Points $$ A(\sqrt{2},4 ) $$ , $$ B(6,-\sqrt{3} ) $$ , and $$ C $$ are collinear. If B is the midpoint of line segment $$ AC $$ , approximately what are the $$ (x,y) $$ coordinates of point $$ C $$ ?

Question

Points $$ A(\sqrt{2},4  ) $$ , $$ B(6,-\sqrt{3}  ) $$ , and $$ C $$ are collinear. If B is the  midpoint of line segment $$ AC $$ , approximately what are the $$ (x,y) $$ coordinates of point $$ C $$ ?

Uniteasy Admin · Accepted Answer

"Collinear" implies the three points lie on the same line. Two points will determine a linear equation if their coordinates are given. In this question, the coordinates of $$A$$ and $$B$$ are given, the linear equation for the line is determined. According to the condition B is the midpoint of line segment AC, we can use the coordinates formula for a midpoint between two points to determine the coordinates of C.
$$x_B = \dfrac{x_A+x}{2} $$n
$$y_B = \dfrac{y_A+y}{2} $$n
Substitute the $$x$$ coordinates of A, B to (1)
$$ \dfrac{\sqrt{2}+x  }{2}=6 $$
We get
$$ x= 12-\sqrt{2}\approx 10.59  $$
Substitute the $$y$$ coordinates of A, B to (2)
$$ \dfrac{4+y}{2}=-\sqrt{3}  $$
Then we get
$$ y = -2\sqrt{3}  -4 \approx -7.46 $$
So D is the correct solution.

Multiple Choice Question (MCQ)

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