STRAIGHT LINE
- 1. RELATION BETWEEN CARTESIAN CO-ORDINATE & POLAR CO-ORDINATE SYSTEM
- 2. DISTANCE FORMULA AND ITS APPLICATIONS :
- 3. SECTION FORMULA:
- 4. CO-ORDINATES OF SOME PARTICULAR POINTS :
- 5. AREA OF TRIANGLE :
- 6. CONDITION OF COLLINEARITY FOR THREE POINTS :
- 7. EQUATION OF STRAIGHT LINE :
- 8. SLOPE OF LINE :
- 9. STANDARD FORMS OF EQUATIONS OF A STRAIGHT LINE :
- 10. ANGLE BETWEEN TWO LINES:
- 11. LENGTH OF PERPENDICULAR FROM A POINT ONA LINE :
- 12. DISTANCE BETWEEN TWO PARALLEL LINES:
- 13. EQUATION OF LINES PARALLEL AND PERPENDICULAR TO A GIVEN LINE :
- 14. STRAIGHT LINE MAKING A GIVEN ANGLE WITH A LINE :
- 15. POSITION OF TWO POINTS WITH RESPECT TO A GIVEN LINE :
- 16. CONCURRENCY OF LINES :
- 17. REFLECTION OF A POINT :
- 18. TRANSFORMATION OF AXES
- 19. EQUATION OF BISECTORS OF ANGLES BETWEEN TWO LINES :
- 20. FAMILY OF LINES :
- 21. GENERAL EQUATION AND HOMOGENEOUS EQUATION OF SECOND DEGREE :
- 22. EQUATIONS OF LINES JOINING THE POINTS OF INTERSECTION OF A LINE AND A CURVE TO THE ORIGIN :
- 23. STANDARD RESULTS :
1. RELATION BETWEEN CARTESIAN CO-ORDINATE & POLAR CO-ORDINATE SYSTEM
and r=√x2+y2,θ=tan−1(yx)
2. DISTANCE FORMULA AND ITS APPLICATIONS :
If A(x1,y1) and B(x2,y2) are two points, then AB=√(x2−x1)2+(y2−y1)2
Note :
(i) Three given points A,B and C are collinear, when sum of any two distances out of AB,BC,CA is equal to the remaining third otherwise the points will be the vertices of triangle.
(ii) Let A,B,C&D be the four given points in a plane. Then the quadrilateral will be:
(b) Rhombus if AB=BC=CD=DA and AC≠BD;AC⊥BD
(c) Parallelogram if AB=DC,BC=AD;AC≠BD;AC⊥̸BD
(d) Rectangle if AB=CD,BC=DA,AC=BD;AC⊥̸BD
3. SECTION FORMULA:
The co-ordinates of a point dividing a line joining the points A(x1,y1) and B(x2,y2) in the ratio m:n is given by:
(b) For external division : (mx2−nx1m−n,my2−ny1m−n)
(c) Line ax+by+c=0 divides line joining points P(x1,y1) & Q(x2,y2) in ratio =−(ax1+by1+c)(ax2+by2+c)
4. CO-ORDINATES OF SOME PARTICULAR POINTS :
Let A(x1,y1),B(x2,y2) and C(x3,y3) are vertices of any triangle ABC, then
(a) Centroid :
(i) The centroid is the point of intersection of the medians (line joining the mid point of sides and opposite vertices).(ii) Centroid divides the median in the ratio of 2:1.
(iii) Co-ordinates of centroid G(x1+x2+x33,y1+y2+y33)
(iv) If P is any internal point of triangle such that area of △APB, △APC and ΔBPC are same then P must be centroid.
(b) Incenter:
Co-ordinates of incenter I (ax1+bx2+cx3a+b+c,ay1+by2+cy3a+b+c)
Where a,b,c are the sides of triangle ABC.
(i) Angle bisector divides the opposite sides in the ratio of remaining sides. e.g. BDDC=ABAC=cb
(ii) Incenter divides the angle bisectors in the ratio (b+c):a,(c+a):b,(a+b):c
(c) Circumcenter :
Also it is a centre of a circle touching all the vertices of a triangle.
(i) If a triangle is right angle, then its circumcenter is mid point of hypotenuse.
(ii) Find perpendicular bisector of any two sides and solve them to find circumcentre.
(d) Orthocenter:
It is the point of intersection of perpendicular drawn from vertices on opposite sides of a triangle and can be obtained by solving the equation of any two altitudes.
Note: If a triangle is right angled triangle, then orthocenter is the point where right angle is formed.
(i) If the triangle is equilateral, then centroid, incentre, orthocenter and circumcenter coincides.
(ii) Orthocentre, centroid and circumcentre are always collinear and centroid divides the line joining orthocentre and circumcentre in the ratio 2:1
(iii) In an isosceles triangle centroid, orthocentre, incentre, circumcentre lies on the same line.
(e) Ex-centers:
The centre of the circle which touches side BC and the extended portions of sides AB and AC is called the ex-centre of ΔABC with respect to the vertex A. It is denoted by I1 and its coordinates are
I1=(−ax1+bx2+cx3−a+b+c,−ay1+by2+cy3−a+b+c)
I2=(ax1−bx2+cx3a−b+c,ay1−by2+cy3a−b+c)
I3=(ax1+bx2−cx3a+b−c,ay1+by2−cy3a+b−c)
5. AREA OF TRIANGLE :
Let A(x1,y1),B(x2,y2) and C(x3,y3) are vertices of a triangle, then Area of △ABC=12|x1y11x2y21x3y31|=12∣[x1(y2−y3)+x2(y3−y1)+x3(y1−y2)]|
To remember the above formula, take the help of the following method:
Remarks:
(i) If the area of triangle joining three points is zero, then the points are collinear.
(ii) Area of Equilateral triangle
If 'a' be the side of equilateral triangle, then its area =(a2√34)
(iii) Area of quadrilateral whose consecutive vertices are (x1,y1),(x2,y2), (x3,y3)&(x4,y4) is 12|x1−x3y1−y3x2−x4y2−y4|
6. CONDITION OF COLLINEARITY FOR THREE POINTS :
y3−y1y2−y1=x3−x1x2−x1 or x1−x2x1−x3=y1−y2y1−y3 or |x1y11x2y21x3y31|=0
7. EQUATION OF STRAIGHT LINE :
A relation between x and y which is satisfied by co-ordinates of every point lying on a line is called equation of the straight line. Here remember that every one degree equation in variable x and y always represents a straight line i.e. ax+by+c=0; a & b ≠0 simultaneously.
(b) Equation of x -axis is y=0
(c) Equation of line parallel to y -axis at a distance b is x=b or x=−b
(d) Equation of y -axis is x=0
8. SLOPE OF LINE :
If a given line makes an angle θ (0∘≤θ<180∘,θ≠90∘) with the positive direction of x -axis, then slope of this line will be tanθ and is usually denoted by the letter m i.e. m=tanθ. Obviously the slope of the x -axis and line parallel to it is zero and y -axis and line parallel to it does not exist.
If A(x1,y1) and B(x2,y2) & x1≠x2 then slope of line AB=y2−y1x2−x1
9. STANDARD FORMS OF EQUATIONS OF A STRAIGHT LINE :
(a) Slope Intercept form : Let m be the slope of a line and c its intercept on y-axis. then the equation of this straight line is written as : y=mx+c
(b) Point Slope form : If m be the slope of a line and it passes through a point (x1,y1), then its equation is written as : y−y1=m(x−x1)
(c) Two point form : Equation of a line passing through two points (x1,y1) and (x2,y2) is written as
y−y1=y2−y1x2−x1(x−x1) or |xy1x1y11x2y21|=0
(d) Intercept form : If a and b are the intercepts made by a line on the axes of x and y, its equation is written as : xa+yb=1
(e) Normal form : If p is the length of perpendicular on a line from the origin and α the angle which this perpendicular makes with positive x -axis, then the equation of this line is written as: xcosα+ysinα=p(p is always positive), where 0≤α<2π.
Any point P on the line will be of the form (h+rcosθ,k+rsinθ) where |r| gives the distance of the point P from the fixed point (h,k)
(g) General form : We know that a first degree equation in x and y,ax+by+c=0 always represents a straight line. This form is known as general form of straight line.
(ii) Intercept made by this line on x -axis =−ca and intercept made by this line on y -axis =−cb
(iii) To change the general form of a line to normal form, first take c to right hand side and make it positive, then divide the whole equation by √a2+b2.
10. ANGLE BETWEEN TWO LINES:
(a) If θ be the angle between two lines :y=m1x+c1 and y=m2x+c2, then tanθ=±(m1−m21+m1 m2)
(b) If equation of lines are a1x+b1y+c1=0 and a2x+b2y+c2=0, then these line are
(ii) Perpendicular ⇔a1a2+b1 b2=0
(iii) Coincident ⇔a1a2=b1 b2=c1c2
(iv) Intersecting ⇔a1a2≠b1 b2
11. LENGTH OF PERPENDICULAR FROM A POINT ONA LINE :
Length of perpendicular from a point (x1,y1) on the line ax+by+c=0 is =|ax1+by1+c√a2+b2|
In particular the length of the perpendicular from the origin on the line ax+by+c=0 is P=|c|√a2+b2
12. DISTANCE BETWEEN TWO PARALLEL LINES:
(Note : The coefficients of x & y in both equations should be same)
(b) The area of the parallelogram =p1p2sinθ, where p1 & p2 are distances between two pairs of opposite sides & θ is the angle between any two adjacent sides. Note that area of the parallelogram bounded by the lines y=m1x+c1,y=m1x+c2 and y=m2x+d1,y=m2x+d2 is given by |(c1−c2)(d1−d2)m1−m2|.
13. EQUATION OF LINES PARALLEL AND PERPENDICULAR TO A GIVEN LINE :
(b) Equation of line perpendicular to line ax+by+c=0 bx−ay+k=0
Here λ,k, are parameters and their values are obtained with the help of additional information given in the problem.
14. STRAIGHT LINE MAKING A GIVEN ANGLE WITH A LINE :
y−y1=m±tanα1∓mtanα(x−x1)
15. POSITION OF TWO POINTS WITH RESPECT TO A GIVEN LINE :
Let the given line be ax+by+c=0 and P(x1,y1),Q(x2,y2) be two points. If the quantities ax1+by1+c and ax2+by2+c have the same signs, then both the points P and Q lie on the same side of the line ax+by+c=0. If the quantities ax1+by1+c and ax2+by2+c have opposite signs, then they lie on the opposite sides of the line.
16. CONCURRENCY OF LINES :
Three lines a1x+b1y+c1=0;a2x+b2y+c2=0 and a3x+b3y+c3=0 are concurrent, if Δ=|a1 b1c1a2 b2c2a3 b3c3|=0
If lines are concurrent then Δ=0 but if Δ=0 then lines may or may not be concurrent lines may be parallel.
17. REFLECTION OF A POINT :
Let P(x,y) be any point, then its image with respect to
(b) y -axis is R(−x,y)
(c) origin is S(−x,−y)
(d) line y=x is T(y,x)
18. TRANSFORMATION OF AXES
(a) Shifting of origin without rotation of axes:
If coordinates of any point P(x,y) with respect to neworigin (α,β) will be (x′,y′) then x=x′+α,y=y′+β or x′=x−α,y′=y−β
Thus if origin is shifted to point (α,β) without rotation of axes, then new equation of curve can be obtained by putting x+α in place of x and y+β in place of y.
(b) Rotation of axes without shifting the origin
Let O be the origin. Let with respect to axes OY and let P≡(x′,y′) with to axes OX' and OY', ∠X′OX=∠YOY′=θ
then x=x′cosθ−y′sinθy=x′sinθ+y′cosθ
y′=−xsinθ+ycosθ
The above relation between (x,y) and (x′,y′) can be easily obtained with the help of following table
New \ Oldx↓y↓x′→cosθsinθy′→−sinθcosθ
19. EQUATION OF BISECTORS OF ANGLES BETWEEN TWO LINES :
If equation of two intersecting lines are a1x+b1y+c1=0 and a2x+b2y+c2=0, then equation of bisectors of the angles between these lines are written as:
a1x+b1y+c1√a21+b21=±a2x+b2y+c2√a22+b22 ..... (1)
(a) Equation of bisector of angle containing origin :
If the equation of the lines are written with constant terms c1 and c2 positive, then the equation of the bisectors of the angle containing the origin is obtained by taking positive sign in (1)
(b) Equation of bisector of acute/obtuse angles:
See whether the constant terms c1 and c2 in the two equation are +ve or not. If not then multiply both sides of given equation by −1 to make the constant terms positive.
Determine the sign of a1a2+b1 b2
If sign of a1a2+b1 b2 For obtuse angle bisector For acute angle bisector + use + sign in eq. (1) use-sign in eq. (1) − use-sign in eq. (1) use + sign in eq. (1)
a1x+b1y+c1√a21+b21=a2x+b2y+c2√a22+b22
20. FAMILY OF LINES :
If equation of two lines be P≡a1x+b1y+c1=0 and Q≡a2x+b2y+c2=0, then the equation of the lines passing through the point of intersection of these lines is: P+λQ=0 or a1x+b1y+ c1+λ(a2x+b2y+c2)=0. The value of λ is obtained with the help of the additional informations given in the problem.
21. GENERAL EQUATION AND HOMOGENEOUS EQUATION OF SECOND DEGREE :
(a) A general equation of second degree ax2+2hxy+by2+2gx+2fy+c=0 represent a pair of straight lines if Δ=abc+2fgh−af2−bg2−ch2=0 or |ahghbfgfc|=0
Obviously these lines are
(i) Parallel, if Δ=0, h2=ab or if h2=ab and bg2=af2
(ii) Perpendicular, if a+b=0 i.e. coeff. of x2+ coeff. of y2=0.
(c) Homogeneous equation of 2nd degree ax2+2hxy+by2=0 always represent a pair of straight lines whose equations are y=(−h±√h2−abb)x≡y=m1x & y=m2x and m1+m2=−2 h b;m1 m2=ab
The condition that these lines are:
(i) At right angles to each other is a+b=0. i.e. co-efficient of x2+ co-efficient of y2=0.
(ii) Coincident is h2=ab.
(iii) Equally inclined to the axis of x is h=0. i.e. coeff. of xy=0.
(d) The combined equation of angle bisectors between the lines represented by homogeneous equation of 2nd degree is given by x2−y2a−b=xyh,a≠b,h≠0.
(e) Pair of straight lines perpendicular to the lines ax2+2hxy+by2= 0 and through origin are given by bx2−2hxy+ay2=0.
(f) If lines ax2+2hxy+by2+2gx+2fy+c=0 are parallel then distance between them is =2√g2−aca(a+b)
22. EQUATIONS OF LINES JOINING THE POINTS OF INTERSECTION OF A LINE AND A CURVE TO THE ORIGIN :
ax2+2hxy+by2+2gx+2fy+c=0…… (i)
and straight line be ℓx+my+n=0....(ii)
Now joint equation of line OP and OQ joining the origin and points of intersection P and Q can be obtained by making the equation (i) homogenous with the help of equation of the line. Thus required equation is given by ax2+2hxy+by2+2(gx+fy)(ℓx+my−n)+c(lx+my−n)2=0
23. STANDARD RESULTS :
(a) Area of rhombus formed by lines a|x|+b|y|+c=0 or ±ax±by+c=0 is 2c2|ab|.
(b) Area of triangle formed by line ax+by+c=0 and axes is c22|ab|.
(c) Co-ordinate of foot of perpendicular (h, k) from (x1,y1) to the line ax+by+c=0 is given by h−x1a=k−y1b=−(ax1+by1+c)a2+b2
(d) Image of point (x1,y1) w.r. to the line ax+by+c=0 is given by h−x1a=k−y1 b=−2(ax1+by1+c)a2+b2
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