
Coordinate
geometry or Analytic geometry 


Line in a coordinate plane

The parametric equations of a line 
Parallel and perpendicular lines 





The parametric equations of a line

If
in a coordinate plane a line is defined by the point P_{1}(x_{1},
y_{1}) and the
direction vector s
then, the position (or
radius)
vector r
of any point P(x,
y) of the line 
r
= r_{1} + t · s,

oo
< t < + oo
and
where, r_{1}
= x_{1}i + y_{1} j
and s
= x_{s}i + y_{s} j, 
represents the vector's equation of the line. 
Therefore,
any point of the line can be reached by the radius vector 
r
= xi + y j
= (x_{1} + x_{s}t) i
+ (y_{1} + y_{s}t) j 
since the scalar quantity t
(called the
parameter) can take any real value from 
oo
to + oo. 
By
writing the scalar components of the above vector's equation obtained is 
x
= x_{1} + x_{s} · t 
y
= y_{1} + y_{s} · t 


the parametric
equations of the line. 



To
convert the parametric equations into the Cartesian coordinates solve
given equations for t.
So 

by
equating 


Therefore,
the parametric equations of a line passing through
two points P_{1}(x_{1},
y_{1}) and P_{2}(x_{2},
y_{2}) 
x
= x_{1} + (x_{2} 
x_{1}) t 
y
= y_{1} + (y_{2} 
y_{1}) t 



Parametric
curves have a direction of motion 
When
plotting
the points of a parametric curve by increasing t, the
graph of the function is traced
out in the direction of motion. 

Example: Write
the parametric equations of the line y
= (1/2)x
+ 3 and sketch its graph. 
Solution: Since 



Let
take the xintercept
as point P_{1}, so 
for
y
= 0 => 0 = (1/2)x
+ 3, x
= 6 therefore,
P_{1}(6,
0). 
Substitute
the values, x_{1}
= 6, y_{1}
= 0, x_{s}
= 2,
and y_{s}
= 1
into the parametric equations of a line 
x
= x_{1} + x_{s} · t,
x
= 6 + 2t 
y
= y_{1} + y_{s} · t,
y = t 

The direction of motion (denoted by red arrows) is given by increasing t. 

Example: Write
the parametric equations of the line through points, A(2,
0) and B(2,
2) and sketch the graph. 
Solution: Plug the coordinates x_{1}
= 2, y_{1}
= 0, x_{2}
= 2, and y_{2}
= 2
into the parametric equations of a line 
x
= x_{1} + (x_{2} 
x_{1}) t,
x
= 2
+ (2 + 2) t
= 2 +
4t,
x
= 2 +
4t, 
y
= y_{1} + (y_{2} 
y_{1}) t,
y = 0
+ (2 
0) t
= 2t,
y
= 2t. 
To
convert the parametric equations into the Cartesian coordinates solve 
x
= 2 +
4t for t
and plug into y
= 2t 
therefore, 





Parallel and perpendicular lines 
Two
lines having slopes m_{1}
and m_{2
}are parallel if 

m_{1}
= m_{2} 

that
is, if they have the same slope. 


To
find the criteria two lines, y
= m_{1}x
and y
= m_{2}x
to be
perpendicular or orthogonal we can
use the principle of similar triangles, OA'A
and OB'B
shown
in the picture. 
Therefore, 
m_{1
}: 1 =
1_{
}: m_{2} 
=> 





This
relation will stay unchanged if we translate the perpendicular
lines, that is, when lines 
y = m_{1}x
+ c_{1
}and y
= m_{2}x
+ c_{2
}are written in the slopeintercept form. 
Two
lines are perpendicular if the slope of one line is the negative
reciprocal of the other. 

Example:
Find the equation of the line that is perpendicular to
the line 

and passes
through 

the point A(2,
5). 










Coordinate
geometry contents 



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