 Click to Chat

1800-1023-196

+91-120-4616500

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

proof of section formula in straight line , where a point devides line sigment in m:n ratio

one year ago

Dear Ankit

Consider ? ABB’

Since BB’||PQ and AP:PB = m:n (see figure given below)

AQ/AB' = PQ/BB' = m/(m+n) (= AP/AB)

? x – x1/x2 – x1 = m/(m+n)

? x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n)

If P is outside AB (less assume it is at P’)

We have (x – x1)/(x2 – x1) = m/(m + n)

? x = (nx1 + mx2)/(m + n) and y = (ny1 + my2)/(m + n)

Similarly if P is at P” then

x = (–mx+ nx1)/(n-m), y = (– my+ ny1)/(n-m)

Note:

m:n can be written as m/n or ?:1. So any point on line joining A and B will be P{(?x+ x1)/(? + 1), (?y+ y1)/(? +1)}. It is useful to assume ?:1 because it involves only one variable.

Regards

Arun (askIITians forum expert)

one year ago
Think You Can Provide A Better Answer ?
Answer & Earn Cool Goodies

Other Related Questions on Algebra

View all Questions »  Course Features

• 731 Video Lectures
• Revision Notes
• Previous Year Papers
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Test paper with Video Solution  Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions