THE STRAIGHT
LINE
General Equation of Lines
General Equation of the first degree is of
the form Ax + By + C = 0, where A, B and C are constants. If A = 0, the line is horizontal, if B = 0, the line is vertical, if C = 0
and A and B have the same sign, the line passes through the origin and
inclines to the left and if C = 0 and A
and B have opposite signs, the line passes through the origin and inclines to
the right.
Standard Equations of Lines
1. Point – slope form 4. Intercept form
2. Two point form 5. Normal form
3. Slope – intercept form
POINT-SLOPE FORM
Slope of the line L
m = y –
y1 .
x – x1
y – y1 = m (x
– x1) [ point–slope form ]
where m is
the given slope and (x1, y1) is the given point
Example 1 : Find the equation of the line
passing through ( 3, –2) with slope 2.
y – y1 = m (x
– x1)
y – (–2) = 2( x
– 3 )
y + 2 = 2x –
6
y = 2x – 6 – 2
y = 2x – 8
2x – y = 8
Example 2 : Find the equation of the
line passing through (–3, 2) with slope ½ .
Example 3 : Find the equation of the
line passing through (1, 3) with slope –
2/3.
Example 4 : Find the equation of the
line passing through (2, 0) with slope
3.
Two point
form
Slope of the line, m = y2 – y1 .
x2 – x1
Equation of the line by point–slope
form :
y –
y1 = m (x – x1)
Substituting m for the
slope :
y –
y1 = y2
– y1 ( x – x1
) [ Two-point
form ]
x2
– x1
Example 1 : Find the equation of the line
passing through (–3, 2) and ( 3, 5).
y – y1 = y2 – y1 ( x – x1 ) [ with
(3, 5) as point 1 ]
x2 – x1
y –
5 =
2 – 5 ( x – 3 )
– 3 – 3
y –
5 =
– 3 ( x
– 3 )
– 6
y –
5 = ½ (x – 3 )
2y –
10 = x – 3
2y –
x = 10 – 3
2y –
x = 7
x –
2y = – 7
x – 2y + 7 = 0
Example 2 : Find the equation of the line
passing through (–2, –2) and ( 4, 5).
Example 3 : Find the equation of the line
passing through (–3, 0) and (2, – 4).
Example 4 : Find the equation of the line
passing through ( 4, 0) and (0, – 6).
Example 5 : Find the equation of the line
passing through (–4, 0) and (0, 3).
Slope–Intercept
form
Slope of line L , m = y – b .
x – 0
y – b = mx – 0
y = mx
+ b [ Slope – intercept form ]
where m is the slope and
b the y-intercept of the line
Example 1 : Find the equation of the line
with slope –2 and a y-intercept
4.
y = mx
+ b
y = –2x
+ 4
2x + y
= 4
Example 2 : Find the equation of the line
with slope 2/3 and a y-intercept –1.
Example 3 : Find the equation of the line with
a y-intercept 3/2 and slope 1/3.
Example 4 : Find the equation of the line
with slope – 2/5 and a y-intercept 2.
Example 5 : Find the equation of the line
with a y-intercept 4/3 and slope – 3.
Example 6 : Find the equation of the line with a y-intercept – 3/2 and slope 2.
Example 7 : Find the equation of the line
with a y-intercept 5/2 and slope –
4.
REDUCTION OF GENERAL EQUATION OF LINES TO
SLOPE-INTERCEPT FORM
Example 1:
Reduce the equation 2x – 3y = 6 to slope-intercept form.
Solution :
2x – 3y = 6
– 3y =
–2x + 6
y = (–2x + 6 )/(–3)
y = 2/3 x
– 2
therefore, m = 2/3
and b = – 2
Example
2: Reduce the equation x + 3y
= 3 to slope-intercept form.
Example
3: Reduce the equation x – 2y
= 4 to slope-intercept form.
Example
4: Reduce the equation 3x + y
= 6 to slope-intercept form.
Example
5: Reduce the equation x – 5y
= 5 to slope-intercept form.
Intercept form
Slope of line L
, m = b – 0 = –
b/a
0 – a
Equation of line L by
point – slope form with (a, 0) as point 1
y – y1 = m (x
– x1)
y –
0 =
– b/a (x – a)
y = –
b/a x + b
transpose
the term involving x and multiply all terms by
a
ay + bx = ab
bx +
ay = ab
divide all
terms by ab
x + y =
1 [ Intercept form ]
a b
where a is the x-intercept and b is the
y-intercept
Example 1:
Determine the equation of the line with x-intercept 2/5 and y-intercept
3/4.
x + y =
1
a b
x + y =
1
2/5 3/4
5x + 4y = 1
2 3
6(5x ) + 6(4y) =
6
2 3
3( 5x) + 2(
4y) =
6
15x + 8y
= 6
Example 2:
Determine the equation of the line with x-intercept –3 and y-intercept
4.
Example 3:
Determine the equation of the line with x-intercept ½ and y-intercept
1/3.
Example 4:
Determine the equation of the line with x-intercept 2/3 and y-intercept –4.
Example 5:
Determine the equation of the line with x-intercept –4/5 and y-intercept
–2.
Example 6:
Determine the equation of the line with x-intercept – 4/3 and
y-intercept 3.
Example 7:
Determine the equation of the line with x-intercept 3/5 and y-intercept – 5.
Normal form
Slope of the line segment ON, m = tan a
Slope of the
line L,
m = –1/tan a
Equation of line L
by point – slope form
y – y1 = m (x
– x1)
y – Psin a =
–1/tan a (x – Pcos a)
y tan a - P sin a tan a =
–x + P cos a, but
tan a =
sin a / cos a
y sin a / cos a - P sin 2a/ cos a = –x + P cos a,
x + y
sin a / cos a = P
sin 2a/ cos a + P cos a,
multiply all terms
by cos a
x cos a + y
sin a = P sin
2a + P cos2a
x cos a + y
sin a
= P(sin 2a + cos2a), but
sin 2a + cos2a = 1
\ x cos a + y
sin a
= P
where p is
the normal distance of the line L from the origin and
a is the normal angle
Example 1: Find the equation of the
normal line with p = 2 and a = 60°.
Solution :
x
cos a
+ y sin a =
P
x
cos 60° + y
sin 60° =
2
½
x +
y(Ö3 /2) = 2
x + yÖ3 =
4
Example 2: Find the equation of
the normal line with p = 3 and a = 30°.
Example 3: Find the equation of
the normal line with p = 2 and a = – 30°.
Example 4: Find the equation of
the normal line with p = Ö2 and a = 225°.
Example 5: Find the equation of the
normal line with p = 3 and a = – 120°.
Example 6: Find the equation of
the normal line with p = Ö8 and a = 135°.
DISTANCE FROM A LINE
TO A POINT
The distance d from
the line L to the point P1(x1,
y1 ) can be determined in the following manner. If the equation of L
in normal form is
x cos a + y sin a – p = 0,
the equation of the
line L1 through P1 and parallel to L is defined by
x cos a + y sin a – ( p + d
) = 0
Since P1(x1,
y1 ) is a point on L1, then we have the equation
x1 cos a + y1 sin a – ( p + d ) = 0
x1 cos a + y1 sin a – p – d = 0
hence,
d = x1
cos a + y1
sin a – p
d
= x1 kA + y1
kB – (– kC )
d = A x1 + By1 + C .
± √ A2 + B2
Theorem : The signed distance from the line
Ax + By + C = 0 to the point
P1(x1
, y1 ) is given by the formula
d = A x1 + By1 + C .
± √ A2
+ B2
where the sign of the radical is
chosen to be opposite the sign of C.
When C = 0, choose the sign of
the radical to be the same as the sign of B. This will make
d positive when P1
lies above L. If C = 0, and B = 0, choose the sign of the radical to be
the same as the sign of A. This will make d positive when P1 is to the right of L.
In accordance with
the definition of d in the above theorem, it should be
noted that
1) d is positive if the origin and
the point P1 lie on opposite sides of the
given line; and
2) d is negative when the origin
and P1 lie on the same side of the line.
Example :
1. Find the distance from the point (-3,1) to the
line 3x + 4y –12 =0.
d = A x1 + By1 + C .
± √ A2 + B2
d =
3( -3 ) + 4 ( 1 ) – 12 .
± √ 32
+ 42
d = – 9 + 4 – 12
= – 17
.
± √ 32 + 42 5
A.
Find the distance of the point from the given line :
1) 3x – 2y = 6, ( 2, 4 )
3) 3x – y = 0,
( 2, – 4)
2) 2x + y = 2 ,
(–1, 8 ) 4) 5x –
12y = 1, (4, 1 )
B. Find the distance of the given lines from
the origin.
1) 2x
– 3y = 6 3) 2y
+ 10 = 0
2) x
+ 2y = 2 4)
3x – 12 = 3
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