co-ordinate geometry.
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- Md.Nazrul Islam
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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.
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...
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...
- Shalabh's Quants
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All 3 points lie in St. Line. So it will have infinite set of solutions for x & y.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.
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,
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- Anurag@Gurome
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Since the three points lie on the same line, so the slope of any two points will be the same.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.
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.
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