Ex 10.1, 13 - Chapter 10 Class 11 Straight Lines (Term 1)

Last updated at Feb. 3, 2020 by Teachoo

Transcript

Ex 10.1, 13
If three point (h, 0), (a, b) & (0, k) lie on a line, show that 𝑎/ℎ + 𝑏/𝑘 = 1 .
Let points be A (h, 0), B (a, b), C (0, k)
Given that
A, B & C lie on a line
Hence the 3 points are collinear
∴ Slope of AB = Slope of BC
We know that
Slope of a line through the points (x1, y1), (x2, y2) is
m = (𝑦_2 − 𝑦_1)/(𝑥_2 − 𝑥_1 )
Slope of line AB through the points A(h, 0), B(a, b)
Here x1 = h & y1 = 0
x2 = a & y2 = b
Putting values
m = (𝑏 − 0)/(𝑎 − ℎ)
m = 𝑏/(𝑎 − ℎ)
Slope of line BC through the points B(a, b) & C(0, k)
Here x1 = a & y1 = b
x2 = 0 & y2 = k
Putting values
m = (𝑘 − 𝑏)/(0 − 𝑎)
m = (𝑘 − 𝑏)/(−𝑎)
Now,
Slope of AB = Slope of BC
𝑏/(𝑎 − ℎ) = (𝑘 − 𝑏)/( − 𝑎)
–a(b) = (k – b) (a – h)
–ab = k(a – h) – b(a – h)
–ab = ka – kh – ab + bh
–ab + ab = ka – kh + bh
0 = ka + bh – kh
ka + bh = kh
Dividing both sides by kh
𝑘𝑎/𝑘ℎ + 𝑏ℎ/𝑘ℎ = 𝑘ℎ/𝑘ℎ
𝑎/ℎ + 𝑏/k = 1
Hence proved

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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