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NotesSlope of a straight line in general equation ax + by + c = 0Slope = - a / bAngle between 2 lines
=
Where ϕ is the angle between two lines and m , m are the slopes of the two linesIntercept of a straight line in general equation ax + by + cz = 0Intercept = - c / bEquation of a straight line in intercept form
= 1Where a and b are intercepts on x-axis and y -axis respectivelyEquation of a straight line in slope - intercept formy =Where m is the slope of the line and c is the intercept that it makes with the y-axisEquation of a straight line in point slope form
=Where A(x ,y ) is a point through which the line passes and m is its slopeDistance between two points A(x1,y1) and B(x2,y2)
Distance =
Where A(x ,y ) and B(x ,y ) are two pointsEquation of a straight line in two - point formy - y =Where A(x ,y ) and B(x ,y ) are two pointsEquation of the line passing through (x , y ) and is perpendicular to
= 0Equation of the line passing through (x , y ) and is parallel to
= 0Area of the triangle formed by ax + by + c = 0 with co-ordinate axisArea =
Whe reCondition of parallelismm = mLines and will be parallel if
i.e., Slope of one line = Slope of the other.Condition of perpendicularitym m = - 1Lines and will be perpendicular
if
orLength of the perpendicular from a given point to a given line
Length of the perpendicular from the point (x , y ) to the line ax + by + c = 0 is =Distance between two parallel lines
d =
Distance between two parallel lines ax + by + c = 0 and ax + by + c = 0 and ax + by + c = 0The locus of the mid point of a systemThe locus of the mid point of a system of parallel chords of a parabola is called its diameter. its equation is y = (2a/m)
Slope of a line The tangent of angle that a line makes with + ve direction of the x - axis in the anticlockwise sense is called slope or gradient of line and isgenerally denoted by m. Thus
(i) Slope of line || to x - axis is m = 0
(ii) Slope of line || to y - axis is (not defined)
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>Coordinate Geometry>Point and Straight Line
JEE
Mathematics
t a n ϕ
∣
∣
− m
1
m
2
1 + m
1
m
2
∣
∣
1 2
+
x
a
y
b
m x + c
y −
y
1
m ( x − )
x
1
1 1
+ ( − ) x
2
x
1
2
( − ) y
2
y
1
2
− − − − − − − − − − − − − − − − − − −
√
1 1 2 2
1 ( x − )
− y
2
y
1
− x
2
x
1
x
1
1 1 2 2
1 1 a x + b y + c = 0
b ( x − ) − a ( y − )
x
1
y
1
1 1 a x + b y + c = 0
a ( x − ) + b ( y − )
x
1
y
1
1
2
∣
∣
c
2
a b
∣
∣
a , b ≠ 0
1 2y = x +
m
1
c
1
y = x +
m
2
c
2
=
m
1
m
2
1 2y = x + c
m
1
y = x + c
m
2
= − 1
m
1
m
2
= −
m
2
1
m
1
1 1+ + c
∣
∣ a x
1
b y
1
∣
∣
+
a
2
b
2
√
−
∣
∣ c
1
c
2
∣
∣
+
a
2
b
2
√
1 2
m = t a n θ
m = ∞
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(iii) Slope of the line equally inclined with the axes is 1 or – 1
(iv) Slope of the line throught the point A (x , y ) and B (x , y ) is (y - y ) / (x - x )
(v) Slope of the line
(vi) Slope of two parallel lines are equal .
(vii) If m & m are slope of two lines then m m = -1Standard form of the equation of line(i) Equation of x - axis is y = 0
(ii) Equation of y - axis is x = 0
(iii) Equation of a straight line || to x-axis at a distance b form it is y = b
(iv) Equation of a straight line || to y - axis at a distance a from it is x = a
(v) Slope form : Equation of a line through the origin and having slope m is y = mx
(vi) Slope Intercept form : Equation of a line with slope m and making an intercept c on the y - axis is y = mx + c
(vii)Point slope form : Equation of a line with slope m and passing through the point (x , y ) is y - y = m (x - x )
(viii) Two point form : Equation of a line passing through the point (x , y ) & (x , y ) is
(ix) Intercept form : Equation of a line making intercepts a and b respectively on x - axis and y - axis is ( x/a) + ( y/b) = 1
(x) Parametric or distance or symmetrical form of the line: Equation of a line passing through (x , y ) and making an angle ,with the +ve direction of x-axis is
Where r is the distance of any point P(x,y) on the line from the point (x , y )
(xi) Normal or Perpendicular form : Equation of a line such that the length of the perpendicular from the origin on it is p and the angle whichthe perpendicular makes with the + ve direction of x-axis is isAngle between two lines(i) Two lines a x + b y + c = 0 & a x + b y + c = 0 are
(a) Parallel if (a /a ) = (b /b ) (c / c )
(b) Perpendicular if a a + b b = 0
(c) Identical or coincident if (a / a ) = (b / b ) = (c / c )
(d) If not above three , then
(ii) Two lines y = m x + c and y = m x + c are
(a) parallel if m = m
(b) perpendicular if m m = -1
(c) If not above two , then
Position of a point with respect to straight lineThe two values L ( x , y ) and L (x , y ) will be of same sign or of opposite sign according to the point A (x , y ) & B (x , y ) lying on same side or onopposite of line L (x, y) respectively .Equation of a line parallel to a given lineEquation of a line parallel (or perpendicular ) to the lineax + by + c = 0 is ax + by + d = 0 (or bx - ay + = 0)Equation of straight line making an angle with other lineEquation of straight line through (x , y ) making an angle with y = mx + c is
Length of perpendicular from a given point to a given line
Length of perpendicular from (x , y ) on ax + by + c = 0 is
Distance between two parallel lines Distance between two parallel lines and = 0
1 1 2 2 2 1 2 1
a + b + c = 0 , b ≠ 0 i s ( − a / b )
1 2 ⊥ 1 2
1 1 1 1
1 1 2 2 =
y − y
1
− y
2
y
1
x − x
1
− x
2
x
1
1 1 θ
0 ≤ θ ≤ π , θ ≠ ( π / 2 ) = = r
x − x
1
c o s θ
y − y
1
s i n θ
⇒ x = + r c o s , θ , y = + r s i n θ
x
1
y
1
1 1
α x c o s α + y s i n α = p
1 1 1 2 2 2
1 2 1 2 ≠ 1 2
1 2 1 2
1 2 1 2 1 2
θ =
t a n
- 1
∣
∣
− a
2
b
1
a
1
b
2
− a
1
a
2
b
1
b
2
∣
∣
1 2
1 2
1 2
θ =
t a n
- 1
∣
∣
− m
1
m
2
1 + m
1
m
2
∣
∣
1 1 2 2 1 1 2 2
λ
1 1 α y − = ( x − )
y
1
m ± t a n α
1 ∓ m t a n α
x
1
1 1+ + + c
∣
∣ a x
1
b y
1
∣
∣
+
a
2
b
2
√
a x + b y +
c
1
a x + b y +
c
2
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is
Condition of concurrency for three straight lines is
Equation of Bisectors of angles between two lines
Family of straight linesThe general equation of family of straight line will be written in one parameter. The equation of straight line which passes through point of intersection of two given lines and can be taken asHomogeneous equation If y = m x and y = m x be the two equation represented by ax + 2hxy + by = 0 , then m + m = -2h/b and m m = a/bGeneral equation of second degree
ax + 2hxy + by + 2gx + 2fy + c = 0 represent a pair straight lines if
If y = m x + c & y = m x + c represents two straight lines then m + m = (-2h / b) , m m = (a/b)Angle between pair of straight lines
The angle between the lines represented by ax + 2hxy + by + 2gx + 2fy + c = 0 or ax + 2hxy + by = 0 is
(i) The two lines given by ax + 2hxy + by = 0 are
(a) Coincident iff h - ab =0
(b) Perpendicular iff a+ b = 0
(ii) The two line given by ax + 2hxy + by + 2gx + 2fy + c = 0 are
(a) Parallel if h - ab = 0
(b) Perpendicular iff a + b = 0
(c) Coincident iff g - ac = 0 , f = bcCombined equation of angle bisector
Combined equation of angle bisector of the angle between the lines ax + 2hxy + by = 0 isDistance Between Two PointsThe distance between two points P(x ,y ) and Q (x , y ) is given by
Let P and Q be two given points whose polar co-ordinates are and respectively.
Area of Triangle (Coordinate Geometry)
The area of a triangle, the co-ordinates of whose vertices are (x , y ), (x , y ) and (x , y ) is
or
Shifting of Origin and Rotation of AxisChange of origin Or Shifting of origin (Translation of axes)
X = x - h and Y = y - k
If origin is shifted to point (h, k) without rotation of axes, then new equation of curve can be obtained by putting x + h in place of x and y + k inplace of y.
−
∣
∣ c
1
c
2
∣
∣
+
a
2
b
2
√
≡ x + y + = 0 , i = 1 , 2 , 3
L
i
a
i
b
i
c
i
= 0
∣
∣
∣
∣
a
1
a
2
a
3
b
1
b
2
b
3
c
1
c
2
c
3
∣
∣
∣
∣
= ±
y −
a
1
x + b
1
c
1
+ a
2
1
b
2
1
√
y +
a
2
x + b
2
c
2
+ a
2
2
b
2
2
√
L
1
L
2
+ λ = 0
L
1
L
2
1 22 2
1 2 1 2
2 2Δ ≡ = 0
∣
∣
∣
∣
a
h
g
h
b
f
g
f
c
∣
∣
∣
∣
1 2 1 2 1 2
2 2 2 2t a n θ =
∣
∣
∣
2 − a b
h
2
√
(
a + b
)
∣
∣
∣
2 2
2
2 2
2
2 2
2 2=
−
x
2
y
2
a − b
x y
h
1 1 2 2
| P Q | =
+ ( − ) x
2
x
1
2
( − ) y
2
y
1
2
− − − − − − − − − − − − − − − − − − −
√
( , )
r
1
θ
1
( , )
r
2
θ
2
| P Q | =
( + − 2 c o s ( − ) ) r
2
1
r
2
2
r
1
r
2
θ
1
θ
2
− − − − − − − − − − − − − − − − − − − − − − −
√
1 1 2 2 3 3
( − ) + ( − ) + ( − )
1
2
∣
∣
x
1
y
2
y
3
y
3
y
1
x
3
y
1
y
2
∣
∣
1
2
∣
∣
∣
∣
∣
∣
∣
∣
x
1
x
2
x
3
y
1
y
2
y
3
1
1
1
∣
∣
∣
∣
∣
∣
∣
∣
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Matrix method :
and
where A' is the transpose matrix of A.Acute and Obtuse Angle Bisector and Position Origini) First re-write the equations of the two lines so that their constant terms are positive.
ii) The bisector of the angle containing the origin and does not containing the origin, then taking +ve and -ve sign in
respectively.
Conditions Acute angle bisector Obtuse angle bisector
a a + b b > 0 - +
a a + b b < 0 + -
1. Bisectors are perpendiculars to each other.2. '+' sign gives the bisector of the angle containing origin.3. If a a + b b > 0 then the origin lies in obtuse angle and if a a + b b < 0, then the origin lies in acute angle.Imp. Points in a TriangleCentroid of a Triangle
Definition : The point of intersection of the medians of a triangle is called the centroid of the triangle and it divides the median internally in theratio 2 : 1.
Theorem : Prove that co-ordinates of the centroid of the triangle whose vertices are (x , y ), (x , y ) and (x , y ) are
Incentre
Definition : The point of intersection of internal angle bisectors of triangle is called the incentre of the triangle.
The co-ordinates of the incentre of a triangle whose vertices are
where, a, b, c are the lengths of sides BC, CA and AB respectively.
i) Excentres of a triangle : This is the point of intersection of the external bisectors of the angles of a triangle.
When and
ii) Circumcentre of a triangle : The circumcentre of a triangle is the point of intersection of the perpendicular bisectors of the sides of a triangle.
The co-ordinates are
or
= = A ( s a y )
X
Y
c o s θ
− s i n θ
s i n θ
c o s θ
x
y
x
y
= =
x
y
c o s θ
s i n θ
− s i n θ
c o s θ
X
Y
A
′
X
Y
= ±
(
x + y +
)
a
1
b
1
c
1
(
+
a
2
1
b
2
1
√
(
x + y +
)
a
2
b
2
c
2
(
+
a
2
2
b
2
2
√
1 2 1 2
1 2 1 2
1 2 1 2 1 2 1 2
1 1 2 2 3 3
( ,
+ + x
1
x
2
x
3
3
+ + y
1
y
2
y
3
3
( ,
+ + a x
1
b x
2
c x
3
a + b + c
+ + a y
1
b y
2
c y
3
a + b + c
≡ ( , I
1
+ + − a x
1
b x
2
c x
3
− a + b + c
+ + − a y
1
b y
2
c y
3
− a + b + c
≡ ( , I
2
− + a x
1
b x
2
c x
3
a − b + c
− + a y
1
b y
2
c y
3
a − b + c
≡ ( , I
3
+ − a x
1
b x
2
c x
3
a + b − c
+ − a y
1
b y
2
c y
3
a + b − c
| B C | = a , | C A | = b | A B | = c
( ,
s i n 2 A + s i n 2 B + s i n 2 C x
1
x
2
x
3
s i n 2 A + s i n e 2 B + s i n e 2 C
s i n 2 A + s i n 2 B + s i n 2 C y
1
y
2
y
3
s i n 2 A + s i n 2 B + s i n 2 C
( ,
c o s A + c o s B + c o s C a x
1
b x
2
c x
3
a c o s A + b c o s B + c c o s C
c o s A + c o s B + c o s C a y
1
b y
2
c y
3
a c o s A + b c o s B + c c o s C
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iii) Orthocentre of a triangle : The orthocentre of a triangle is the point of intersection of altitudes. Its co-ordinates are
or
where andPoint of Intersection
For a lines represented by
ax + 2h xy + by + 2gx + 2fy + c = 0
is given by
LocusThe locus of a moving point is the path traced out by that point under one or more given conditions.
Equation of a Locus
A relation f(x, y) = 0 between x and y which is satisfied by each point on the locus and such that each point satisfying the equation is on the locus
is called the equation of the locus.
How to find the locus of a point
Let (x , y ) be the co-ordinates of the moving point say P. Now apply the geometrical conditions on x , y . This gives a relation between x andy . Now equation of the locus.
Corollary 1 : If x and y are not there in the equation, the co - ordinates of P may also be taken as (x, y).
Corollary 2 : If co-ordinates and equation are not given in the question, suitable choice of origin and axes may be made.
Corollary 3 : To find the locus of the point of intersection of two straight lines, eliminate the parameter or parameters from the given lines. If more than one parameter, then additional condition or conditions will also be given.Area Between Curves
The shaded area = Area of curvilinear trapezoid APQB - Area of curvilinear trapezoid CPQD.
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( ,
t a n A + t a n B + t a n C x
1
x
2
x
3
t a n A + t a n B + t a n C
t a n A + t a n B + t a n C y
1
y
2
y
3
t a n A + t a n B + t a n C
( ,
s e c A + s e c B + s e c C a x
1
b x
2
c x
3
a s e c A + b s e c B + c s e c C
s e c A + s e c B + s e c C a y
1
b y
2
c y
3
a s e c A + b s e c B + c s e c C
| B C | = a , | C A | = b | A B | = c
2 2
( x , y ) = ( ,
b g − f h
− a b h
2
a f − g h
− a b h
2
1 1 1 1 1
1
= f ( x ) d x − g ( x ) d x = = f ( x ) − g ( x ) d x
∫
b
a
∫
b
a
∫
b
a
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0Gurashish
0Karan Bansal
0Satyam Anand
0Satyam Anand
0Rishabh Jain
0Keshav Arora
0Rudresh Kumar shukla
0SUNNY DIGRA
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