Ex 7.4, 7 (Optional) - Chapter 7 Class 10 Coordinate Geometry (Term 1)
Last updated at June 29, 2018 by Teachoo
Last updated at June 29, 2018 by Teachoo
Transcript
Ex 7.4, 7 Let A (4, 2), B(6, 5) and C(1, 4) be the vertices of ฮ 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 = ((๐ฅ_1 + ๐ฅ_2)/2, (๐ฆ_1 + ๐ฆ_2)/2) Here ๐ฅ_1=6 ๐ฆ_1=5 ๐ฅ_2=1 ๐ฆ_2=4 Coordinates of D = ((๐ฅ_1 + ๐ฅ_2)/2, (๐ฆ_1 + ๐ฆ_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 ((๐_1 ๐ฅ_2 + ๐_2 ๐ฅ_1)/(๐_1 + ๐_2 ), (๐_1 ๐ฆ_2+ ๐_2 ๐ฆ_1)/(๐_1 + ๐_2 )) Put ๐_1=2 ๐_2=1 ๐ฅ_1=4 ๐ฅ_2=7/2 ๐ฆ_1=2 ๐ฆ_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 = ((๐ฅ_1 + ๐ฅ_2)/2, (๐ฆ_1 + ๐ฆ_2)/2) Point E Since BE is the median, point E is the mid-point of AB Coordinates of E = Mid Point of AC = ((๐ฅ_1 + ๐ฅ_2)/2, (๐ฆ_1 + ๐ฆ_2)/2) Here, ๐ฅ_1=4 ๐ฅ_2=6 ๐ฆ_1=2 ๐ฆ_2=5 Coordinates of F = ((4 + 6)/2,(2 + 5)/2) = (10/2,7/2) = (5, 7/2) = ((๐ฅ_1 + ๐ฅ_2)/2, (๐ฆ_1 + ๐ฆ_2)/2) Here, ๐ฅ_1=4 ๐ฅ_2=1 ๐ฆ_1=2 ๐ฆ_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 = ((๐_1 ๐ฅ_2 + ๐_2 ๐ฅ_1)/(๐_1 + ๐_2 ), (๐_1 ๐ฆ_2+ ๐_2 ๐ฆ_1)/(๐_1 + ๐_2 )) Put ๐_1=2 ๐_2=1 ๐ฅ_1=1 ๐ฅ_2=5 ๐ฆ_1=4 ๐ฆ_2=7/2 Point Q Applying section formula, Coordinates of Q = ((๐_1 ๐ฅ_2 + ๐_2 ๐ฅ_1)/(๐_1 + ๐_2 ), (๐_1 ๐ฆ_2+ ๐_2 ๐ฆ_1)/(๐_1 + ๐_2 )) Put ๐_1=2 ๐_2=1 ๐ฅ_1=6 ๐ฅ_2=5/2 ๐ฆ_1=5 ๐ฆ_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 (๐๐/๐,๐๐/๐) 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 (๐ฅ_1, ๐ฆ_1), B (๐ฅ_2, ๐ฆ_2) and C (๐ฅ_3, ๐ฆ_3) are the vertices of ฮ 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 = ((๐ฅ_2 +ใ ๐ฅใ_3)/2, (๐ฆ_2 + ๐ฆ_3)/2) Now, O divides AD in the ratio 2 : 1 Using Section Formula, Coordinates of O are ((๐_1 ๐_2 + ๐_2 ๐_1)/(๐_1 + ๐_2 ), (๐_1 ๐_2 + ๐_2 ๐_1)/(๐_1 + ๐_2 )) Where A (๐_1, ๐_1) and D (๐_2, ๐_2) Here ๐_1=๐ฅ_1 ๐_1= ๐ฆ_1 ๐_2=(๐ฅ_2 + ๐ฅ_3)/2 ๐_2=(๐ฆ_2 + ๐ฆ_3)/2 ๐_1=2 ๐_2=1 Coordinates of O = (((2) ((๐ฅ_2 + ๐ฅ_3)/2) + (1) (๐ฅ_1 ))/(2 + 1),((2) ((๐ฆ_2 + ๐ฆ_3)/2) + (1) (๐ฆ_1 ))/(2 + 1)) =((๐ฅ_2 +ใ ๐ฅใ_3 + ๐ฅ_1)/3, (๐ฆ_2 + ๐ฆ_3 + ๐ฆ_1)/3) = ((๐ฅ_1 +ใ ๐ฅใ_2 + ๐ฅ_3)/3, (๐ฆ_1 + ๐ฆ_2 + ๐ฆ_3)/3) Note :- If we would have used section formula in CF or BE, we would have got the same result.
Ex 7.4 (Optional)
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