If `A(4,-4)` and `B(9,6)` lies on `y^2=4x` and a point C on arc `AOB(O= origin)` such that the area of `triangleACB` is maximum then point c is (1) `(1/4,1)` (2) `(1,1/4)` (3) `(1,1/2)` (4) `(1/2,1)`

Find area of the triangle formed by the points (1,2,3), (-1,-1,8) and (-4,4,6) .

The points (-2,3,4), (1,1,2) and (4,-1,0) are

Let A(4,-4) and B(9,6) be points on the parabola y^(2)=4x. Let C be chosen on the on the arc AOB of the parabola where O is the origin such that the area of DeltaACB is maximum. Then the area (in sq. units) of DeltaACB is :

The point on the curve y^2=x where tangent makes 45o angle with x-axis is (1//2,\ 1//4) (b) (1//4,\ 1//2) (c) (4,\ 2) (d) (1,\ 1)

The line x+2y=1 cuts the ellipse x^(2)+4y^(2)=1 1 at two distinct points A and B. Point C is on the ellipse such that area of triangle ABC is maximum, then find point C.

Plot the points A(4,1),B(-3,2),C(2,-3),D(-4,1),E(-4,-4) .

y^(2)=4x be the parabola and the points A(9,6),B(4,-4) lie on the parabola such that area of /_ABC is maximum (given C lies at the arc contain origin).Find the maximum area of /_ABC( A) (101)/(2) (B) (69)/(2) (C) (125)/(4) (D) (521)/(3)

Two vertices of triangle are (-2,-1) and (3,2) and third vertex lies on the line x+y=5. If the area of the triangle is 4 square units, then the third vertex is a. (-0,50 or (4,1) b. (5,0) of (1,4) c. (5,0) or (4,1) d. (o,5) or (1,4)

If sqrt(y)=4x, then (x^(2))/(y) is (a) 2 (b) (1)/(16) (c) (1)/(4) (d) 4

The straight lines 4ax+3by+c=0, where a+b+c are concurrent at the point a )(4,3) b) (1/4,1/3) c) (1/2,1/3) d) none of these