In geometry of ncert class 10th how i solve the questions in simplest way . Please help me to give solution of the questions of exercise 7.4 question no.7(3) . Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC.(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1
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Let A (4, 2), B (6, 5) and C (1, 4) be the vertices of ∆ ABC.(iii) Find the coordinates of point Q and R on medians BE and CF respectively such that BQ : QE = 2:1 and CR : RF = 2 : 1. Q.7(3)
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(iii) Coordinates of E can be calculated as follows:
Coordinates of E = ( (4+1)/2, (2+4)/2 ) = (5/2, 6/2) = (5/2 , 3)
So, E is (5/2 , 3)
Point Q and P would be coincident because medians of a triangle intersect each other at a common point called centroid. Coordinate of Q can be given as follows:
Coordinates of Q =( [2(5/2) + 1(6)]/(2 + 1), [2(3) + 1(5)]/(2 + 1) ) = (11/3, 11/3)
F is the mid- point of the side AB
Coordinates of F = ( (4+6)/2, (2+5)/2 ) = (5, 7/2)
Point R divides the side CF in ratio 2:1
Coordinates of R = ( [2(5) + 1(1)]/(2 + 1), [2(7/2) + 1(4)]/(2 + 1) ) = (11/3, 11/3)