Quote:
The line is (1,1) to (100,1000), so the slope of line m is \(\frac{1000-1}{100-1}=\frac{999}{99}=\frac{111}{11}\)
Thus the slope form of the equation can be written as .. y=mx....y=\(\frac{111}{11}\)x
Thus, for all points on line, the slope should be same, so if we take a point (x,y), the slope from (x,y) to (1,1) should be \(\frac{111}{11}\)
So, \(\frac{y-1}{x-1}=\frac{111}{11}.....11y-11=111x-111......11y=111x-100.....11y=11x+100x-100......11(y-x)=100(x-1)\)
Since x and y are integers and 11 and 100 are co-prime, x-1 should be a multiple of 11...
Thus x-1 can be 0,11,22,33,44,55,66,77,88,99. The corresponding x values will be 1,12,23,34,45,56,67,78,89,100, so total 10 values.
But 1 and 100 are already the end points of the line, and we are looking at the points in between, so 10-2=8 values
D
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