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Straight Line - Notes, Concept and All Important Formula

STRAIGHT LINE

1. RELATION BETWEEN CARTESIAN CO-ORDINATE & POLAR CO-ORDINATE SYSTEM

If (x,y) are Cartesian co-ordinates of a point P, then : x=rcosθ, y=rsinθ
and r=x2+y2,θ=tan1(yx)



2. DISTANCE FORMULA AND ITS APPLICATIONS :

If A(x1,y1) and B(x2,y2) are two points, then AB=(x2x1)2+(y2y1)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:

(a) Square if AB=BC=CD=DA & AC=BD;ACBD
(b) Rhombus if AB=BC=CD=DA and ACBD;ACBD
(c) Parallelogram if AB=DC,BC=AD;ACBD;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:

(a) For internal division : (mx2+nx1m+n,my2+ny1m+n)
(b) For external division : (mx2nx1mn,my2ny1mn)
(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 :

CO-ORDINATES OF SOME PARTICULAR POINTS - 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 SOME PARTICULAR POINTS - Incenter
The incenter is the point of intersection of internal bisectors of the angles of a triangle. Also it is a centre of a circle touching all the sides of a triangle.
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.

Note:
(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 :

CO-ORDINATES OF SOME PARTICULAR POINTS - Circumcenter
It is the point of intersection of perpendicular bisectors of the sides of a triangle. If O is the circumcenter of anytriangle ABC, then OA2=OB2=OC2
Also it is a centre of a circle touching all the vertices of a triangle.

Note:
(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:

CO-ORDINATES OF SOME PARTICULAR POINTS - 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.

Remarks:
(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:

CO-ORDINATES OF SOME PARTICULAR POINTS - 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+cx3a+b+c,ay1+by2+cy3a+b+c)

Similarly ex-centers of ABC with respect to vertices B andC are denoted by I2 and I3 respectively, and
I2=(ax1bx2+cx3ab+c,ay1by2+cy3ab+c)
I3=(ax1+bx2cx3a+bc,ay1+by2cy3a+bc)



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(y2y3)+x2(y3y1)+x3(y1y2)]|

To remember the above formula, take the help of the following method:

AREA OF TRIANGLE using trick methods

Remarks:

(i) If the area of triangle joining three points is zero, then the points are collinear.

(ii) Area of Equilateral triangle

If altitude of any equilateral triangle is P, then its area =P23.
If 'a' be the side of equilateral triangle, then its area =(a234)

(iii) Area of quadrilateral whose consecutive vertices are (x1,y1),(x2,y2), (x3,y3)&(x4,y4) is 12|x1x3y1y3x2x4y2y4|




6. CONDITION OF COLLINEARITY FOR THREE POINTS :

Three points (x1,y1),(x2,y2) and (x3,y3) are collinear if any one of the given point lies on the line passing through the remaining two points. Thus the required condition is
y3y1y2y1=x3x1x2x1 or x1x2x1x3=y1y2y1y3 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.

(a) Equation of a line parallel to x -axis at a distance a is y=a or y=a
(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 :

SLOPE OF LINE

If a given line makes an angle θ (0θ<180,θ90) 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) & x1x2 then slope of line AB=y2y1x2x1




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 : yy1=m(xx1)

(c) Two point form : Equation of a line passing through two points (x1,y1) and (x2,y2) is written as 

yy1=y2y1x2x1(xx1) 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 0α<2π.

STANDARD FORMS OF EQUATIONS OF A STRAIGHT LINE - Parametric form
(f) Parametric form : To find the equation of a straight line which passes through a given point A( h,k) and makes a given angle θ with the positive direction of the x-axis. P(x,y) is any point on the line LAL'. Let AP=r then xh=rcosθ,yk=r sinθ & xhcosθ=yksinθ=r is the equation of the straight line LAL'. 
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.

(i) Slope of this line =ab= coeff. of x coeff. of y
(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=±(m1m21+m1 m2)

(b) If equation of lines are a1x+b1y+c1=0 and a2x+b2y+c2=0, then these line are

(i) Parallel a1a2=b1 b2c1c2
(ii) Perpendicular a1a2+b1 b2=0
(iii) Coincident a1a2=b1 b2=c1c2
(iv) Intersecting a1a2b1 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+ca2+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:

(a) The distance between two parallel lines ax+by+c1=0 and ax+by+c2=0 is =|c1c2|a2+b2
(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 |(c1c2)(d1d2)m1m2|.




13. EQUATION OF LINES PARALLEL AND PERPENDICULAR TO A GIVEN LINE :

(a) Equation of line parallel to line ax+by+c=0 ax+by+λ=0
(b) Equation of line perpendicular to line ax+by+c=0 bxay+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 :

Equations of lines passing through a point (x1,y1) and making an angle α, with the line y=mx+c is written as:
yy1=m±tanα1mtanα(xx1)



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

Note:
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 :

REFLECTION OF A POINT

Let P(x,y) be any point, then its image with respect to

(a) x -axis is Q(x,y)
(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:

TRANSFORMATION OF AXES - 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=xα,y=yβ

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 

TRANSFORMATION OF AXES - 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', XOX=YOY=θ

 then x=xcosθysinθy=xsinθ+ycosθ

and x=xcosθ+ysinθ
y=xsinθ+ycosθ

The above relation between (x,y) and (x,y) can be easily obtained with the help of following table

New \ Oldxyxcosθsinθysinθcosθ




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+c1a21+b21=±a2x+b2y+c2a22+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)  

i.e. if a1a2+b1 b2>0, then the bisector corresponding to +sign gives obtuse angle bisector
a1x+b1y+c1a21+b21=a2x+b2y+c2a22+b22



20. FAMILY OF LINES :

If equation of two lines be Pa1x+b1y+c1=0 and Qa2x+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+2fghaf2bg2ch2=0 or |ahghbfgfc|=0

(b) If θ be the angle between the lines, then tanθ=±2h2aba+b
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±h2abb)xy=m1x & y=m2x and m1+m2=2 h b;m1 m2=ab

These straight lines passes through the origin and for finding the angle between these lines same formula as given for general equation is used.
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 x2y2ab=xyh,ab,h0.

(e) Pair of straight lines perpendicular to the lines ax2+2hxy+by2= 0 and through origin are given by bx22hxy+ay2=0.

(f) If lines ax2+2hxy+by2+2gx+2fy+c=0 are parallel then distance between them is =2g2aca(a+b)




22. EQUATIONS OF LINES JOINING THE POINTS OF INTERSECTION OF A LINE AND A CURVE TO THE ORIGIN :

EQUATIONS OF LINES JOINING THE POINTS OF INTERSECTION OF A LINE AND A CURVE TO THE ORIGIN
Let the equation of curve be:
ax2+2hxy+by2+2gx+2fy+c=0 (i)
and straight line be x+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)(x+myn)+c(lx+myn)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 hx1a=ky1b=(ax1+by1+c)a2+b2

(d) Image of point (x1,y1) w.r. to the line ax+by+c=0 is given by hx1a=ky1 b=2(ax1+by1+c)a2+b2



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