F our points 6-3- B -3-5- C 4-2 and D x - 3 x are given in such a way that - DBC - ABC -1-2- find x-

A(6,3),B( 3,5),C(4, 2)and D(x,3x)or (Δ ABC) =1/2[x_1(y_2 y_3)+x_2(y_3 y_1)+x_3(y_1 y_2)] =1/2[ 3( 2 3x)+4(3x 5)+x(5+x)] =1/2[6+9x+12x 20+5x+2x] =1/2[28x 14] ...

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Byju's Answer

Standard XII

Mathematics

Area of Triangle Using Coordinates

F our points ...

Question

$F o u r p o i n t s A (6, 3), B (- 3, 5), C (4, - 2) a n d D (x, 3 x) a r e g i v e n i n s u c h a w a y t h a t \frac{Δ D B C}{Δ A B C} = \frac{1}{2}, f i n d x .$

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Solution

$A (6, 3), B (- 3, 5), C (4, - 2) a n d D (x, 3 x) o r (Δ A B C) = \frac{1}{2} [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})] = \frac{1}{2} [- 3 (- 2 - 3 x) + 4 (3 x - 5) + x (5 + x)] = \frac{1}{2} [6 + 9 x + 12 x - 20 + 5 x + 2 x] = \frac{1}{2} [28 x - 14] = 7 [2 x - 1] o r (Δ A B C) = \frac{1}{2} [6 (5 + 2) - 3 (- 2 - 3) + 4 (3 - 5)] = \frac{1}{2} [42 + 15 - 8] = \frac{49}{2} \frac{o r (Δ A B C)}{o r (Δ A B C)} = \frac{1}{2} \frac{7 (2 x - 1)}{49} = \frac{1}{2} \Rightarrow \frac{14 (2 x - 1)}{49} = \frac{1}{2} \Rightarrow \frac{(28 x - 14)}{49} = \frac{1}{2} 56 x - 28 = 49 \Rightarrow 56 x = 28 + 49 \Rightarrow 56 x = 77 x = \frac{11}{8}$

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Four pointsA(6,3),B(−3,5),C(4,−2)and D(x,3x)are given in such a way thatΔ DBCΔ ABC=12,find x.

$F o u r p o i n t s A (6, 3), B (- 3, 5), C (4, - 2) a n d D (x, 3 x) a r e g i v e n i n s u c h a w a y t h a t \frac{Δ D B C}{Δ A B C} = \frac{1}{2}, f i n d x .$