ABCD is a parallelogram with AB = q and AD = 4.
ABM is a straight line with AB : BM= 1:1.
ADN is a straight line with AD : DN = 3:2.
(a)Write MN, in terms of p and q, in its simplest form.
Answers
Given :-
- ABCD is a parallelogram with AB = q and AD = 4.
- ABM is a straight line with AB : BM= 1:1.
- ADN is a straight line with AD : DN = 3:2.
To Find :- Write MN, in terms of p and q, in its simplest form. ?
Solution :-
→ AB = q (given)
and,
→ AB : BM = 1 : 1
so,
→ AM = AB + BM = q + q = 2q .
now,
→ AD = p
and,
→ AD : DN = 3 : 2
so,
→ p/DN = 3/2
→ DN = (2p/3)
then,
→ AN = AD + DN = p + (2p/3) = (5p/3)
Now, given that, the diagram is not to scale . Then, we can say that, this is not a normal geometry question . It is a vector geometry problem .
therefore,
→ MN = AM + AN ( Triangle law of vector addition. )
→ MN = -MA + AN (negative vector.)
→ MN = (-2q) + (5p/3)
→ MN = [(5p/3) - (2q)] (Ans.)
Learn more :-
in triangle ABC seg DE parallel side BC. If 2 area of triangle ADE = area of quadrilateral DBCE find AB : AD show that B...
https://brainly.in/question/15942930
2) In ∆ABC seg MN || side AC, seg MN divides ∆ABC into two parts of equal area. Determine the value of AM / AB
https://brainly.in/question/37634605
