Let A≡(3,−4),B≡(1,2)˙ Let P≡(2k−1,2k+1) be a variable point such that PA+PB is the minimum. Then k is 7/9 (b) 0 (c) 7/8 (d) none of these
Let 0≡(0,0),A≡(0,4),B≡(6,0)˙ Let P be a moving point such that the area of triangle POA is two times the area of triangle POB . The locus of P will be a straight line whose equation can be
See Fig.3.14. and write the following:(i) The coordinates of B.(ii) The coordinates of C.(iii) The point identified by the coordinates (3, 5).(iv) The point identified by the coordinates (2, 4)˙ (v) The abscissa of the point D. (vi) The ordinate of the points H. (vii) The coordinates of the points L. (viii) The coordinates of the point M.
In each of the following find the value of k for which the points are collinear.
If x1,x2,x3 as well as y1,y2,y3 are in GP with the same common ratio, then the points (x1,y1),(x2,y2), and (x3,y3)˙ lie on a straight line lie on an ellipse lie on a circle (d) are the vertices of a triangle.
The locus of the moving point whose coordinates are given by (et+e−t,et−e−t) where t is a parameter, is xy=1 (b) x+y=2 x2−y2=4 (d) x2−y2=2
Given that A1,A2,A3,An are n points in a plane whose coordinates are x1,y1),(x2,y2),(xn,yn), respectively. A1A2 is bisected at the point P1,P1A3 is divided in the ratio A:2 at P2,P2A4 is divided in the ratio 1:3 at P3,P3A5 is divided in the ratio 1:4 at P4 , and so on until all n points are exhausted. Find the final point so obtained.