Question

Find the value for $x$, so that the three points,$\{(2,7),(6,1),(x,0)\}$ are collinear.

• A. $7$
• B. $4\frac{1}{2}$
• C. $10$
• D. $6\frac{2}{3}$

Answer 1

The correct option is D $6\frac{2}{3}$

$A(2,7)=(x_1,y_1)$;

$B(6,1)=(x_2,y_2)$;

$c(x,0)=(x_3,y_3)$;

If ABC is collinear there

Area of $\Delta$le ABC$=0$

$\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0$

$\frac{1}{2}\left[2\right(1-0)+6(0-7)+x(7-1\left)\right]=0$

$\frac{1}{2}[2-42+6x]=0$

$2-42+6x=0$

$40-6x=0$

$6x=40$

$x=\frac{40}{6}$

$=\frac{20}{3}$.