The sides of a rhombus ABCD are parallel to the lines, x−y+2=0 and 7x−y+3=0. If the diagonals of the rhombus intersect at P(1,2) and the vertex A (different from the origin) is on the y−axis, then the ordinate of A is
A
74
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B
2
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C
52
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D
72
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Solution
The correct option is C52 Let the point A be (0,a) As we know, diagonals of rhombus will be parallel to the bisector of the lines which are parallel to sides of rhombus So, bisector of the lines x−y+2=0 and 7x−y+3=0 are x−y+2√2=±7x−y+35√2 ⇒5x−5y+10=±(7x−y+3) So, bisectors are 2x+4y−7=0and12x−6y+13=0 ∴ Slope of diagonals =−12and2 As, slope of AP=2−a ⇒2−a=−12or2−a=2 ⇒a=52ora=0 a=0 can not be possible because vertex A should be different from the origin. Hence, a=52