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Ex 7.4, 7 Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of Δ ABC. (i) The median from A meets BC at D. Find the coordinates of the point D. Since Median is AD D is the mid-point of BC Coordinates of D = Coordinates of mid-point of BC = ((𝑥_1 + 𝑥_2)/2, (𝑦_1 + 𝑦_2)/2) Here 𝑥_1=6 𝑦_1=5 𝑥_2=1 𝑦_2=4 Coordinates of D = ((𝑥_1 + 𝑥_2)/2, (𝑦_1 + 𝑦_2)/2) = ((6 + 1)/2, (5 + 7)/2) = (𝟕/𝟐,𝟗/𝟐) Ex 7.4, 7 (ii) Find the coordinates of the point P on AD such that AP : PD = 2 : 1 Point P divides AD in the ratio 2 : 1 Applying section formula, Coordinates of P are ((𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ), (𝑚_1 𝑦_2+ 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 )) Put 𝑚_1=2 𝑚_2=1 𝑥_1=4 𝑥_2=7/2 𝑦_1=2 𝑦_2=9/2 Coordinates of Point P = ((2 (7/2) + 1(4))/(2 + 1), (2 (9/2) + 1 (2))/(2 + 1)) = ((7 + 4)/3,(9 + 2)/3) = (𝟏𝟏/𝟑,𝟏𝟏/𝟑) Ex 7.4, 7 (iii) Find the coordinates of points Q and R on medians BE and CF respectively such that BQ : QE = 2 : 1 and CR : RF = 2 : 1. To find points Q and R, we need to first find points E and F Point F Since CF is the median, point F is the mid-point of AB Coordinates of F = Mid Point of AB Point E Since BE is the median, point E is the mid-point of AB Coordinates of E = Mid Point of AC = ((𝑥_1 + 𝑥_2)/2, (𝑦_1 + 𝑦_2)/2) Here, 𝑥_1=4 𝑥_2=6 𝑦_1=2 𝑦_2=5 Coordinates of F = ((4 + 6)/2,(2 + 5)/2) = (10/2,7/2) = (5, 7/2) = ((𝑥_1 + 𝑥_2)/2, (𝑦_1 + 𝑦_2)/2) Here, 𝑥_1=4 𝑥_2=1 𝑦_1=2 𝑦_2=4 Coordinates of E = ((4 + 1)/2,(2 + 4)/2) = (5/2,6/2) = (5/2, 3) Now finding Points Q & R Point R Applying section formula, Coordinates of R = ((𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ), (𝑚_1 𝑦_2+ 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 )) Put 𝑚_1=2 𝑚_2=1 𝑥_1=1 𝑥_2=5 𝑦_1=4 𝑦_2=7/2 Point Q Applying section formula, Coordinates of Q = ((𝑚_1 𝑥_2 + 𝑚_2 𝑥_1)/(𝑚_1 + 𝑚_2 ), (𝑚_1 𝑦_2+ 𝑚_2 𝑦_1)/(𝑚_1 + 𝑚_2 )) Put 𝑚_1=2 𝑚_2=1 𝑥_1=6 𝑥_2=5/2 𝑦_1=5 𝑦_2=3 Coordinates of R = ((2(5)+ 1(1))/(2 + 1), (2 (7/2) + 1(4))/(2 + 1)) = ((10 + 1)/3,(7 + 4)/3) = (𝟏𝟏/𝟑,𝟏𝟏/𝟑) Coordinates of Q = (((2) (5/2)+ (1) (6))/(2 + 1), (2 (3) + 1(5))/(2 + 1)) = ((5 + 6)/3,(6 + 5)/3) = (𝟏𝟏/𝟑,𝟏𝟏/𝟑) Thus, coordinates of Q and R are (𝟏𝟏/𝟑,𝟏𝟏/𝟑) Ex 7.4, 7 (iv) What do you observe? [Note : The point which is common to all the three medians is called the centroid and this point divides each median in the ratio 2 : 1.] Coordinates of P, Q, and R are same, thus it is a common point to all the medians. Since Centroid of a triangle divides each median in ratio 2 : 1. So, point (11/2 ", " 11/2) is called centroid of triangle Ex 7.4, 7 (v) If A (𝑥_1, 𝑦_1), B (𝑥_2, 𝑦_2) and C (𝑥_3, 𝑦_3) are the vertices of Δ ABC, find the coordinates of the centroid of the triangle. Centroid of a triangle divides each median in ratio 2 : 1 Let us draw median AD Let O be the point which divides AD in the ratio 2 : 1 So, O is the centroid. Finding coordinates of point D Since AD is the median, D is the mid-point of BC Coordinates of D = ((𝑥_2 +〖 𝑥〗_3)/2, (𝑦_2 + 𝑦_3)/2) Now, O divides AD in the ratio 2 : 1 Using Section Formula, Coordinates of O are ((𝑚_1 𝑎_2 + 𝑚_2 𝑎_1)/(𝑚_1 + 𝑚_2 ), (𝑚_1 𝑏_2 + 𝑚_2 𝑏_1)/(𝑚_1 + 𝑚_2 )) Where A (𝑎_1, 𝑏_1) and D (𝑎_2, 𝑏_2) Here 𝑎_1=𝑥_1 𝑏_1= 𝑦_1 𝑎_2=(𝑥_2 + 𝑥_3)/2 𝑏_2=(𝑦_2 + 𝑦_3)/2 𝑚_1=2 𝑚_2=1 Coordinates of O = (((2) ((𝑥_2 + 𝑥_3)/2) + (1) (𝑥_1 ))/(2 + 1),((2) ((𝑦_2 + 𝑦_3)/2) + (1) (𝑦_1 ))/(2 + 1)) =((𝑥_2 +〖 𝑥〗_3 + 𝑥_1)/3, (𝑦_2 + 𝑦_3 + 𝑦_1)/3) = ((𝒙_𝟏 +〖 𝒙〗_𝟐 + 𝒙_𝟑)/𝟑, (𝒚_𝟏 + 𝒚_𝟐 + 𝒚_𝟑)/𝟑) Note :- If we would have used section formula in CF or BE, we would have got the same result.

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Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.