Three vertices of a parallelogram are (a+b,a-b), (2a+b,2a-b) and (a-b,a+b) find the fourth vertex. (PLZ EXPLAIN STEPWISE)
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Answer is (-b,b) according to the formula( sum of opposite vertices-remainig vertex)
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A(a+b,a-b),B(2a+b,2a-b), C(a-b,a+b)
let D(x,y)
in a parallelogram diagonals bisects each other
mid point of BD = mid point of AC
[(2a+b+x)/2,(2a-b+y)/2] = [(a+b+a-b)/2,(a-b+a+b)/2]
now x -co ordinates are equal and y- co ordinates are equal
i)(2a+b+x)/2 = (a+b+a-b)/2
2a+b+x = 2a
x=2a-2a-b
x= -b
ii)(2a-b+y)/2 = (a-b+a+b)/2
2a-b+y =2a
y =2a-2a+b
y =b
required vertex D(x,y) = (-b,b)
let D(x,y)
in a parallelogram diagonals bisects each other
mid point of BD = mid point of AC
[(2a+b+x)/2,(2a-b+y)/2] = [(a+b+a-b)/2,(a-b+a+b)/2]
now x -co ordinates are equal and y- co ordinates are equal
i)(2a+b+x)/2 = (a+b+a-b)/2
2a+b+x = 2a
x=2a-2a-b
x= -b
ii)(2a-b+y)/2 = (a-b+a+b)/2
2a-b+y =2a
y =2a-2a+b
y =b
required vertex D(x,y) = (-b,b)
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