Learn in your speed, with individual attention - Teachoo Maths 1-on-1 Class
Important Coordinate Geometry Questions
Question 2 Deleted for CBSE Board 2024 Exams
Question 3 Deleted for CBSE Board 2024 Exams
Question 4 Important Deleted for CBSE Board 2024 Exams
Question 5 Important Deleted for CBSE Board 2024 Exams
Question 6 Deleted for CBSE Board 2024 Exams You are here
Question 7 Important Deleted for CBSE Board 2024 Exams
Question 8 Deleted for CBSE Board 2024 Exams
Important Coordinate Geometry Questions
Last updated at May 29, 2023 by Teachoo
Question 6 (Method 1) The vertices of a ฮ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4. Calculate the area of the ฮ ADE and compare it with the area of ฮ ABC. (Recall Theorem 6.2 and Theorem 6.6). Given ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4 ๐จ๐ซ/๐จ๐ฉ=๐/๐ ๐จ๐ฌ/๐จ๐ช=๐/๐ 4 AD = AB 4 AD = AD + BD 4 AD โ AD = BD 3 AD = BD ๐ด๐ท/๐ต๐ท=1/3 Thus point D divides AB in the ratio 1 : 3. Using section formula, coordinates of D 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 )) 4 AE = AC 4 AE = AE + CE 4 AE โ AE = CE 3 AE = CE ๐ด๐ธ/๐ถ๐ธ=1/3 Thus point E divides AC in the ratio 1 : 3. Using section formula, coordinates of E 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=1, ๐_2=3 ๐ฅ_1=4, ๐ฅ_2=1 ๐ฆ_1=6, ๐ฆ_2=5 Point D = ((1(1) + (3)(4))/(1 + 3), (1(5) + 3(6))/(1 + 3)) = ((1 + 12)/4,(5 + 18)/4) = (13/4,23/4) Put ๐_1=1, ๐_2=3 ๐ฅ_1=4, ๐ฅ_2=7 ๐ฆ_1=6, ๐ฆ_2=2 Point E = ((1(7) + (3)(4))/(1 + 3), (1(2) + 3(6))/(1 + 3)) = ((7 + 12)/4,(2 + 18)/4) = (19/4,20/4) Now, finding Area of ฮ ADE and ฮ ABC Area ฮ ADE Area of โADE = 1/2 [๐ฅ_1 (๐ฆ_2โ๐ฆ_3 )+๐ฅ_(2 ) (๐ฆ_3 โ๐ฆ_1 )+๐ฅ_(3 ) (๐ฆ_1 โ๐ฆ_2 )] Put ๐ฅ_1=4, ๐ฅ_2=13/4 ๐ฅ_3=19/4, ๐ฆ_1=6 ๐ฆ_2=23/4, ๐ฆ_3=20/4 ๐ฆ_2=23/4, ๐ฆ_3=20/4 Ar (โADE) = 1/2 [4 (23/4โ20/4)+13/4 (20/4โ6)+19/4 (6 โ23/4)] = 1/2 [4 (3/4)+13/4 ((20 โ 24)/4)+19/4 ((24 โ 23)/4)] = 1/2 [4 (3/4)+13/4 ((โ4)/4)+19/4 (1/4)] = 1/2 [12/4โ52/16+19/16] = 1/2 [(48 โ 52 + 19)/16] = ๐๐/๐๐ sq. units. Area ฮ ABC Area of โABC = 1/2 [๐ฅ_1 (๐ฆ_2โ๐ฆ_3 )+๐ฅ_(2 ) (๐ฆ_3 โ๐ฆ_1 )+๐ฅ_(3 ) (๐ฆ_1 โ๐ฆ_2 )] Put ๐ฅ_1=4, ๐ฅ_2=1 ๐ฅ_3=7, ๐ฆ_1=6 ๐ฆ_2=5, ๐ฆ_3=2 Ar (โABC) = 1/2 [4 (5โ2)+1 (2 โ 6)+7 (6 โ5)] = 1/2 [4 (3)+1 (โ4)+7 (1)] = 1/2 [12 โ 4+7] = 15/2 sq. units. Thus, (๐๐ (โ๐ด๐ท๐ธ))/(๐๐ (โ๐ด๐ต๐ถ))=(15/32)/(15/2) = 15/32ร2/15 = 1/16 Thus, area of โADE is ๐๐/๐๐ sq. units and (๐๐ (โ๐ด๐ท๐ธ))/(๐๐ (โ๐ด๐ต๐ถ))=๐/๐๐ Question 6 (Method 2) The vertices of a ฮ ABC are A(4, 6), B(1, 5) and C(7, 2). A line is drawn to intersect sides AB and AC at D and E respectively, such that ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4. Calculate the area of the ฮ ADE and compare it with the area of ฮ ABC. (Recall Theorem 6.2 and Theorem 6.6). Given ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/4 ๐จ๐ซ/๐จ๐ฉ=๐/๐ 4 AD = AB 4 AD = AD + BD 4 AD โ AD = BD 3 AD = BD ๐ด๐ท/๐ต๐ท=1/3 ๐จ๐ฌ/๐จ๐ช=๐/๐ 4 AE = AC 4 AE = AE + CE 4 AE โ AE = CE 3 AE = CE ๐ด๐ธ/๐ถ๐ธ=1/3 Thus, ๐ด๐ท/๐ด๐ต=๐ด๐ธ/๐ด๐ถ=1/3 โด DE divides AB and AC in the ratio 1 : 3 Applying theorem 6.2, DE โฅ BC From theorem 6.2 If a line divides any two sides of a triangle in the same ratio, then it is parallel to the third side. Now, In โADE and โABC โ A = โ A โ ADE = โ ABC โด โADE ~ โABC From theorem 6.6 The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides. Thus, (๐๐ (โ๐ด๐ท๐ธ))/(๐๐ (โ๐ด๐ต๐ถ))=(๐ด๐ท/๐ด๐ต)^2 (Common angle) (Corresponding angles) (By AA similarity criterion) (๐๐ (โ๐ด๐ท๐ธ))/(๐๐ (โ๐ด๐ต๐ถ))=(1/4)^2=1/16 ๐๐ (โ๐ด๐ท๐ธ)=1/16 ๐๐ (โ๐ด๐ต๐ถ) Now, Finding Area ฮ ABC ๐๐ (โ๐ด๐ต๐ถ)=1/2 [๐ฅ_1 (๐ฆ_2โ๐ฆ_3 ) + ๐ฅ_2 (๐ฆ_3โ๐ฆ_1 ) + ๐ฅ_3 (๐ฆ_1โ๐ฆ_2 )] Put ๐ฅ_1=4, ๐ฆ_1=6 ๐ฅ_2=1, ๐ฆ_2=5 ๐ฅ_3=7, ๐ฆ_3=2 Ar (โABC) = 1/2 [4 (5โ2)+1 (2 โ 6)+7 (6 โ5)] = 1/2 [4 (3)+1 (โ4)+7 (1)] = 1/2 [12 โ 4+7] = 15/2 sq. units. Thus, From (1) ar (โADE) = 1/16 ar (โABC) = 1/16ร15/2=15/32 sq. units. Thus, area of โADE is ๐๐/๐๐ sq. units and (๐๐ (โ๐ด๐ท๐ธ))/(๐๐ (โ๐ด๐ต๐ถ))=๐/๐๐