Question

In quadrilateral \(ABCD\),
\[ \overrightarrow{AB}=\vec a, \qquad \overrightarrow{BC}=\vec b, \qquad \overrightarrow{DA}=\vec a-\vec b \] \(M\) is the midpoint of \(BC\) and \(X\) is a point on \(DM\) such that
\[ \overrightarrow{DX}=\frac45\overrightarrow{DM}. \] Then the points \(A,X,C\)

• A. form an equilateral triangle
• B. are collinear
• C. form an isosceles triangle
• D. form a right angled triangle

Hint:

To prove collinearity using vectors, show that one position vector is a scalar multiple of another.

Answer 1

The Correct Option is B

Step 1: Take \(A\) as the origin.
Let the position vector of \(A\) be \(\vec0\).
Given,
\[ \overrightarrow{AB}=\vec a \] therefore, position vector of \(B\) is
\[ \vec B=\vec a \]
Also,
\[ \overrightarrow{BC}=\vec b \] so position vector of \(C\) is
\[ \vec C=\vec a+\vec b \]
Now,
\[ \overrightarrow{DA}=\vec a-\vec b \] which means
\[ \vec A-\vec D=\vec a-\vec b \]
Since \(\vec A=\vec0\),
\[ -\vec D=\vec a-\vec b \] \[ \vec D=\vec b-\vec a \]

Step 2: Find the position vector of midpoint \(M\).
Since \(M\) is the midpoint of \(BC\),
\[ \vec M=\frac{\vec B+\vec C}{2} \]
\[ =\frac{\vec a+(\vec a+\vec b)}{2} \]
\[ =\frac{2\vec a+\vec b}{2} \]
\[ =\vec a+\frac{\vec b}{2} \]

Step 3: Find the position vector of \(X\).
Given,
\[ \overrightarrow{DX}=\frac45\overrightarrow{DM} \]
Thus,
\[ \vec X = \vec D+\frac45(\vec M-\vec D) \]
Substitute \(\vec D=\vec b-\vec a\) and \(\vec M=\vec a+\frac{\vec b}{2}\):
\[ \vec X = (\vec b-\vec a) + \frac45\left( \vec a+\frac{\vec b}{2}-(\vec b-\vec a) \right) \]
\[ = (\vec b-\vec a) + \frac45\left( 2\vec a-\frac{\vec b}{2} \right) \]
\[ = \vec b-\vec a+\frac85\vec a-\frac25\vec b \]
\[ = \frac35\vec a+\frac35\vec b \]
\[ = \frac35(\vec a+\vec b) \]
But,
\[ \vec C=\vec a+\vec b \]
Hence,
\[ \vec X=\frac35\vec C \]
Therefore, \(\vec X\) is a scalar multiple of \(\vec C\).
So, points \(A,X,C\) are collinear.

Step 4: Final conclusion.
Hence,
\[ \boxed{\text{The points }A,X,C\text{ are collinear}} \]