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co-ordinate geometry.

Expert replies
Source: — Problem Solving |

by gmatmath » Sat Apr 07, 2012 9:07 pm
If the 3 points lie on the same line, they are collinear.
For 3 points to be collinear, the slope of 2 points taken at a time will be equal.
A(5,3), B(x,y), C(3,2)
slope for 2 points(x1,y1) and (x2,y2) is given by (y2-y1)/(x2-x1)

Slope m1 = (y-3)/(x-5)
slope m2 = (2-y)/(3-x)

m1 = m2 as they lie on the same line.

(y-3)/(x-5) = (2-y)/(3-x)
==> (y-3)(3-x) = (2-y)(x-5)
==> 3y - 9 -xy +3x = 2x - 10 - xy + 5y
==> x + 1 = 2y-------(eqn 1)
slope of AC = m3 = (2-3)/(3-5)
m3 = -1/-2
m3 = 1/2
from eqn 1, x = 2y - 1

giving values for y, we can get values for x.

y = 0, x = -1
y = 1, x = 1
y = 2, x = 3

and so on...
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by Shalabh's Quants » Sun Apr 08, 2012 5:09 am
Md.Nazrul Islam wrote:In a co-ordinate system , if three points (5,3)(x,y) and (3,2)lie on a same line ,fine the value of X.
All 3 points lie in St. Line. So it will have infinite set of solutions for x & y.

Lets find out Eqn of St. Line...

y-y'=[(y"-y')/(x"-x")]*(x-x')

=> y-3=[(2-3)/(3-5)]*(x-5)
.
.
.
=> It reduces to x=2y+1;

As it is a linear eqn with 2 variables, hence it will have infinite solutions.

=> This eqn. will seek any value of x & yield corresponding infinite values of y.
Shalabh Jain,
e-GMAT Instructor
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by Anurag@Gurome » Sun Apr 08, 2012 7:53 pm
Md.Nazrul Islam wrote:In a co-ordinate system , if three points (5,3)(x,y) and (3,2)lie on a same line ,fine the value of X.
Since the three points lie on the same line, so the slope of any two points will be the same.
Now slope of a line passing through (x1, y1) and (x2, y2) = (y2 - y1)/(x2 - x1)

Slope of line passing through (5, 3) and (x , y) = (y - 3)/(x - 5)
Slope of line passing through (x, y) and (3 , 2) = (2 - y)/(3 - x)
Now, Slope of line passing through (5, 3) and (x , y) = Slope of line passing through (x, y) and (3 , 2)
(y - 3)/(x - 5) = (2 - y)/(3 - x)
(y - 3)(3 - x) = (2 - y)(x - 5)
3y - 9 - xy + 3x = 2x - xy - 10 + 5y
x - 2y = -1 ... Equation 1

Similarly, slope of line through (5, 3) and (3, 2) = Slope of line passing through (x, y) and (3 , 2)
(2 - 3)/(3 - 5) = (2 - y)/(3 - x)
(-1/-2) = (2 - y)/(3 - x)
1/2 = (2 - y)/(3 - x)
3 - x = 2(2 - y)
3 - x = 4 - 2y
x - 2y = -1 ... Equation 2

Both equations 1 and 2, give the same equation, so we can have infinite solutions to this, and hence infinite values for x.
Anurag Mairal, Ph.D., MBA
GMAT Expert, Admissions and Career Guidance
Gurome, Inc.
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