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15. Three lines x +2y +3-0; x +2y-70 and2x-y-4=0 form the three sides of twosquares. The equation to the fourth side of eachsquare is1) 2x -y +14 0 and 2x-y+602) 2x-y+14 = 0 and 2x-y-6-03) 2x-y-14 0 and 2x-y-602x-y-14 0 and 2x-y+6 0 |
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Answer» 2 answers·Mathematics Best Answer Let's get them in point slope form... y=(-x-3)/2, y=(7-x)/2, y=2x+4 notice that the first two have the same slope of -1/2 this means that they are parallel and are one pair of opposite sides... thus the missing line must have the slope of the 3rd line above...which is 2x. y=2x+b and must meet lines one and two above.. Now let's see how far apart dy is between the opposite sides... y=(-x-3)/2, y=(7-x)/2 we can set both x=0 zero to see their separation in height... -3/2 and 7/2 so dy is (7--3)/2=5 this means that our parallel sides must have a dx of 5.... our missing line y=2x+b must differ in x by 5 from y=2x+4 when the y values are equal, we'll just set them equal to each to each other... 2x+b=2x+4 let x1=0 and x2=5 b=10+4=14 So now we know that the missing line is: y=2x+14 so our line is y=2x+14 |
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