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For What Value of K(K>0) is the Area of the Triangle with Vertices (-2, 5), (K, -4) and (2k+1, 10) Equal to 53 Square Units? - Mathematics

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For what value of k(k>0) is the area of the triangle with vertices (-2, 5), (k, -4) and (2k+1, 10) equal to 53 square units?

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Solution

`"Let " A(x_1=-2,y_1=5),B(x_2=k,y_2=-4) and C (x_3 = 2k+1,y_3=10)` be the vertices of the  triangle, So

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

`⇒53=1/2[(-2)(-4-10)+k(10-5)+(2k+1)(5+4)]`

`⇒53=1/2[28+5k+9(2k+1)]`

⇒ 28+5k+18k+9=106

⇒37+23k=106

⇒23k=106-37=69

`⇒k=69/23=3`

Hence , k=3.

Concept: Area of a Triangle
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APPEARS IN

RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 12
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