12.) write the points in detreminnat form.. and equate it to zero.
1.If the sum of the distances of a point from two perpendicular lines in a plane is 1,then its locus is
(a)a square
(b)a circle
(c)a straight line
(d)two intersecting line
2.If a,b,c are unequal and different from 1 such that the points (a3/a-1,a2-3/a-1),(b3/b-1,b2-3/b-1) and(c3/c-1,c2-3/c-1) are collinear,then
(a)bc+ac+ab+abc=0
(b)a+b+c=abc
(c)bc+ac+ab=abc
(d)bc+ac+ab-abc=3(a+b+c)
3.An equation of a line through the point (1,2) whose distance from the point (3,1) has the greatest value is
(a)y=2x
(b)y=x+1
(c)x+2y=5
(d)y=3x-1
4.Let 0<A<pi/2 be fixed angle.If P=(cos@,sin@) and Q=(cos(A-@),sin(A-@)) then Q is obtained from P by
(a)clockwise rotation around the origin through an angle A
(b)anticlockwise rotation around the origin through an angle A
(c)reflection in the line through the origin with slope tanA
(d)reflection in the line through the origin with slope tan(A/2)
Please give explainations also.THANK YOU
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3 Answers
106Q1.
Let the two perpendicular lines be x=0 and y=0 and let the point be (h,k)
So, lhl+lkl = 1 is the reqd condition
i.e. lxl+lyl=1 whose graph is a square with diagonals on x=0 and y=0 and of length 2..
Q3.
any line be y=mx+c
It passes thru (1,2) so, 2=m+c ... (i)
Its distance from (3,1) = l1-3m-cl/√1+m2 = l2m+1l/√1+m2
This is max when d[(2m+1)2/(1+m2)]/dx = 0
Solving it we get m=-1/2,2
Obviously when m=2 it is max
So, the equation is y=2x
