Prove that (a,0),(0,b) and (1,1) are collinear,if 1/a+1/b=1.
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Thanks Steve..:-))
@Bosnian:- I already told that I have no enough time for wasting.
After posting the question here,I can do the remaining questions in my book.understood:-))
Steve,yesterday I asked a question and u answered that but u r confused and asked me how do u got 25 as answer?did u remember that?and I answered how I got the answer as 25.can u please check it out whether it is right or not?
Forgot to say thank u to Bosnian..:)
I like your 25 stones.
Really..Anyway thanks Steve:)
To prove that the points (a, 0), (0, b), and (1, 1) are collinear, we need to determine if the slope between any two pairs of points is the same.
Let's start by finding the slope between the points (a, 0) and (0, b):
The slope, m1, between (a, 0) and (0, b) is given by:
m1 = (b - 0) / (0 - a) [using the slope formula: (y2 - y1) / (x2 - x1)]
Simplifying, we have:
m1 = b / (-a) [since 0 - a = -a and b - 0 = b]
Next, let's find the slope between the points (0, b) and (1, 1):
The slope, m2, between (0, b) and (1, 1) is given by:
m2 = (1 - b) / (1 - 0)
Simplifying, we have:
m2 = (1 - b) / 1 [since 1 - 0 = 1]
Now, we have the slopes m1 = b / (-a) and m2 = (1 - b) / 1.
To prove that the points are collinear, we need to show that m1 = m2:
So, we equate m1 and m2:
b / (-a) = (1 - b) / 1
Cross-multiplying, we get:
b * 1 = (-a) * (1 - b)
Simplifying, we have:
b = -a + ab
Rearranging, we get:
ab + a = b
Factoring out 'a' on the left side, we have:
a(b + 1) = b
Dividing both sides by (b + 1), we get:
a = b / (b + 1)
Now, since we are given that 1/a + 1/b = 1, we can substitute the value of a:
1 / (b / (b + 1)) + 1 / b = 1
Simplifying further, we get:
(b + 1) / b + 1 / b = 1
Combining the fractions on the left side, we have:
((b + 1) + 1) / b = 1
Simplifying, we get:
(b + 2) / b = 1
Cross-multiplying, we have:
b + 2 = b
This equation is not possible, as it leads to a contradiction (b + 2 can never be equal to b). Therefore, the given points (a, 0), (0, b), and (1, 1) are not collinear.
Hence, we have proven that (a, 0), (0, b), and (1, 1) are not collinear, given that 1/a + 1/b = 1.
I do in this way....
If A(a,0),B(0,b),C(1,1) are collinear,
Area(triangle ABC)=0
1/2[X1(y2-y3)+X2(y3-y1)+x3(y1-y2)]=0
[a(b-1)+0(1-0)+1(0-b)=0
[ab-a-b]=0
ab=a+b
Dividing throughout by ab we get
1=1/a+1/b
Therefore,1/a+1/b=1
Thus proved.....Is it right???
The line containing (a,0),(0,b) has equation
x/a + y/b = 1
This is called the intercept form of a line.
Clearly, (1,1) lies on that line if 1/a + 1/b = 1