AP EAMCET · Maths · Vector Algebra

In quadrilateral $A B C D, \vec{A B} = \vec{a}, \vec{B C} = \vec{b}, \vec{D A} = \vec{a} - \vec{b}, M$ is the midpoint of $B C$ and $X$ is a point on $D M$ such that $\vec{D X} = \frac{4}{5} \vec{D M}$ . Then the points $A, X$ and $C$

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

Verified Solution

Answer & Solution

Correct Answer

(B) are collinear

Step-by-step Solution

Detailed explanation

In ∆ABC, AB→+BC→=AC→ ⇒AC→=a→+b→ ...i In ∆ADC, AD→+DC→=AC→ ⇒DC→=AC→-AD→ ⇒DC→=a→+b→+a→-b→=2a→ M is the midpoint of BC,…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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In quadrilateral ABCD, AB→=a→, BC→=b→, DA→=a→-b→, M is the midpoint of BC and X is a point on DM such that DX→=45DM→. Then the points A, X and C

Step-by-step Solution

In quadrilateral $A B C D, \vec{A B} = \vec{a}, \vec{B C} = \vec{b}, \vec{D A} = \vec{a} - \vec{b}, M$ is the midpoint of $B C$ and $X$ is a point on $D M$ such that $\vec{D X} = \frac{4}{5} \vec{D M}$ . Then the points $A, X$ and $C$