_1* m_2 =0 which contradicts the fact that m_1 * m_2 = -1. In consequence, none of the lines are horizontal.
Let's assume temporarily that l_1 is vertical. In this case, m_1 is undefined and therefore m_1* m_2 is also undefined. This contradicts the fact that m_1 * m_2 = -1. In consequence, none of the lines are vertical.
b In this part, we will show that the lines intersect each other. As before, let's do it by indirect reasoning.
Original Conclusion
Opposite Conclusion
l_1 and l_2 intersect each other
l_1 and l_2 do not intersect each other
Let's assume temporarily that l_1 and l_2 do not intersect. This means that l_1∥ l_2 and so m_1=m_2. In consequence, m_1* m_2 = m_1^2, and we are also given that m_1* m_2 = -1.
m_1* m_2 = m_1^2 m_1 * m_2 = -1
↘ ↙
m_1^2 = -1
The latter equation has no real solutions, implying that m_1 is not a real number. This is a contradiction since m_1 is the slope of l_1. Therefore, the lines must intersect each other.
c Let's consider the lines l_1: y= bax and l_2: y=- abx. These two lines intersect each other at the origin.
Next, let's consider a point on l_1 and l_2. To find our points, we can substitute x=a into each equation. Also, we will draw the segment connecting these two points.
Similarly, we will find the distance from B to C, and from A to B. We will summarize the computations in the following table.
Points
d = sqrt((x_2-x_1)^2 + (y_2-y_1)^2)
Distance
C(0,0) and B(a,-a^2/b )
BC = sqrt((a-0)^2 + (-a^2/b-0)^2)
BC = sqrt(a^2 + a^4/b^2)
A(a,b) and B(a,-a^2/b )
AB = sqrt((a-a)^2 + (-a^2/b-b)^2)
AB = b + a^2/b
Let's square each of the distances we've found.
Distance
Distance Squared
AC = sqrt(a^2 + b^2)
AC^2 = a^2 + b^2
BC = sqrt(a^2 + a^4/b^2)
BC^2 = a^2 + a^4/b^2
AB = b + a^2/b
AB^2 = b^2 + 2a^2 + a^4/b^2
From the latter table, we can see that AC^2 + BC^2 = AB^2 and so, by the Converse of the Pythagorean Theorem, we conclude that △ ABC is a right triangle. This means that ∠ ACB is a right angle which implies that the lines are perpendicular.
