1.

Find the value of k if the area is \(\frac{7}{2}\) sq. units and the vertices are (1,2), (3,5), (k,0).(a) \(\frac{8}{3}\)(b) –\(\frac{8}{3}\)(c) –\(\frac{7}{3}\)(d) –\(\frac{8}{5}\)I have been asked this question in an online quiz.The question is from Area of a Triangle in portion Determinants of Mathematics – Class 12

Answer»

Right OPTION is (b) –\(\frac{8}{3}\)

Explanation: Given that the vertices are (1,2), (3,5), (k,0)

Therefore, the area of the TRIANGLE with vertices (1,2), (3,5), (k,0) is given by

Δ=\(\frac{1}{2}\begin{Vmatrix}1&2&1\\3&5&1\\k&0&1\end{Vmatrix}\)=\(\frac{7}{2}\)

Expanding ALONG R3, we get

\(\frac{1}{2}\) {k(2-5)-0+1(5-6)}=\(\frac{1}{2}\) {-3k-1}=\(\frac{7}{2}\)

⇒-\(\frac{1}{2}\) (3k+1)=\(\frac{7}{2}\)

3k=-8

k=-\(\frac{8}{3}\).


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