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East Campus, CB 117 361-698-1579 Math Learning Center West Campus, HS1 203 361-698-1860 ALL ABOUT LINES 1. Different forms of writing an equation of a line a) Slope-Intercept Form example: y = 𝑚𝑚x + 𝑏𝑏 y − 𝑦𝑦1 = 𝑚𝑚(x − 𝑥𝑥1 ) b) Point-Slope Form example: 3 𝑦𝑦 = 𝑥𝑥 − 3 4 3 𝑦𝑦 − 3 = (𝑥𝑥 − 8) 4 c) Standard Form* 𝐴𝐴x + 𝐵𝐵y = 𝐶𝐶 example: 3𝑥𝑥 − 4𝑦𝑦 = 12 *In standard form A, B, and C must be integers (no decimals or fractions) and A must be positive. 2. Vertical lines – any equation in the form of x = constant is a vertical line with a slope of undefined example: 𝑥𝑥 = 3 (x is always 3, y can be anything) x y 3 1 3 -2 3. Horizontal lines – any equation in the form of y = constant is a horizontal line with a slope of zero example: 𝑦𝑦 = −2 (y is always -2, x can be anything) x y -1 -2 2 -2 4. Finding the slope of a line 𝑦𝑦 −𝑦𝑦 𝑦𝑦 −𝑦𝑦 a) if given two points – use the formula 𝑚𝑚 = 𝑥𝑥2 −𝑥𝑥1 or 𝑚𝑚 = 𝑥𝑥1 −𝑥𝑥2 , where (𝑥𝑥1 , 𝑦𝑦1 ) and (𝑥𝑥2 , 𝑦𝑦2 ) are given 2 points example: if given (2, 3) and (-1, 4) 𝑚𝑚 = 1 4−3 −1−2 1 = 1 −3 =− 2 1 3 b) if given an equation – use the slope-intercept form 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑏𝑏 by solving for y and the coefficient of x will be the slope. example: if given 3𝑥𝑥 − 4𝑦𝑦 = 12, isolate the variable y by first subtracting the 3x −3𝑥𝑥 − 3𝑥𝑥 −4𝑦𝑦 = −3𝑥𝑥 + 12 divide by -4 to get y alone −4𝑦𝑦 −4 = 3 −3 −4 𝑥𝑥 + 12 −4 𝑦𝑦 = 𝑥𝑥 − 3 the coefficient of x is 4 3 4 , so the slope is 3 4 5. Finding the x-intercept Let the y variable equal zero and then solve for x- then the ordered pair will be (𝑥𝑥, 0) example: if given 3𝑥𝑥 − 4𝑦𝑦 = 8, let y = 0 8 Therefore the ordered pair is � , 0� 3 3𝑥𝑥 − 4(0) = 8 3𝑥𝑥 = 8 so 𝑥𝑥 = 8 3 East Campus, CB 117 361-698-1579 Math Learning Center West Campus, HS1 203 361-698-1860 6. Finding the y-intercept Let the x-variable equal zero and then solve for y- then the ordered pair will be (0, 𝑦𝑦) example: if given 3𝑥𝑥 − 4𝑦𝑦 = 8, let x = 0 3(0) − 4𝑦𝑦 = 8 −4𝑦𝑦 = 8 so 𝑦𝑦 = −2 Therefore the ordered pair is (0,-2) 7. Finding the equation of a line a) if the slope and a point on the line is known use the formula 𝑦𝑦 − 𝑦𝑦1 = 𝑚𝑚(𝑥𝑥 − 𝑥𝑥1 ), where (𝑥𝑥1 , 𝑦𝑦1 ) is a given point and m is the slope. example: if given (2, 3) and 𝑚𝑚 = −4, 𝑦𝑦 − 3 = −4(𝑥𝑥 − 2) 𝑦𝑦 − 3 = −4𝑥𝑥 + 8 so 𝑦𝑦 = −4𝑥𝑥 + 11 b) if the slope and y –intercept is known – use the formula 𝑦𝑦 = 𝑚𝑚𝑚𝑚 + 𝑏𝑏, where m is the slope and b is the yintercept example: if given (0, 3) and 𝑚𝑚 = −4, since 𝑏𝑏 = 3 (the y-intercept) 𝑦𝑦 = −4𝑥𝑥 + 3 8. Parallel lines – two lines that are parallel have the same slope 2 example: if the equation 𝑦𝑦 = 𝑥𝑥 − 2 was given, a line that is parallel to it would have a slope of 3 9. Perpendicular lines – two lines that are perpendicular have opposite reciprocal slopes 2 3 as well. 2 3 example: if the equation 𝑦𝑦 = 𝑥𝑥 − 2 was given, a line that is perpendicular to it would have a slope of − . 3 2 10. Graphing a line a) Plugging in points – assign a number to one of the variables (it doesn’t matter which) and solve for the other variable to get an ordered pair. Do this at least twice, then plot the plots and draw the line that connects these two points. example: 3𝑥𝑥 − 4𝑦𝑦 = 8 3(0) − 4𝑦𝑦 = 8 −4𝑦𝑦 = 8 𝑦𝑦 = −2 3(4) − 4𝑦𝑦 = 8 12 − 4𝑦𝑦 = 8 −4𝑦𝑦 = −4 𝑦𝑦 = 1 x y 0 -2 4 1 b) Using a point and the slope – start by plotting the given point. Then use the slope to find another point using the “rise and run” approach where 𝑚𝑚 = 2 2 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑟𝑟𝑟𝑟𝑟𝑟 example: 𝑦𝑦 = 𝑥𝑥 − 2, the slope is . Start by plotting the y-intercept which is (0, −2)and from that 3 3 point (if slope is +) rise up 2 units and go right 3 units to get the point (3,0) {if the slope was (-) go down 2 units and go right 3 units to get the point (3, −4)} x y 0 -2 3 0