Question

Find the value of $x$, 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 $6\frac{2}{3}$

The given points are $(2,7),\left(6.1\right)$ and $(x,0)$

of the points are collinear , they will lie on the same line ,i.e, they will not form triangle .

Area of $△$ABC = 0

$⟹\frac{1}{2}\left[x_1\right(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2\left)\right]=0=\frac{1}{2}\left[2\right(1-0)+6(0-7)+x(7-1\left)\right]=0\frac{1}{2}[2-42+6x]=0-40+6x=0⟹x=\frac{40}{6}=6\frac{2}{3}=6.666$