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Linear Equations Solutions Intermediate Algebra, Joseph Lee (1-3) Write the equation of the line, in slope-intercept form, that has the given slope and passes through the given point. 1. m = 3, (0, −2) Solution: For any equation with slope m and y-intercept (0, b), the equation is given as y = mx + b. Therefore, the equation of this line is y = 3x − 2. 2 2. m = − , (0, 4) 3 Solution: For any equation with slope m and y-intercept (0, b), the equation is given as y = mx + b. Therefore, the equation of this line is 2 y = − x + 4. 3 1 3. m = 1, 0, 2 Solution: For any equation with slope m and y-intercept (0, b), the equation is given as y = mx + b. Therefore, the equation of this line is 1 y =x+ . 2 (4-5) Identify the slope and y-intercept of the line. Write the equation of the line in slope-intercept form. 4. Note: Passes though (0, −1) and (1, 1). Solution: The y-intercept of the line is (0, −1). To calculate the slope, observe m= rise 2 = = 2. run 1 Thus, the equation of this line is y = 2x − 1. 5. Note: Passes though (0, 2) and (4, −1). Linear Equations Solutions Intermediate Algebra, Joseph Lee Solution: The y-intercept of the line is (0, 2). To calculate the slope, observe m= rise −3 = . run 4 Thus, the equation of this line is 3 y = − x + 2. 4 (6-7) Write the equation of the line, in slope-intercept form, that passes through the given points. 6. (0, 3), (4, −1) Solution: The y-intercept of the line is (0, 3). To calculate the slope, observe m= y2 − y1 −4 −1 − 3 = = −1. = x2 − x1 4−0 4 Thus, the equation of this line is y = −x + 3. 7. (0, −4), (5, 6) Solution: The y-intercept of the line is (0, −4). To calculate the slope, observe m= 10 y2 − y1 6 − (−4) = = = 2. x2 − x1 5−0 5 Thus, the equation of this line is y = 2x − 4. (8-13) Write the equation of the line, in slope-intercept form, that has the given slope and passes through the given point. 8. m = 3, (2, 5) Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 = m(x − x1 ). Thus, the equation of the line is y − 5 = 3(x − 2) y − 5 = 3x − 6 y = 3x − 1. Linear Equations Solutions Intermediate Algebra, Joseph Lee 1 9. m = − , (−4, 1) 2 Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 = m(x − x1 ). Thus, the equation of the line is 1 y − 1 = − (x − (−4)) 2 1 y − 1 = − (x + 4) 2 1 y−1= − x−2 2 1 y = − x − 1. 2 2 10. m = − , (6, −5) 3 Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 = m(x − x1 ). Thus, the equation of the line is 2 y − (−5) = − (x − 6) 3 2 y+5= − x+4 3 2 y = − x−1 3 11. m = 4 , (−10, −8) 5 Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 = m(x − x1 ). Thus, the equation of the line is y − (−8) = y+8= 4 (x + 10) 5 y+8= 4 x+8 5 y= 12. m = 1 , (−5, 3) 2 4 (x − (−10)) 5 4 x 5 Linear Equations Solutions Intermediate Algebra, Joseph Lee Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 = m(x − x1 ). Thus, the equation of the line is y−3= 1 (x − (−5)) 2 y−3= 1 (x + 5) 2 y−3= 1 5 x+ 2 2 y= 1 11 x+ 2 2 3 13. m = − , (6, −5) 4 Solution: The equation of a line with slope m that passes through a point (x1 , y1 ) is given by y − y1 = m(x − x1 ). Thus, the equation of the line is 3 y − (−5) = − (x − 6) 4 3 9 y+5= − x+ 4 2 1 3 y = − x− 4 2 (14-19) Write the equation of the line that passes through the given points. Write the equation in the following: (a) slope-intercept form (b) standard form. 14. (3, 4), (5, 10) Solution: To write the equation, we must find the slope of this line. m= y2 − y1 10 − 4 6 = = = 3. x2 − x1 5−3 2 We may choose either point to write the equation. y − 4 = 3(x − 3) y − 4 = 3x − 9 (a) y = 3x − 5 To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and make the coefficient of the x term positive. y = 3x − 5 −3x + y = −5 (b) 3x − y = 5 Linear Equations Solutions Intermediate Algebra, Joseph Lee 15. (4, −3), (−6, 2) Solution: To write the equation, we must find the slope of this line. y2 − y1 2 − (−3) 5 1 = = =− . x2 − x1 −6 − 4 −10 2 m= We may choose either point to write the equation. 1 y − (−3) = − (x − 4) 2 1 y+3= − x+2 2 1 y = − x−1 2 (a) To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and make the coefficient of the x term positive. 1 y = − x−1 2 1 x + y = −1 2 (b) x + 2y = −2 16. (−6, 4), (3, 1) Solution: To write the equation, we must find the slope of this line. m= y2 − y1 1−4 −3 1 = = =− . x2 − x1 3 − (−6) 9 3 We may choose either point to write the equation. 1 y − 4 = − (x − (−6)) 3 1 y − 4 = − (x + 6) 3 1 y−4= − x−2 3 (a) 1 y = − x+2 3 To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and make the coefficient of the x term positive. 1 y = − x+2 3 1 x+y = 2 3 (b) x + 3y = 6 Linear Equations Solutions Intermediate Algebra, Joseph Lee 17. (−4, 0), (2, −12) Solution: To write the equation, we must find the slope of this line. y2 − y1 −12 − 0 −12 = = = −2. x2 − x1 2 − (−4) 6 m= We may choose either point to write the equation. y − 0 = −2(x − (−4)) y = −2(x + 4) y = −2x − 8 (a) To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and make the coefficient of the x term positive. y = −2x − 8 2x + y = −8 (b) 18. (5, 2), (−3, −2) Solution: To write the equation, we must find the slope of this line. m= −4 1 −2 − 2 y2 − y1 = = . = x2 − x1 −3 − 5 −8 2 We may choose either point to write the equation. (a) y−2= 1 (x − 5) 2 y−2= 1 5 x− 2 2 y= 1 1 x− 2 2 To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and make the coefficient of the x term positive. y= 1 1 x− 2 2 1 1 − x+y = − 2 2 (b) x − 2y = 1 19. (−7, 6), (2, 3) Solution: To write the equation, we must find the slope of this line. m= y2 − y1 3−6 −3 1 = = =− . x2 − x1 2 − (−7) 9 3 Linear Equations Solutions Intermediate Algebra, Joseph Lee We may choose either point to write the equation. 1 y − 6 = − (x − (−7)) 3 1 y − 6 = − (x + 7) 3 1 7 y−6= − x− 3 3 (a) 1 11 y = − x+ 3 3 To write the equation in standard form, we must move the x’s to the left side, eliminate any fractions, and make the coefficient of the x term positive. 11 1 y = − x+ 3 3 1 11 x+y = 3 3 (b) x + 3y = 11 (20-25) Determine whether the lines are parallel, perpendicular, or neither. 1 x−2 3 1 y = x+4 3 20. y = Solution: The first line has a slope m1 = 13 , and the second line has a slope m2 = 13 . Since m1 = m2 , the lines are parallel. 21. y = −2x + 4 1 y =− x−5 2 Solution: The first line has a slope m1 = −2, and the second line has a slope m2 = − 12 . Since m1 6= m2 and m1 · m2 6= −1, the lines are neither parallel nor perpendicular. 22. 2x − 6y = 5 3x + y = 5 Solution: First, we will write both equations in slope-intercept form. (L1 ) 2x − 6y = 5 −6y = −2x + 5 1 5 x− 3 6 3x + y = 5 y= (L2 ) y = −3x + 5 Linear Equations Solutions Intermediate Algebra, Joseph Lee The first line has a slope m1 = 13 , and the second line has a slope m2 = −3. Since m1 · m2 = −1, the lines are perpendicular. 23. 3x − 5y = 15 5x − 3y = 10 Solution: First, we will write both equations in slope-intercept form. (L1 ) 3x − 5y = 15 −5y = −3x + 15 y= (L2 ) 3 x−3 5 5x − 3y = 10 −3y = −5x + 10 y= 10 5 x− 3 3 The first line has a slope m1 = 35 , and the second line has a slope m2 = 35 . Since m1 6= m2 and m1 · m2 6= −1, the lines are neither parallel nor perpendicular. 24. 4x − 2y = 8 −6x + 3y = 4 Solution: First, we will write both equations in slope-intercept form. (L1 ) 4x − 2y = 8 −2y = −4x + 8 y = 2x − 4 (L2 ) − 6x + 3y = 4 3y = 6x + 4 4 3 The first line has a slope m1 = 2, and the second line has a slope m2 = 2. Since m1 = m2 , the lines are parallel. y = 2x + 25. x = 3 y = −1 Solution: The equation x = 3 is a vertical line, and the equation y = −1 is a horizontal line. Any two vertical and horizontal lines are perpendicular. (26-30) Write the equation of the line that is parallel to the given line and passes through the given point. Write the equation in the following: (a) slope-intercept form (b) standard form. Linear Equations Solutions Intermediate Algebra, Joseph Lee 26. (3, 2), y = 3x − 4 Solution: Since the line is parallel to y = 3x − 4, we know the slope must be the same: m = 3. y − y1 = m(x − x1 ) y − 2 = 3(x − 3) y − 2 = 3x − 9 (a) y = 3x − 7 −3x + y = −7 (b) 3x − y = 7 2 27. (−6, 2), y = − x + 7 3 2 Solution: Since the line is parallel to y = − 32 x + 7, we know the slope must be the same: m = − . 3 y − y1 = m(x − x1 ) 2 y − 2 = − (x − (−6)) 3 2 y − 2 = − (x + 6) 3 2 y−2= − x−4 3 (a) 2 y = − x−2 3 2 x + y = −2 3 (b) 2x + 3y = −6 28. (−3, −5), 4x − y = 8 Solution: First, we will write 4x − y = 8 in slope-intercept form. 4x − y = 8 −y = −4x + 8 y = 4x − 8 Linear Equations Solutions Intermediate Algebra, Joseph Lee Since the line is parallel to y = 4x − 8, we know the slope must be the same: m = 4. y − y1 = m(x − x1 ) y − (−5) = 4(x − (−3)) y + 5 = 4(x + 3) y + 5 = 4x + 12 (a) y = 4x + 7 −4x + y = 7 (b) 4x − y = −7 29. (4, −5), 6x + 4y = 4 Solution: First, we will write 4x − y = 8 in slope-intercept form. 6x + 4y = 4 4y = −6x + 4 3 y = − x+1 2 Since the line is parallel to y = − 32 x + 1, we know the slope must be the same: m = − 23 . y − y1 = m(x − x1 ) 3 y − (−5) = − (x − 4) 2 3 y+5= − x+6 2 (a) 3 y = − x+1 2 3 x+y = 1 2 (b) 3x + 2y = 2 30. (3, −4), 3x − 5y = 15 Solution: First, we will write 3x − 5y = 15 in slope-intercept form. 3x − 5y = 15 −5y = −3x + 15 3 y = x−3 5 Linear Equations Solutions Intermediate Algebra, Joseph Lee Since the line is parallel to y = 35 x − 3, we know the slope must be the same: m = 35 . y − y1 = m(x − x1 ) y − (−4) = y+4= (a) y= 3 (x − 3) 5 3 9 x− 5 5 3 29 x− 5 5 3 29 − x+y = − 5 5 (b) 3x − 5y = 29 (31-35) Write the equation of the line that is perpendicular to the given line and passes through the given point. Write the equation in the following: (a) slope-intercept form (b) standard form. 31. (3, 2), y = 3x − 4 Solution: Since the line is perpendicular to y = 3x − 4, we know the slope must be its opposite reciprocal: m = − 31 . y − y1 = m(x − x1 ) 1 y − 2 = − (x − 3) 3 1 y−2= − x+1 3 (a) 1 y = − x+3 3 1 x+y = 3 3 (b) x + 3y = 9 2 32. (−6, 2), y = − x + 7 3 Solution: Since the line is perpendicular to y = − 32 x + 7, we know the slope must be its opposite reciprocal: Linear Equations Solutions Intermediate Algebra, Joseph Lee m = 32 . y − y1 = m(x − x1 ) (a) y−2= 3 (x − (−6)) 2 y−2= 3 (x + 6) 2 y−2= 3 x+9 2 y= 3 x + 11 2 3 − x + y = 11 2 (b) 3x − 2y = −22 33. (3, −6), 5x + 2y = 10 Solution: First, we will write 5x + 2y = 10 in slope-intercept form. 5x + 2y = 10 2y = −5x + 10 5 y = − x+5 2 Since the line is perpendicular to y = − 25 x + 5, we know the slope must be its opposite reciprocal: m = 52 . y − y1 = m(x − x1 ) y − (−6) = y+6= (a) y= 2 (x − 3) 5 2 6 x− 5 5 2 36 x− 5 5 36 2 − x+y = − 5 5 (b) 2x − 5y = 36 Linear Equations Solutions Intermediate Algebra, Joseph Lee 34. (−4, 6), 2x − 8y = 16 Solution: First, we will write 2x − 8y = 16 in slope-intercept form. 2x − 8y = 16 −8y = −2x + 16 y= 1 x−2 4 Since the line is perpendicular to y = 41 x − 2, we know the slope must be its opposite reciprocal: m = −4. y − y1 = m(x − x1 ) y − 6 = −4(x − (−4)) y − 6 = −4(x + 4) y − 6 = −4x − 16 (a) (b) y = −4x − 10 4x + y = −10 35. (5, 1), 10x − 5y = 15 Solution: First, we will write 10x − 5y = 15 in slope-intercept form. 10x − 5y = 15 −5y = −10x + 15 y = 2x − 3 Since the line is perpendicular to y = 2x − 3, we know the slope must be its opposite reciprocal: m = − 12 . y − y1 = m(x − x1 ) 1 y − 1 = − (x − 5) 2 1 5 y−1= − x+ 2 2 (a) 7 1 y = − x+ 2 2 1 7 x+y = 2 2 (b) x + 2y = 7 (36-37) Write the equation of the line with the given properties. 36. parallel to y = −3 through the point (5, 2) Linear Equations Solutions Intermediate Algebra, Joseph Lee Solution: Since y = −3 is a horizontal line, any line parallel to it is also horizontal. So the horizontal line that passes through (5, 2) is the line give by y = 2. 37. perpendicular to y = 1 through the point (−3, −5) Solution: Since y = 1 is a horizontal line, any line perpendicular to it is vertical. So the vertical line that passes through (−3, 5) is the line give by x = −3.