question_answer

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

Answer:

Given, the vertices are \[(k+1,1,)\text{ (}4,-3)\]and \[(7,-k)\] and the area of the triangle is 6 square units.

Therefore,

Area \[=\frac{1}{2}[{{x}_{1}}({{y}_{2}}-{{y}_{3}})+{{x}_{2}}({{y}_{3}}-{{y}_{1}})+{{x}_{3}}({{y}_{1}}-{{y}_{2}})]\]

\[6=\frac{1}{2}[(k+1)(-3+k)+4(-k-1)+7(1+3)]\]

\[12=(k+1)(k-3)+4(-k-1)+28\]

\[12={{k}^{2}}-3k+k-3-4k-4+28\]

\[{{k}^{2}}-6k+9=0\]

\[{{k}^{2}}-3k-3k+9=0\]

\[k(k-3)-3(k-3)=0\]

\[(k-3)(k-3)=0\]

\[\therefore k=3,3\]

Hence, value of k is \[3\].