Coordinate Geometry

Pedros wrote:If point A coordinates are ( -7 , -3 ) point B coordinates are (8 ,4 ) if C is along the line A-B so that AC = 1/2 CB. What are the coordinates of point C.

1- ( -2 , - 2/3 )
2- ( 2 , -2/3 )
3- ( 2/3 , -2 )
4 ( 0 , 2 )
5 ( 2 , 2/3 )

Take inkling from

https://www.beatthegmat.com/ab-2bc-repre ... 53772.html

That post relies heavily on some complicated formula.. Does anyone have a more easy way to do this problem by mere logic.. What's the MGMAT EXPLANATION?

Then , it is better to plot the points on a graph on the scratch pad
A(-7, -3) and B is (8,4)
AB will be a bit > 15
and the X coordinate of C will be app. 5 unit away from A and app. 10 units from B
so the x coordinate of C should be around -2 (-7+5) i.e moving app. 5 units towards B
only option (1) has x coordinate as -2

This is heavily dependent on the choice options if there was two answers with x coordinate as - 2 it may onfuse. But the example given in sanju90's thread , again the graph method can be used.
I notice GMAT expects you to know only basic maths but for 700+ Qs will this hold?

pink_08 wrote:That post relies heavily on some complicated formula.. Does anyone have a more easy way to do this problem by mere logic.. What's the MGMAT EXPLANATION?

the section formula will yield A as the answer.

Easier approach: (please employ this method only when you are running short of time / have forgotten the formula)

you have to observe that point A lies in the third quadrant and point B lies in the first.

and the question says, the point C is nearer to A. Therefore, try to find out a point whose X and Y coordinates are closer to -7 and -3 respectively. the closest one will be A.

But again, this guess work will hit the bulls eye in this question and may not work in other questions.

pink_08 wrote:That post relies heavily on some complicated formula.. Does anyone have a more easy way to do this problem by mere logic.. What's the MGMAT EXPLANATION?

the formula is not that tough. It's kind of a mechanical process, if you have the coordinates of two points and the ratio in which the third point divides the line segment joining the first two.

Here, we have (-7,-3) and (8,4) as the coordinates and the ratio in which C divides A&B is 1:2.

|------------1---------|-----------------2--------------|
(-7,-3) ________(x,y)_________________(8,4)


once you get this picture, do the following :

1)multiply the right-ratio with left-coordinate ,
2)multiply left-ratio with right coordinate
3) get their sum
4) divide it with sum of ratio.

Here :

The X coordinate of C according to process above will be :

1) -7 * 2
2) 8 * 1
3) sum = -6
4) division: -6/(2 +1) = -2

Similarly, the Y coordinate :

1)-3*2
2)4*1
3)sum : -2
4)division : -2/(2+1) = -2/3

Hope this is easier now!

harshavardhanc wrote:
the formula is not that tough. It's kind of a mechanical process, if you have the coordinates of two points and the ratio in which the third point divides the line segment joining the first two.

Here, we have (-7,-3) and (8,4) as the coordinates and the ratio in which C divides A&B is 1:2.

|------------1---------|-----------------2--------------|
(-7,-3) ________(x,y)_________________(8,4)


once you get this picture, do the following :

1)multiply the right-ratio with left-coordinate ,
2)multiply left-ratio with right coordinate
3) get their sum
4) divide it with sum of ratio.

Here :

The X coordinate of C according to process above will be :

1) -7 * 2
2) 8 * 1
3) sum = -6
4) division: -6/(2 +1) = -2

Similarly, the Y coordinate :

1)-3*2
2)4*1
3)sum : -2
4)division : -2/(2+1) = -2/3

Hope this is easier now!

Its not only easier, Its......wonderful. Thanks harshavardhanc.

guys you All got it right, the OA is indeed A

The point on segment PQ that is twice as far from P as from Q is:
A) (3,1)
B) (2,1)
C) (2, -1)
D) (1.5, 0.5)
E) (1,0)


Attachments:
https://gmatclub.com/forum/coordinate-pl ... 75135.html

I tried using the formula for this question. but was getting wrong answer. can you please explain how to use the formula for this?

Wow !! I like harsh style..would really help in such Qs and I can save my 1 min at least.

sang5650 wrote:The point on segment PQ that is twice as far from P as from Q is:
A) (3,1)
B) (2,1)
C) (2, -1)
D) (1.5, 0.5)
E) (1,0)


Attachments:
https://gmatclub.com/forum/coordinate-pl ... 75135.html

I tried using the formula for this question. but was getting wrong answer. can you please explain how to use the formula for this?

I think I got my mistake. The ratios are to be interchanged.. Awesome trick!

mx2+nx1/m+n
My2 +ny1 /m+n

shoudl also work directly
the x y cooridnates are given and the ratio is m:n

Pedros wrote:If point A coordinates are ( -7 , -3 ) point B coordinates are (8 ,4 ) if C is along the line A-B so that AC = 1/2 CB. What are the coordinates of point C.

1- ( -2 , - 2/3 )
2- ( 2 , -2/3 )
3- ( 2/3 , -2 )
4 ( 0 , 2 )
5 ( 2 , 2/3 )

Pedros wrote:If point A coordinates are ( -7 , -3 ) point B coordinates are (8 ,4 ) if C is along the line A-B so that AC = 1/2 CB. What are the coordinates of point C.

1- ( -2 , - 2/3 )
2- ( 2 , -2/3 )
3- ( 2/3 , -2 )
4 ( 0 , 2 )
5 ( 2 , 2/3 )

If AC = 1/2CB, then AC = 1/3(AB).

This means that (the difference between the x values of A and C) = 1/3 * (the difference between the x values of A and B).

Thus, if the coordinates of C are (x,y):
x-(-7) = 1/3 [8- (-7)]
x + 7 = 5
x = -2

Only answer choice A has x = -2.

The correct answer is A.

sanju09 wrote:

Pedros wrote:If point A coordinates are ( -7 , -3 ) point B coordinates are (8 ,4 ) if C is along the line A-B so that AC = 1/2 CB. What are the coordinates of point C.

1- ( -2 , - 2/3 )
2- ( 2 , -2/3 )
3- ( 2/3 , -2 )
4 ( 0 , 2 )
5 ( 2 , 2/3 )

Take inkling from

https://www.beatthegmat.com/ab-2bc-repre ... 53772.html

If C lies in the line AB then point c must satisfy the equation of line AB .

y+3=(4+3)/(8+7) * (x+7)

or y+3= 7(x+7)/15 .

only A satisfy the equation .

Pedros here is the solution to your problem!
Check the pdf file were I included all the steps.

summary:
1) find the total distance from point A to point B
2) Use the equation given to be substituted on the new equation knowing that the distance AB must equal to the sume of AC +BC
3) use triagles to find how many units you must add to (-7-3) and your problem will be solved.

cheers!
T

Thanks harshavardhan! that is an amazing method!!!!!!