(a) Find the coordinates of the point which divides the line joining (7, -5) and (-2, 7) externally in the ration 3 : 2. (b) Without using calculators or ma...
Question 1 Report
(a) Find the coordinates of the point which divides the line joining (7, -5) and (-2, 7) externally in the ration 3 : 2.
(b) Without using calculators or mathematical tables, evaluate \(\frac{2}{1 + \sqrt{2}}\) - \(\frac{2}{2 + \sqrt{2}}\), leaving the answer in the form p + q\(\sqrt{n}\), where p, q and n are integers.
Answer Details
(a) For external division of \((x_1,y_1)=(7,-5)\) and \((x_2,y_2)=(-2,7)\) in the ratio \(m:n = 3:2\):
\[x = \frac{m x_2 - n x_1}{m - n} = \frac{3(-2) - 2(7)}{3 - 2} = \frac{-6 - 14}{1} = -20\] \[y = \frac{m y_2 - n y_1}{m - n} = \frac{3(7) - 2(-5)}{3 - 2} = \frac{21 + 10}{1} = 31\]The point is \((-20,\ 31)\).
(b) Rationalise each term.
\[\frac{2}{1+\sqrt2}\times\frac{\sqrt2-1}{\sqrt2-1} = \frac{2(\sqrt2-1)}{(\sqrt2)^2-1^2}=\frac{2(\sqrt2-1)}{1}=2\sqrt2-2\] \[\frac{2}{2+\sqrt2}\times\frac{2-\sqrt2}{2-\sqrt2} = \frac{2(2-\sqrt2)}{4-2}=\frac{2(2-\sqrt2)}{2}=2-\sqrt2\]Therefore
\[\frac{2}{1+\sqrt2}-\frac{2}{2+\sqrt2} = (2\sqrt2-2)-(2-\sqrt2) = 3\sqrt2 - 4\]In the form \(p+q\sqrt n\): \(p=-4,\ q=3,\ n=2\), i.e. \(-4+3\sqrt2\).
Download The App On Google Playstore
Everything you need to excel in your exams
Personalized AI Learning Chat Assistant
200,000+ Exam Questions Across IGCSE, JAMB, WAEC & NECO
Over 3,900 Lesson Notes
Offline Support - Learn Anytime, Anywhere
Green Bridge Timetable
Literature Summaries & Potential Questions
Track Your Performance & Progress
In-depth Explanations for Comprehensive Learning