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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 |
Answer» y=x+2 slope=1 y=7x+3 slope=7 O(1,2)=`((0+m)/2,(n+alpha)/2)=((x+a)/2,(b+y)/2)` m=2 x+a=2 `n+alpha=4 b+y=4` `m_(ab)=(b-alpha)/(a-0)=1=b-alpha=a` `m_(bc)=(b-n)/(a-2)=7=b-n=7(a-2)` AB=BC `sqrt((b-alpha)^2+(a-0)^2)=sqrt((b-n^2+(a-2)^2)` solving this `a=5/2,5/3` `b=a+alpha=5/2+alpha,5/3+alpha` `B(5/2,5/2+alpha),B(5/3,5/3+alpha)` A(0,alpha) O(1,2) `m_(OA)*m_(OB)=-1` `(2-alpha)/(1-0)*((5/2)+alpha-2)/((5/2)-1)=-1` solving this `alpha=(0,3)` A=`(0,alpha)` A=(0,0),(0,3) |
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