#### Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Click to Chat

1800-5470-145

+91 7353221155

CART 0

• 0
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

# Let P be the point that divides the line segment AB in the ration AP:PB=m:n. If O is the origin show that OP=(n/m+n)OA + (m/m+n)OB

10 years ago

Dear student,

Solution:

Step 1: Assume that the point P is dividing the line segment AB in the ratio m: n initially.

Step 2: Assume that P is not unique.

Step 3: Then there is another point Q on AB which divides AB in the ratio m: n internally. From the above diagram we understand

= and =

n AP = m PB and n AQ = m QB

n AP = m(AB – AP) and n AQ = m (AB – AQ)

n AP = m AB – m AP and n AQ = m AB – m AQ

n AP + m AP  = m AB and n AQ + m AQ = m AB

(m + n) AP = m AB and (m + n) AQ = m AB

AP = , AQ =

From above derivation we get AP = AQ

Therefore P and Q are the same point. It is opposite to our assumption. Therefore given (if P is a point which partitions the line segment AB in the ratio m: n, than P is unique) is proved.