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# Solution - Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units. - CBSE Class 10 - Mathematics

ConceptArea of a Triangle

#### Question

Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units.

#### Solution

Let A(k + 1, 1), B(4,-3) and c(7,-k)  are the vertices of the triangle.

Given that the area of the triangle is 6sq. units.

Area of the triangle is given by

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

6=1/2[x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)]

12=(k+1)(-3+k)+4(-k-1)+7(1+3)

12=3k+

12=-3k+k^2-3+k-4k-4+28

12=k^2-6k+21

k^2-6k+21-12=0

k^2-6k+9=0

(k-3)^2=0

k=3,3

Is there an error in this question or solution?

#### Reference Material

Solution for question: Find the values of k so that the area of the triangle with vertices (k + 1, 1), (4, -3) and (7, -k) is 6 sq. units. concept: null - Area of a Triangle. For the course CBSE
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