St. Lines



  • Capture.PNG
    Ans:
    a1.PNG



  • @Seven-Three

    One diagonal is a member of both the family of lines (x+y-1)+k1(2x+3y-2)=0 and (x-y+2)+k2(2x-3y+5)
    Hence it must pass through the point of intersection of x+y-1= 0 and 2x+3y-2 = 0 .
    Also it should also pass through the point of intersection of x-y+2 = 0 and 2x-3y+5 = 0 .
    Solving both pairs of equations, we get points as (1,0) and ( -1, 1) .
    Hence equation of one diagonal of the rhombus which passes through the above two points is
    y-0 = -1/ 2 ( x – 1 ) ; 2y = –x + 1 …..........(1)
    One of the vertex has coordinates (3,2) , which does not satisfy the above equation. So, it must lie on the other diagonal which is perpendicular to the diagonal 2y = –x +1 .
    Hence, it’s equation can be of form y = 2x + c. Since it passes through (3, 2) we get the equation as y = 2x-4 ….......... (2)
    Now, solve (1) and (2) to find the point of intersection. We get the coordinates as ( 9/5 , -2/5) .
    Hence the other vertex on 2nd Diagonal is the point with coordinates x = 3 / 5 and y = -14 / 5 ( using mid point formula)
    distance between (3,2) and ( 3/5 , –14/5) is 12 / sqrt(5) = d2
    Area = ½ d1d2= 12 sqrt(5) , we get d1 = 10 .
    Hence, length of semi-longer diagnonal is 5.


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