Question 7 - End-of-Chapter Exercises - Chapter 1 Class 9 - Orienting Yourself: The Use of Coordinates (Ganita

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Question 7 Use your method (from Problem 6) to check if the points R (– 5, – 1), B (– 2, – 5) and C (4, – 12) are on the same straight line. Now plot both sets of points and check your answers. If three points are on the exact same straight line (collinear), then Distance of the two shorter segments added together will perfectly equal the distance of the longest segment. Let’s find distance of all 3 points Distance RB Here, R (–5, –1) and B (–2, –5) So, RB = √((−𝟐−(−𝟓))^𝟐+(−𝟓−(−𝟏))^𝟐 ) = √(〖(−2+5)〗^2+〖(−5+1)〗^2 ) = √(3^2+〖(−4)〗^2 ) = √(𝟑^𝟐+𝟒^𝟐 ) = √(9+16) = √25 = √(5^2 ) = 5 Distance RC Here, R (–5, –1) and C (4, –12) So, RC = √((𝟒−(−𝟓))^𝟐+(−𝟏𝟐−(−𝟏))^𝟐 ) = √(〖(4+5)〗^2+〖(−12+1)〗^2 ) = √(9^2+〖(−11)〗^2 ) = √(𝟗^𝟐+〖𝟏𝟏〗^𝟐 ) = √(81+121) = √𝟐𝟎𝟐 Distance BC Here, B (–2, –5) and C (4, –12) So, BC = √((𝟒−(−𝟐))^𝟐+(−𝟏𝟐−(−𝟓))^𝟐 ) = √(〖(4+2)〗^2+〖(−12+5)〗^2 ) = √(6^2+〖(−7)〗^2 ) = √(𝟔^𝟐+𝟕^𝟐 ) = √(36+49) = √𝟖𝟓 Thus, RB = 5, RC = √202 and BC = √85 Here, √202 ≠ 5 + √85 Here, √202 ≠ 5 + √85 Since the sum is not equal Thus, all 3 points are not collinear Now, let’s plot R (– 5, – 1), B (– 2, – 5) and C (4, – 12) And from Question 6 - M (– 3, – 4), A (0, 0) and G (6, 8) We can observe that Joining MA, AG and MG forms a straight line But joining RB, BC and RC makes a triangle