The given question is a medium difficulty 2-mark question that appeared in the CBSE Class 10 Math 2016 board paper. Chapter : Coordinate Geometry. Concept: Computing the ratio in which two points divide a line segment and then applying the section formula to compute the coordinates of those two points.

Question 6: Let P and Q be the points of trisection of the line segment joining the points A(2, -2) and B(-7, 4) such that P is nearer to A. Find the coordinates of P and Q.

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Because points P and Q trisect the line segment AB, P and Q will divide AB into 3 equal parts such that AP = PQ = QB.

Hence, point P divides AB in the ratio 1 : 2

∴ X coordinate of P = \\frac{\text{1(-7) + 2(2)}}{\text{1 + 2}}) = \\frac{\text{-3}}{\text{3}}) = -1

Y coordinate of P = \\frac{\text{1(4) + 2(-2)}}{\text{1 + 2}}) = 0

Coordinates of point P are (-1, 0)

Point Q divides AB in the ratio 2 : 1

∴ X coordinate of Q = \\frac{\text{2(-7) + 1(2)}}{\text{1 + 2}}) = -4

Y coordinate of Q = \\frac{\text{2(4) + 1(-2)}}{\text{1 + 2}}) = 2

Coordinates of point Q are (-4, 2)

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