Question

The point D(_1,3),E(1,7) and F(3K_4,3+2k) on a straight line.Find the value of K?​

Answer 1

Answer:

Step-by-step explanation:

Given that

$\sf{D(-1,3)\,\,;\,\,E(1,7)\,\,;\,\,F(3k-4,3+2k)}$

lies on a straight line

so, the area of the triangle formed by these points will be zero

$\therefore\left|\begin{array}{ccc}-1&3&1\\1&7&1\\3k-4&3+2k&1\end{array}\right|=0$

$\implies-1\left|\begin{array}{cc}7&1\\3+2k&1\end{array}\right|-3\left|\begin{array}{cc}1&1\\3k-4&1\end{array}\right|+1\left|\begin{array}{cc}1&7\\3k-4&3+2k\end{array}\right|=0$

$\implies-1(7-3-2k)-3(1-3k+4)+1(3+2k-21k+28)=0$

$\implies-1(4-2k)-3(5-3k)+1(31-19k)=0$

$\implies-4+2k-15+9k+31-19k=0$

$\implies-8k+12=0$

$\implies\,k=\dfrac{12}{8}$

$\implies\,k=\dfrac{3}{2}$