Question 6 - Important Coordinate Geometry Questions - Chapter 7 Class 10 Coordinate Geometry
Last updated at April 16, 2024 by Teachoo
Important Coordinate Geometry Questions
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Important Coordinate Geometry Questions
Last updated at April 16, 2024 by Teachoo
Question 6 (Method 1) The vertices of a Ξ 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 π΄π·/π΄π΅=π΄πΈ/π΄πΆ=1/4. Calculate the area of the Ξ ADE and compare it with the area of Ξ ABC. (Recall Theorem 6.2 and Theorem 6.6). Given π΄π·/π΄π΅=π΄πΈ/π΄πΆ=1/4 π¨π«/π¨π©=π/π π¨π¬/π¨πͺ=π/π 4 AD = AB 4 AD = AD + BD 4 AD β AD = BD 3 AD = BD π΄π·/π΅π·=1/3 Thus point D divides AB in the ratio 1 : 3. Using section formula, coordinates of D are ((π_1 π₯_2 + π_2 π₯_1)/(π_1 + π_2 ),(π_1 π¦_2 + π_2 π¦_1)/(π_1 + π_2 )) 4 AE = AC 4 AE = AE + CE 4 AE β AE = CE 3 AE = CE π΄πΈ/πΆπΈ=1/3 Thus point E divides AC in the ratio 1 : 3. Using section formula, coordinates of E are ((π_1 π₯_2 + π_2 π₯_1)/(π_1 + π_2 ),(π_1 π¦_2 + π_2 π¦_1)/(π_1 + π_2 )) Put π_1=1, π_2=3 π₯_1=4, π₯_2=1 π¦_1=6, π¦_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 π_1=1, π_2=3 π₯_1=4, π₯_2=7 π¦_1=6, π¦_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 Ξ ADE and Ξ ABC Area Ξ ADE Area of βADE = 1/2 [π₯_1 (π¦_2βπ¦_3 )+π₯_(2 ) (π¦_3 βπ¦_1 )+π₯_(3 ) (π¦_1 βπ¦_2 )] Put π₯_1=4, π₯_2=13/4 π₯_3=19/4, π¦_1=6 π¦_2=23/4, π¦_3=20/4 π¦_2=23/4, π¦_3=20/4 Ar (βADE) = 1/2 [4 (23/4β20/4)+13/4 (20/4β6)+19/4 (6 β23/4)] = 1/2 [4 (3/4)+13/4 ((20 β 24)/4)+19/4 ((24 β 23)/4)] = 1/2 [4 (3/4)+13/4 ((β4)/4)+19/4 (1/4)] = 1/2 [12/4β52/16+19/16] = 1/2 [(48 β 52 + 19)/16] = ππ/ππ sq. units. Area Ξ ABC Area of βABC = 1/2 [π₯_1 (π¦_2βπ¦_3 )+π₯_(2 ) (π¦_3 βπ¦_1 )+π₯_(3 ) (π¦_1 βπ¦_2 )] Put π₯_1=4, π₯_2=1 π₯_3=7, π¦_1=6 π¦_2=5, π¦_3=2 Ar (βABC) = 1/2 [4 (5β2)+1 (2 β 6)+7 (6 β5)] = 1/2 [4 (3)+1 (β4)+7 (1)] = 1/2 [12 β 4+7] = 15/2 sq. units. Thus, (ππ (βπ΄π·πΈ))/(ππ (βπ΄π΅πΆ))=(15/32)/(15/2) = 15/32Γ2/15 = 1/16 Thus, area of βADE is ππ/ππ sq. units and (ππ (βπ΄π·πΈ))/(ππ (βπ΄π΅πΆ))=π/ππ Question 6 (Method 2) The vertices of a Ξ 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 π΄π·/π΄π΅=π΄πΈ/π΄πΆ=1/4. Calculate the area of the Ξ ADE and compare it with the area of Ξ ABC. (Recall Theorem 6.2 and Theorem 6.6). Given π΄π·/π΄π΅=π΄πΈ/π΄πΆ=1/4 π¨π«/π¨π©=π/π 4 AD = AB 4 AD = AD + BD 4 AD β AD = BD 3 AD = BD π΄π·/π΅π·=1/3 π¨π¬/π¨πͺ=π/π 4 AE = AC 4 AE = AE + CE 4 AE β AE = CE 3 AE = CE π΄πΈ/πΆπΈ=1/3 Thus, π΄π·/π΄π΅=π΄πΈ/π΄πΆ=1/3 β΄ DE divides AB and AC in the ratio 1 : 3 Applying theorem 6.2, DE β₯ 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 βADE and βABC β A = β A β ADE = β ABC β΄ βADE ~ βABC 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, (ππ (βπ΄π·πΈ))/(ππ (βπ΄π΅πΆ))=(π΄π·/π΄π΅)^2 (Common angle) (Corresponding angles) (By AA similarity criterion) (ππ (βπ΄π·πΈ))/(ππ (βπ΄π΅πΆ))=(1/4)^2=1/16 ππ (βπ΄π·πΈ)=1/16 ππ (βπ΄π΅πΆ) Now, Finding Area Ξ ABC ππ (βπ΄π΅πΆ)=1/2 [π₯_1 (π¦_2βπ¦_3 ) + π₯_2 (π¦_3βπ¦_1 ) + π₯_3 (π¦_1βπ¦_2 )] Put π₯_1=4, π¦_1=6 π₯_2=1, π¦_2=5 π₯_3=7, π¦_3=2 Ar (βABC) = 1/2 [4 (5β2)+1 (2 β 6)+7 (6 β5)] = 1/2 [4 (3)+1 (β4)+7 (1)] = 1/2 [12 β 4+7] = 15/2 sq. units. Thus, From (1) ar (βADE) = 1/16 ar (βABC) = 1/16Γ15/2=15/32 sq. units. Thus, area of βADE is ππ/ππ sq. units and (ππ (βπ΄π·πΈ))/(ππ (βπ΄π΅πΆ))=π/ππ