Parallel and Perpendicular Lines

The figure shows two parallel horizontal lines. What are their slopes? Good. Actually, the slopes of parallel lines are always equal. Let me show you the proof. Yes. If now, we rotate the lines to the same extent, what do you think about their steepness and their slopes? The figure shows two parallel horizontal lines. What are their slopes? y slope = 0 slope = 0 x By observation, it seems that their steepness are always the same, so they have the same slopes. Both of the slopes are 0.

Parallel Lines The figure shows two straight lines L1 and L2, whose inclinations are 1 and 2 respectively. y L1 L2 1 2 If L1 // L2, then ∴ i.e. slope of L1= slope of L2 x 1 = 2 (corr. s, L1 // L2) tan 1 = tan 2

From the above result, we have If L1 // L2, then m1 = m2. The converse of the above result is also true: If m1 = m2, then L1 // L2.

Determine whether two lines AB and CD are parallel. y A(–1, 3) B(1, 6) D(8, 5) C(6, 2) ) 1 ( 3 6 of Slope - = AB 2 3 = 6 8 5 of Slope 2 CD - = 2 3 = x ∵ Slope of AB = slope of CD ∴ AB // CD

Follow-up question The figure shows four points P(6, 3), Q(2, 8), R(2, 5) and S(6, 6). Prove that PQ is parallel to RS. y R(2, 5) P(6, 3) x Solution S(6, 6) of Slope PQ = 6) ( 2 3 8 - 4 11 - = Q(2, 8) of Slope = RS 2 6 5 - 4 11 - = ∵ Slope of PQ = slope of RS ∴ PQ // RS

Perpendicular Lines Consider a point A(2, 1) in the figure. Rotate 90 Consider a point A(2, 1) in the figure. In fact, the slopes of two perpendicular lines are also related. We have learnt the relationship between the slopes of parallel lines. 1 O 1 3 2 x y 2 A Rotate A anti-clockwise about O through 90 to A. A Let me show you. Then, the coordinates of A are (1, 2). Slope of OA = 2 1 - 2 1 = Slope of OA = 1 2 - 2 - = Slope of OA  slope of OA = 2) ( 2 1 -  ∴ 1 - =

The converse of the above result is also true: In general, we have: If L1  L2, then m1  m2 = –1. Proof The converse of the above result is also true: If m1  m2 = –1, then L1  L2. Note: The results are not applicable when one of the straight lines is vertical.

In the figure, AB ⊥ CD. Find x. 2 1 1) ( 3 of Slope AB - = 3 4 - = y B(2, 1) A(1, 3) C(1, 0) D(3, x) 1) ( 3 of Slope x CD - = 4 x = ∵ AB ⊥ CD x ∴ 1 of slope Slope - =  CD AB 1 4 3 - =  x 3 = x

Follow-up question In the figure, AB ⊥ CD. Find x. Solution ) 1 ( 4 2  x y A(–1, –1) B(4, 2) C(3, –1) D(x, 2) Solution ) 1 ( 4 2 of Slope - = AB 5 3 = 3 ) 1 ( 2 CD of Slope - = x 3 - = x ∵ AB  CD ∴ Slope of AB  slope of CD = –1 1 3 5 - =  x 15 5 9 + - = x 5 6 = x