Coordinate Geometry - Distance & Section Formula for Exams

Coordinate Geometry - Distance and Section Formula - Problems on Circles - Case Studies

Q. Find the coordinates of the point that divides the segment joining (1, 2) and (3, 8) in the ratio 1:3.

Solution

Using the section formula: (mx2 + nx1)/(m+n), (my2 + ny1)/(m+n) where m=1, n=3. Coordinates = ((1*3 + 3*1)/(1+3), (1*8 + 3*2)/(1+3)) = (2, 6).

Correct Answer: A — (2, 6)

Q. What is the area of a triangle with vertices at (1, 1), (4, 5), and (7, 2)?

Solution

Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)| = 0.5 * |1(5 - 2) + 4(2 - 1) + 7(1 - 5)| = 0.5 * |3 + 4 - 28| = 0.5 * 21 = 10.5.

Correct Answer: B — 12

Q. What is the equation of a circle with center at (0, 0) and radius 5?

Solution

The equation of a circle is (x - h)² + (y - k)² = r². Here, h=0, k=0, r=5, so x² + y² = 5² = 25.

Correct Answer: A — x² + y² = 25

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