ABCD is a rhombus. The slope of AC is 1 and is among the family of lines (x+2y−5)+λ(3x+y−5)=0, where λ∈R. One of the vertex of the rhombus is (−2,3). If the area of rhombus is 10√2 sq. units, then which of the following is (are) CORRECT?
A
Length of smaller diagonal is 4
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B
Length of larger diagonal is 4√2
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C
One of the vertex is (2,−1)
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D
Perimeter of rhombus is 2√57
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Solution
The correct option is D Perimeter of rhombus is 2√57 The diagonal AC is the member of family of lines (x+2y−5)+λ(3x+y−5)=0.
So it will pass through the intersection point of lines x+2y−5=0 and 3x+y−5=0 i.e., (1,2).
Since, slope of AC is 1,
therefore equation of AC is, x−y+1=0
The point (−2,3) does not lie on x−y+1=0, so it will lie on another diagonal BD.
Since, diagonals of rhombus are perpendicular to each other, therefore equation of BD is, x+y−1=0
Point of intersection of AC and BD is (0,1).
⇒B:(x,y)=(2,−1) BD=4√2=d2 Area of rhombus=12(d1×d2)=10√2 ⇒d1=5 Side of rhombus=√d214+d224=√572 Perimeter=2√57