मराठी

Four Points a (6, 3), B (−3, 5), C(4, −2) and D (X, 3x) Are Given in Such a Way that δ D B G δ a B G = 1 2 , Find X - Mathematics

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प्रश्न

Four points A (6, 3), B (−3, 5), C(4, −2) and D (x, 3x) are given in such a way that `(ΔDBG) /(ΔABG)=1/2,` find x 

 

थोडक्यात उत्तर

उत्तर

GIVEN: four points A (6, 3), B (−3, 5) C (4, −2) and D(x, 3x) such that `(ΔDBC)/(ΔABC) `

TO FIND: the value of x

PROOF:

We know area of the triangles formed by three points(x1y1),(x2y2),and (x3y3) is given by `=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)`  

Now 

Area of triangle DBC taking D (x,3x),b(-3,5),c(4-2) 

`triangleDBC=>1/2[x(5-(-2))+(-3)((-2)-3x)+(4)(3x-5)]` 

`triangle DBC=>1/2[7x+6+9x+12x-20]` 

`triangle =>1/2[28x-14]` 

`triangle=>1/2[14(2x-1]) ` 

`triangle =>[7(2x-1)]`   .......(1)

 Area of triangle ABC taking,A(6,3),B(-3,5),C(4,-2) 

`=>1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]` 

`=>1/2[6(5-(-2))+(-3)((-2)- 3)+(4)(3-5)]` 

`=>1/2[6(7)+(-3)(-5)+(4)(-2)]` 

`=>1/2[42+15-8]` 

`=>49/2`                  .......(2) 

Also it is given that 

`(triangle DBC)/(triangleABC)=1/2` 

Substituting the value from (1) and (2) we get 

`(triangle DBC)/(triangleABC)=1/2`

`+-(7(2x-1))/(49/2)=1/2`

`2xx7(2x-1)/49=1/2or-2xx7((2x-1))/49=1/2` 

`(2x-1)=1/2xx7/2or (-2x+1)=1/2xx7/2` 

`2x=7/4+1 or 2x=7/4-1` 

`2x=11/4 or 2x=(-3)/4 `

`x=11/8 or x=(-3)/8`

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पाठ 6: Co-Ordinate Geometry - Exercise 6.5 [पृष्ठ ५४]

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आरडी शर्मा Mathematics [English] Class 10
पाठ 6 Co-Ordinate Geometry
Exercise 6.5 | Q 27 | पृष्ठ ५४
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