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Two sides of a rhombus ABCD are parallel...

Two sides of a rhombus ABCD are parallel to the lines `y = x + 2` and `y = 7x + 3` If the diagonals of the rhombus intersect at the point `(1, 2)` and the vertex `A` is on the y-axis, then vertex `A` can be
If `alpha,beta,gamma` are Acute angles and `costheta=sinbeta//sinalpha,cosvarphi=singammasinalpha " and" cos(thetavarphi)=sinbetasingamma`, then the value of `tan^2alpha-tan^2beta-tan^2gamma` is equal to

A

(0,3)

B

`(0,5/2)`

C

(0,0)

D

(0,6)

Text Solution

Verified by Experts

The correct Answer is:
B

Let point A be `(0,lambda)`
Slope of `AP=(lambda-2)/(0-1)=2-lambda`
AB and AC are parallel to the lines `y=x+2` and `y=7x+3`respectively.
Thus, slope of AB and AC are 1 and 7, respectively.
From the figure,
`tantheta=((2-lambda)-1)/(1+(2-lambda))=(7-(2-lambda))/(1+7(2-lambda))`
`rArr((1-lambda)/(3-lambda))=((5+lambda)/(15-7lambda))`
`rArr(lambda-1)(7lambda-15)=(5+lambda)(3-lambda)`
`rArr 8lambda^20lambda=0`
`A=(0,0) and (0,5//2)` .
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