Fit a straight line passing through points (0,1), (1,2), (2,3), (3,4,5), (4,6), (5,7,5).

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written 4.6 years ago by

teamques10 ★ 70k

Answer:

We shall solve this problem by direct method.

x	y	$x^2$	$y^2$	xy
0	1	0	1	0
1	2	1	4	2
2	3	4	9	6
3	4.5	9	20.25	13.5
4	6	16	36	24
5	7.5	25	56.25	37.5
N=6 ; 15	24.0	55	126.5	83

Let the equation of the line be y =ax+b

Then normal equations are

$\sum y= Na+ b \sum x \\ \therefore 24= 6a+ 15 b \; \; \; \; \dots(1) $

$ and \; \sum xy = a \sum x +b \sum x^2 \\ \therefore 83= 15a + 55 b \; \; \; \; \dots (2) $

Now, multiply equation (1) by 15 and equation (2) by 6 and then subtract (1) from (2)

$\dfrac{498 \; = \; 90 a \; + \; 330b \; \\ 360\; = \; 90a\; + \; 225 b } { \therefore \; b=1.32} $

$\therefore a = 0.70 $

Hence the equation of the line will be

y= 0.70 x +1.32

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