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Unformatted text preview: Chapter 1 Review of Straight Lines 1.1 Find the slope and y intercept of the following straight lines: (a) y = 4 x 5 (b) 3 x 4 y = 8 (c) 2 x = 3 y (d) y = 3 (e) 5 x 2 y = 23 Detailed Solution: (a) This is the slopeintercept equation for the line. The slope is 4 and the y intercept is 5. (b) Solving the equation for y yields y = 3 4 x 2. The slope is 3 4 and the y intercept is 2. (c) Solving the equation for y gives y = 2 3 x = 2 3 + 0. The slope is 2 3 and the y intercept is 0. (d) y = 3 can be written as y = 0 · x + 3. The slope is 0 and the y intercept is 3. (e) Solving the equation for y yields y = 5 2 x 23 2 . The slope is 5 2 and the y intercept is 23 2 . 1.2 Find the equations of the following straight lines (a) Through the points (2,0) and (1,5). (b) Through (3,1) with slope 1/2. (c) Through (10,2) with y intercept 10. (d) The straight line shown in Figure 1.1. v.2005.1  September 4, 2009 1 Math 102 Problems Chapter 1 1 y = 8 + 18 x  9 x 2 Figure 1.1: Figure for problem 1.2(d) Detailed Solution: (a) y x 2 = 5 1 2 = 5 so y = 5( x 2) = 5 x + 10. (b) y + 1 x 3 = 1 2 leads to y = 1 2 x 5 2 . (c) y 10 x = 2 10 10 = 4 5 . After simplifying we get y = 4 5 x + 10. (d) The quadratic shown has roots r 1 , 2 = ( 18 ± √ 18 2 4 · 8 · 9) / 2( 9) = 2 / 3 , 4 / 3 , The line goes through the larger of these roots, (4 / 3 , 0), and its yintercept is (0 , 1), as can be seen on the graph. Therefore, the slope is m = 1 / (4 / 3) = 3 / 4. The equation is then y = (3 / 4) x + 1. 1.3 Find the equations of the following straight lines: (a) Slope 4 and y intercept 3. (b) Slope 3 and x intercept 2 / 3. (c) Through the points (2 , 7) and ( 1 , 11). v.2005.1  September 4, 2009 2 Math 102 Problems Chapter 1 (d) Through the point (1 , 3) and the origin. (e) Through the intersection of the lines 3 x + 2 y = 19 and y = 4 x + 7 and through the point (2 , 7). (f) Through the origin and parallel to the line 2 x + 8 y = 3. (g) Through the point ( 2 , 5) and perpendicular to the line y = 1 2 x + 6. Detailed Solution: (a) The slopeintercept equation for a straight line is y = mx + b where m is the slope and b is the y intercept for the line, so the equation is y = 4 x + 3. (b) Since the slope of the line is 3, its slopeintercept equation has the form y = 3 x + b . To find the value of b , note that the line passes through the point ( 2 / 3 , 0) ( x intercept). Substituting x = 2 / 3, y = 0 into the equation gives: 0 = 3 parenleftbigg 2 3 parenrightbigg + b Solve for b 0 = 2 + b b = 2 So the equation of the line is y = 3 x + 2. (c) The slope of the line is 11 ( 7) 1 2 = 6 Using the pointslope equation with the point (2 , 7) yields: y ( 7) x 2 = 6 y + 7 = 6( x 2) y = 6 x + 12 7 y = 6 x + 5 (d) The line passes through (0 , 0) and (1 , 3). The slope of the line is 3 1 = 3 The y intercept is 0 since the line passes through the origin. Therefore, the equation of the line is y = 3 x ....
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This note was uploaded on 05/19/2011 for the course MATH 102 Math 102 taught by Professor Allard during the Winter '09 term at The University of British Columbia.
 Winter '09
 Allard

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