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AP EAMCET · Maths · Vector Algebra

\(A B C D\) is a parallelogram such that \(L\) is mid-point of \(B C\), then \(\mathbf{A L}\) is equal to

  1. A \(\mathrm{DC}+\frac{1}{2} \mathrm{AD}\)
  2. B \(\frac{1}{2} A D+B C\)
  3. C \(\frac{1}{2} \mathrm{AD}+\mathrm{DL}\)
  4. D \(\frac{1}{2} \mathrm{AD}+\mathrm{BL}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{DC}+\frac{1}{2} \mathrm{AD}\)

Step-by-step Solution

Detailed explanation

\(A B C D\) is a parallelogram such that \(L\) is mid-point on \(B C\) Now for \(\mathbf{A L}\) Consider, \(\triangle A B L\) \[ \mathbf{A B}+\mathbf{B L}=\mathbf{A L} \] or…