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More on Parallel and Perpendicular Lines More on Parallel and Perpendicular Lines By Ma. Louise De Las Peñas et’s look at more problems on parallel and perpendicular lines. The following results, presented Lin the article, “About Slope” will be used to solve the problems. Theorem 1. Let L1 and L2 be two distinct nonvertical lines with slopes m1 and m2, respectively. Then L1 is parallel to L2 if and only if m1 = m2. Theorem 2. Let L1 and L2 be two distinct nonvertical lines with slopes m1 and m2, respectively. Then L1 is perpendicular to L2 if and only if m1m2 = -1. Example 1. The perpendicular bisector of a segment is the line which is perpendicular to the segment at its midpoint. Find equations of the three perpendicular bisectors of the sides of a triangle with vertices at A(-4,1), B(4,5) and C(4,-3). Solution: Consider the triangle with vertices A(-4,1), B(4,5) and C(4,-3), and the midpoints E, F, G of sides AB, BC, and AC, respectively. TATSULOK FirstSecond Year Year Vol. Vol. 12 No.12 No. 1a 2a e-Pages 1 In this problem, our aim is to fi nd the equations of the lines which pass through the midpoints of the sides of the triangle and at the same time are perpendicular to the sides of the triangle. Figure 1 The fi rst step is to fi nd the midpoint of every side. The next step is to fi nd the slope of the lines containing every side. The last step is to obtain the equations of the perpendicular bisectors. These equations will have slopes which are negative reciprocals of the slopes of the lines containing the sides by Theorem 2. To obtain the midpoint of every side, the midpoint formula is applied. The midpoint of the line segment x + x y + y with endpoints (x , y ) and (x , y ) is ( x, y) where x = 1 2 and y = 1 2 . To obtain the slope of a 1 1 2 2 2 2 y2 – y1 line containing points (x1, y1) and (x2, y2), we have the formula: slope = m = x2 – x1 (-4) + 4 1 + 5 5 – 1 4 1 For side AB: midpoint is E , = (0,3). The slope m = = = . The ( 2 2 ) AB 4 – (–4) 8 2 perpendicular bisector of AB will have slope -2. The perpendicular bisector intersects the y axis at (0,-1). Thus the equation is y = -2x + 3. 4 + 4 5 + –3 –3 –5 For side BC: midpoint is F , = (4,1). The slope m = is undefi ned. Note that BC ( 2 2 ) BC 4 – (–4) is a vertical line. The perpendicular bisector of BC will be a horizontal line. In particular, the line passing through (4,1). Hence the equation of this line is y = 1. (–4) + 4 1 + –3 – 3 – 1 4 1 For side AC: midpoint is F , = (0,-1). The slope m = = – = – . The ( 2 2 ) AC 4 – (–4) 8 2 perpendicular bisector of AC will have slope 2. The perpendicular bisector intersects the y axis at (0,-1). Thus the equation is y = 2x -1. Reference: Precalculus, De Las Peñas, Francisco, Marcelo, Editors, Offi ce of Research and Publications, Ateneo de Manila University 2006. TATSULOK FirstSecond Year Year Vol. Vol. 12 No.12 No. 1a 2a e-Pages 2 Answer the following problem: Find the equation(s) of the line(s) parallel to the line x + 4y + 3 = 0 and forms a right triangle with the coordinate axes with an area of 16 square units. TATSULOKTATSULOK FirstSecond YearYear Year Vol.Vol. Vol. 12 No.12 No. 1a 2a e-Pages 3 ANSWER Let l be the line parallel to x + 4y + 3 = 0, which forms a right triangle with the coordinate axes. Let a be the x-intercept of l and b be the y-intercept of l. The area of the triangle is 32 1 ab = 16 or b = , a ≠ 0 (1) 2 a The slope of l = – b . a Now note that l is parallel to x + 4y + 3 = 0. By Theorem 1, both lines will have the same slopes. The line x + 4y + 3 = 0 can be written as y = – x – 3 . Clearly, its slope is – 1 . Thus the slope of l is – b = – 1 . 4 4 4 a 4 Consequently, we obtain b = 1 or b = a . (2) a 4 4 Combining (1) and (2), we have 32 = a or a2 = 128 a 4 Solving for a we obtain a = ±8 2 . 32 If a = 8 2 , b = = 2 2 . On the other hand, if a = –8 2 , b = –2 2 . 8 2 This implies that there are two solutions! There will be two lines parallel to x + 4y + 3 = 0. One line will have y intercept 2 2 , the other line will have y intercept - 2 2 . b 1 b 1 The equations of the lines are l : y = – = – + 2 2 , l : y = – = – - 2 2 . 1 a 4 2 a 4 The lines are shown in Figure 2, l1 cuts through the fi rst quadrant, and l2 through the third quadrant. FigureFigure 2 2 TATSULOK FirstSecond Year Year Vol. Vol. 12 No.12 No. 1a 2a e-Pages 4.
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