Test: Coordinate Geometry (April 2) - CAT MCQ

Explanation:

Coordinates are given for  A(1 , 2) , B(1 , 0) and C(4 , 0)

Let coordinates of D be (x,y).
Since diagonals of a parallelogram bisect each other. at point O

Therefore O is the mid point of diagonal AC 

Explanation:

ABC is a triangle with A(7 ,- 3), B(5 , 3) and C(3 , - 1) 

 Let median on BC bisects BC at D.          (AD is given as the median)

Explanation:

If given points are collinear, the area of triangle formed by these three points is 0.

Explanation:

Let point A be (p,−5) and point B (2, 7) and distance between A and B = 13 units

Explanation:

Let A (1, 7), B (4, 2), C(−1,−1) and D(−4,4) are the vertices of a quadrilateral ABCD.

Explanation:

Since x−coordinate is negative and y−coordinate is positive.Therefore, the point (−3,5) lies in II quadrant.

The sum is according to distance formula
= √(x₂-x₁)²+(y₂-y₁)²
= √(0-p cos 25)² + (p cos 65-0)
= √(p cos 25)² + (p cos  65)²
= √(p cos 25)² + (p sin 25)²
= √(p² cos²25) + (a² sin²25)
= √p²(cos²25 + sin²25)                  we know that  sin²Θ + cos²Θ = 1
= √p²(1)
= √p²
= p Units