A line passes through point p(1,1) and q(100,1000). How many other poi

Question

A line passes through point p(1,1) and q(100,1000). How many other points with integer co-ordinates are on the line and strictly between points p and q ?

a) 0
b) 2
c) 3
d) 8
e) 9

manpreetsingh86 · Accepted Answer

Let's say point p is at the origin and q is at the (99,999) (this does not change the problem). 
The line pq has the equation y = \frac{999}{99}x =  \frac{111}{11}x  .

Now the co-ordinates of y will be integer, when x takes the value of 11,22,33,44,55,66,77,88.

hence answer is D.

issamL · Answer

Hi folks, here is how i solved the problem :

From the two point coordinates, we obtain the following equation Y= (111/11)*X - 100/11; This equation can be written as follow : (111*X - 100)/11 or More better (100(X-1) + 11 X)/11 .

Now the problem here is that we should obtain an INTEGER Y, in other words (100(X-1) + 11 X) should be divisible by 11.

11X is obviously divisible by 11 so we can neglect it, now our unique friend is 100(X-1) which should be divisible by 11.

X ranges from 1 to 100 , thus the possible values for X to be divisible by 11 are : 12, 23,34,45,56,67,78,89 . So we have 8 values, thus Answer (D) is correct.

Hope it helps

chetan2u · Answer

hi...
 the Y coord is increasing by(1000-1) or999 as x increases by (100-1) or 99..
 so the ratio of increase is 99:999 or 11:111...
 it means with every increase of x by 11 , y increases by 111..
 so the value of 12(1+11) will have y as 112(1+111).. 23(12+11) will have 223(112+111) and so on till 89 will have 889..
 value of x as 1 and 100 is not taken so 8 values 12,23,34,45,56,67,78,89 with corresponding int values for Y

chetan2u · Answer

ans D.....
 p(1,1) and q(100,1000)... so the ratio in which the increase is ,increase of x by 10 results in increase of 100 in y coord..
 so for x, there are 8 such points for which y is also int...

DesiGmat · Answer

Can you please elaborate?

Cadaver · Answer

Line equation y = mx + c....Using the inform...m = 111/11....so y = 111/11x + c......put y=x= 1....c = -100/11
Line eqn :-11y = 111x - 100......x = (11y+100)/111...limit x till 100....only 8 values permissible

Shree9975 · Answer

Hi,

I didnt get the solution.Why can't 99 also be the value ??
When x is 99 ,y can still be an integer..??
Can some1 please explain the problem

jayanthjanardhan · Answer

the slope we will get as 11/111.
Then equation of the line y=11/111x+c
if we take for every mulitple of 111, the value of y is integer, then we will have 9 values.

Can someone pls tell me where i am going wrong?

jeopardo · Answer

Hi,

The error is that c is not an integer. Am I right in suggesting that you have assumed c to be an integer?

AryamaDuttaSaikia · Answer

The slope is y2-y1/x2-x1. So, it is 1000-1/100-1 = 999/99 = 111/11

Also, the value of C needs to be calculated (to check whether it is integer). In this case, the value of C is not an integer = -100/11.

Hope this helps.

kanigmat011 · Answer

Hi Aryama,

Can you extend your explanation further
I mean now we have got the equation of the line as

y= (11/111)x -100/11

Now how to go ahead with this from here
What values one should place here to get integer value of y

Thanks
Kani

AryamaDuttaSaikia · Answer

Slope is calculated by using the formula :

m = y2-y1/x2-x1

Slope intercept format (Equation of the line)

Y = mx+c

In this equation substitute the values of " m "and one of the given coordinates (x,y) to get the value of C.

c= -100/11

So y becomes y=(111/11)x-100/11

= (111x-100) /11

X =1 So we get Y = 1 (given co-ordinates). We want values between (1,1) and (100, 1000)

The next values of X that make Y as integer is 11 (divisor is 11) more than the first value.

12,23 (12+11),34,45,56,67,78,89 = Total 8 values. 
HOPE IT HELPS!!

Fdambro294 · Answer

Concept: to find the INTEGER Values of Coordinates that would occur on a Line, take the Simplified Version of the Slope and figure out how many Integer moves you can make until you get to the End Point (here it is: (100 ; 1,000) )

Slope = Rise/ Run = (1,000 - 1) / (100 - 1) = 999 / 99 = 111 / 11 = (+)Positive Slope

to keep the Points as Integer Coordinates, we can move 111 Units UP along the Y-Axis and 11 Units to the RIGHT along the X-Axis to get to the NEXT Integer Coordinate on the Line. We can keep doing this until we get to Point (100 ; 1,000)

Starting from Point (1 , 1)

(12 ; 112) will be an Integer Coordinate that falls on this line

the 11 Units to the Right of X Coordinate 12 and 111 Units UP from Y Coordinate 112, we will land on another Integer Coordinate that is on the Line:

(23 ; 123)

if you keep doing this, you will find that there are 8 Points that have Integer Coordinates

-D-

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A line passes through point p(1,1) and q(100,1000). How many other poi

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