straight line nie

8/8/2019 Straight Line NIE

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Straight Line-NIE Seminar November 2010.

Straight Lines

Straight Lines is a fundamental and an important topic under Co-ordinate Geometry.

Several methods have been developed by mathematicians to uniquely locate the position of thepoints in space. The easiest and most widely used one is the Cartesian coordinate system, whichis based on mutually perpendicular axes. In this chapter you will learn about this system: locatingthe points in this system and finding equations of line passing through these points.

If we say a line passes through two given points, we are imposing two conditions on this line.The conditions can be imposed in several other manners also e.g. we may say that a line passesthrough one point and is perpendicular to another line. These two conditions are sufficient touniquely known our line. In this chapter you will also learn various manners of imposingconditions and finding the equation of line under those conditions.

Having known the line uniquely in space we shall try to do lots timings with it e.g. divide it in agiven ratio, find distances of other points or lines from it, find the angles, which this line makewith some straight line.

Topics Covered:

y Distance between two pointsy Centroid Incentre and Circum Centrey Area of a triangley

Standard equations of the Straight Liney Different Forms of line

y Angle between two straight lines

y Examples on Angle Between two straight linesy Length of the Perpendicular from a Point on a Liney Distance between two parallel lines

y Family of linesy Concurrency of Straight Linesy Position of two points with respect to a given liney Angle Bisectors

y Combined equations of the angle bisectors of the lines


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Distance between two points

The distance between two points P(x1, y1) and Q(x2, y2) is (see the figuregiven below).

Length PQ = (x2 x1)2 + (y2 y1)

2

Proof:

Let P(x1, y1) and Q(x2, y2) be the two points and let the distance betweenthem be d. Draw PA, QR parallel to y-axis and PR parallel to x-axis.

Angle QRP = 90o

d2 = PR2 + RQ2

d2 = (x2 x1)2 + (y2 y1)

2

d = (x2 x1)2 + (y2 y1)

2.

Section Formula

Let us say we want to know the co-ordinates of point which divides a linesegment between two points A(x1, y1) and B(x2, y2) in the ratio m : n.

The coordinates of such a point are given by (nx1 + mx2/m+n, ny1 +my2/m+n) (for internal division)


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Note:

This is called section formula.

Let P divide the line segment AB in the ratio m : n. If P is inside AB then itis called internal division; if it is outside AB then it is called externaldivision.

However in each case AP/BP [or AP'/BP' or AP"/BP"] = m/n.

Note:

m:n can be written as m/n or :1. So any point on line joining A and B willbe P(x

2+x

1/+1.y

2+y

1 /+1). It is useful to assume :1 because it

involves only one variable.

illustration:

Find the ratio in which line segment A(2, 1) and B(5, 2) is divided by x-axis.

Solution:

Let x-axis intersect line at point P(xp, 0) such that AP/BP = /1 yP = 0 = y2+1.ya/+1 = 2+(1)/+1 = 1/2 AP/BO = 1/2

Illustration:

Prove that altitudes of a triangle are concurrent and prove that the co-ordinates of the point of con-currency are

(x1 tan A + x2 tan B + x3 tan C/tan A + tan B + tan C, y1 tan A + y2 tan B +

y3 tan C/tan A + tan B + tan C),

Solution:

In triangle A(x1, y1), B(x2, y2) and C(x3, y3), draw AD perpendicular to BC.


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Our effort now should be to find the co-ordinates of the point D.

To do that, we need to find BC/CD. (figure is given above)

tan B = AD/BD and tan C = AD/CD

BD/DC = tan C/tan B

Now we apply section formulae.

xD = x2 tan B + x3 tan C/tan B + tan C (i)

yD = y2 tan B + y3 tan C/tan B + tan C (ii)

We know that orthocenter will lie on AD. We need to find this point and itsco-ordinates.

We should select a point H1 on AD and take the ratio AH1/H1D in such amanner so that xH1 and yH1 calculated form (i) should be symmetric in x1,x2, x3, tan A, tan B and tan C. Think before you proceed

Let AH1/H1D = tan B + tan C/tan A

xH1 = x1 tan A + x2 tan B + x3 tan C/tan A + tan B +

tan C

and yH1 = y1 tan A + y2 tan B + y3 tan C/tan A + tan B + tan C

Since the result is symmetric, this point H1 will lie on other altitude as welli.e. the altitudes are concurrent


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xH = xH1 and yH = yH1

Illustration:

Prove analytically that in a right angled triangle the midpoint of thehypotenuse is equidistant from the three angular points.

Solution:

While proving a problem analytically take most convenient co-ordinates ofknown points.

In the present case triangle is assumed as AOB with coordinates as shownin figure given below, C is midpoint of AB.

So co-ordinates of C will be (a/2, b/2)

Now AB = a2 + b2CA = CB = AB/2 (C is mid point of AB)

= a2 + b2

and, we know that the distance between two points C and O is given by

CO = (a/2 0)2 + (b/2 0)2 = a2 + b2/2

Hence CA = CB = CO

Coordinates of the point P dividing the join of two points A(x1, y1) and


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B(x2, y2) internally in the given ratio 1 : 2 i.e., AP/BP = 1/2 areP(2x1+1x2/2+1, 2y1+1y2/2+1).

Coordinates of the point P dividing the join of two points A(x1, y1) and B(x2,y2) externally in the ratio 1 : 2 i.e., Ao/BP = 1/2 are P(2x1+1x2/2

1, 2y1+1y2/21).

Centroidof Triangle

The centroid of a triangle is the point of concurrency of the medians. The centroid G of thetriangle ABC, divides the median AD, in the ratio of 2 : 1.

Illustration:

Find the centroid of the triangle the coordinates of whose vertices are given by A(x1, y1), B(x2,y2) and C(x3, y3) respectively.

Solution:

AG/AD = 2/1

Since D is the midpoint of BC, coordinates of D are (x2+x3/2, y2+y3/2)

Using the section formula, the coordinates of G are

(2(x2+x3/2)+1.x1/2+1, 2(y2+y3/2)+1.y1/2+1)

Coordinates of G are (x1+x2+x3/3, y1+y2+y3/3).

Incentre of TriangleThe incentre I of a triangle is the point of concurrency of the bisectors of the angles of thetriangle.


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Illustration:

Find the incentre of the triangle the coordinates of whose vertices are given by A(x1, y1), B(x2,y2), C(x3, y3).

Solution:

By geometry, we know that BD/DC = AB/AC (since AD bisects A).

If the lengths of the sides AB, BC and AC are c, a and b respectively, then BD/DC = AB/AC =c/b.

Coordinates of D are (bx2+cx3/b+c, by2+cy3/b+c)

IB bisects B. Hence ID/IA = BD/BA = (ac/b+c)/c = a/c+b.

Let the coordinates of I be (x, y).

Then x = ax1+bx2+cx3/a+b+c, y = ay1+by2+cy3/a+b+c.

Circum Centre of Triangle

This the point of concurrency of the perpendicular bisectors of the sides of the triangle. This isalso the centre of the circle, passing through the vertices of the given triangle.

Orthocentre of Triangle

This is the point of concurrency of the altitudes of the triangle.

Excentre

Excentre of a triangle is the point of concurrency of bisectors of two exterior and third interiorangle. Hence there are three excentres I1, I2 and I3 opposite to three vertices of a triangle.

If A(x1, y1), B(x2, y2) and C(x3, y3) are the vertices of a triangle ABC,


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coordinates of centre of ex-circle opposite to vertex A are given as

I1(x, y) = (ax1+bx2+cx3/a+b+c/a+b+c, ay1+by2+cy3/a+b+c).

Similarly co-ordinates of centre of I2(x, y) and I3(x, y) are

I2(x, y) = (ax1bx2+cx3/ab+c, ay1by2+cy3/ab+c)

I3(x, y) = (ax1+bx2cx3/a+bc, ay1+by2cy3/a+bc)

Area of a triangle

Let (x1, y1), (x2, y2) and (x3, y3) respectively be the coordinates of the vertices A, B, C of atriangle ABC. Then the area of triangle ABC, is

|1/2[x1(y2 y3) + x2(y3 + y1) + x3(y1 y2)]| = .


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It follows that the three points (x1, y1), (x2, y2) and (x3, y3) will be collinear if

= 0.

Area of a polygon of n sides

First of all we plot the points and see their actual order. Let A1(x1, y1),A(x2, y2), , An(xn, yn) be the vertices of the polygon in anticlockwise order.

Then area of the polygon = .

Illustration:

Calculate area of a triangle shown in figures given below.


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Solution:

Using the just derived formula

Area of a triangle

1/2 [3(2 4) + (1) (4 6) + 5 (6 2)]

= 1/2 [ 6 2 + 20] = 6

Similarly, area of the triangle shown in the figure given above.

Area of a ?ABC is

= 1/2 [3(4 2) + (5) (2 6) + (1) (6 4)]

=1/2 [6 20 + 2] = 6

Caution:

Thus we observe that the area of a triangle is positive vertices are taken in the anticlockwisedirection and negative when the vertices are taken the clockwise direction.


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Note:

Area of a triangle can also be expressed as = 1/2

= 1/2 [x1 y2 y1 x2 + x2 y3 y2 x3 + x3 y1 y3 x1]

This form is important. It can be used to find area of a quadrilateral, pentagon, hexagon andpolygons.

Important:

If three points P1, P2 and P3 are collinear then the determinant at must vanish i.e. the area oftriangle formed must be zero.

Note:

If the vertices are in clockwise order then take modulus.

Illustration:

Prove that the area of the triangle with vertices at (p 4, p + 5),(p + 3, p 2) and (p, p) remains constant as p varies.

Solution:

The area of the triangle is

which remains constant for all values of p.


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Straight Line

Any equation of first degree of the form Ax + By + C = 0, where A, B, C areconstants always represents a straight line (at least one out of A and B isnon zero).

SlopeIf is the angle at which a straight line is inclined to the positive direction of the x-axis, then m

= tan, (0 < < 180o) is the slope of the line.

Standard equations of the Straight Line

Slope Intercept From:

y = mx + c, where

m = slope of the line

c = y intercept

Intercept Form:x/a + y/b = 1

x intercept = ay intercept = b


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Slope point form

(a) One point on the straight line

(b) The gradient of the straight line i.e., the slope m of the line

Equation:

y y1 = m(x x1), where (x1, y1) is a point on the straight line.

Illustration:

Pause:Equation of line in figure (ii) is x = 3, because x-co-ordinate of each point on the line is 3.Equation of line in figure (iv) is y = 2, because y-co-ordinate of each point on the line is 2.

Although every line satisfied the above given basic definition, a line can be represented in manyforms, some of which are given hereunder.

Two points form:Let there be two points A(x1, y1) and B(x2, y2) in a co-ordinate plane. If any point P(x, y) lies onthe line joining A and b then m = tan = yy1/xx2 = y2 y1/x2 x1, (see figure given below).


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y y1 = y2 y1/x2 x1 (x x1) which is the equation of the given line.Equation of line can also be written asy y2 = y2 y1/x2 x1 (x x2) or

= 0.

Examples based on straight line

Illustration:

Reduce 3x 4y + 5 = 0 to all other forms.

Solution:

(a) Slope intercept form

y = 3/4 x + 5/4where, m = slope = 3/4c = 5/4 (y intercept)

(b) Intercept form3x 4y = 5 3x/5 + 4y/5 = 1x/5/3 + y/5/4 = 1 (intercept form)

x intercept = 5/3y intercept = 5/4

(c) point slope form

Let x = 1, then y = 3/4 + 5/4 = 2y 2 = 3/4 (x 1)


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(d) parametric form

x1/cos = r, where tan = 3/4

x1/4/5 = y2/3/5 = r

(e) Normal form

3x 4y = 53x + 4y = 5Dividing by ((3)2 + 4) 3/5 x + 4/5 y = 5/5 = 1 x cos + y sin = p, p > 0Where cos = 3/5, sin = 4/5, p = 1

Illustration:The straight line drawn through the point P(0, 3) and making an angle of 30owith positive x-axis, meets the line x + y = 6 at Q. Find the length PQ.Solution:Method 1.Equation of the line through the point P is

x0/cos 30 = y 3/sin 30 = r xQ = 3/r r, yQ = 3 + r/2, r = distance PQ

Point Q lies on x + y = 6

r3/2 + (3 + r/2) = 6 r = 6/3+1

Method 2.Equation of the line through the point P is y = 1/3 x + 3[because here m = tan 30o = 1/3, c = 3]


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Solving this line, with x + y = 6, we get

xQ = 33/3+1, yQ = 33+6/3+1distance PQ = (xQ xP)

2 + (yQ yP)2

= 6/3+1Illustration:Find the equation of the line whose perpendicular distance from the origin is4 units and the angle which the normal makes with positive direction of x-axis is 15o.

Solution:Here, we are given p = 4 and = 15o.Now cos 15o = 3+1/22and sin 15o = 31/22By the normal form, the equation of the line is

x cos 15o + y sin 15o = 4 or 3+1/22 x + 31/22 y = 4or (3 + 1) x + (3 1)y = 82.

This is the required equation.

Illustration:Given a line 2x 3y + 5 = 0. Write various forms of the line.Solution:

Slope intercept form:y = 2x/3 + 5/3, C = 5/3

and m = (coeffcient of x/coefficient of y) = 2/3Intercept Form:

x/(5/3) + y/(5/3) = 1, a = 5/2; b = + 5/3.Normal Form:

sin = 3/13, cos = 213 2/13 p = 5/13.

Illustration:

Find the equation to the straight line which passes through the point (5, 4)and is such that the position of it between the axes is divided by the givepoint in the ratio 1 : 2.

Solution:Let the required straight line be (x/a) + (y/b) = 1.


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Using the given conditions, P (2a+1.0/2+1, 2.0+1.b/2+1) is the point whichdivides (a, 0) and (0, b) internally in the ratio 1 : 2.

But P is (5/4)Hence 5 = 2a/3, 4 = b/3 a = 15/2, b = 12.

Hence the required equation is x/(15/2) + y/2 < < . tan = 8 = slope of line.We know that the equation of the straight line passing through the point (x1,y1) having slope m is y y1 = m(x x1).Therefore the equation of the required line is y 2 = 8 (x 1) 8 x + y 8 2 = 0.Illustration:Find the equation of the line joining the points (1, 3) and (4, 2).Solution:Equation of the line passing through the points (x1, y1) and (x2, y2) is

y y1 = y1y2/x1x2 (x x1)Hence equation of the required line will be

y 3 = 3+2/14 (x + 1) x + y 2 = 0.Illustration:Represent the straight-line y = x + 2 in the parametric form.Solution:Slope of the given line is = 1 = tan /4.Equation of the straight line can be written as y 2 = x.or y2/1/2 = x/1/2= r.Any point on the line is (r/2, 2 + r/2).

The point (x, y) is at a distance r from the point (0, 2).Illustration:A line joining two points A(2, 0) and B(3, 1) is rotated about A in theanticlockwise direction through an angle of 15o. Find the equation of the linein the new position. If B goes to C, what will be the coordinates of C, in thenew position?Solution:Slope of BAB(m) = 1 m = tan = 1 = 45o.

= tan (60o) (because angle between AB and AC = 15o).Also AB = AC = 2 and A is (2, 0).


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Hence equation of the line AC isx2/cos60o = y0/sin 60o

or x2/1/2 = y/3/2 = r = 2 C is (2 + 2.1/2, 0 + 2.3/2) i.e. C is (2 + 1/2, 3/2).

Angle between two straight lines

Angle bisector of two lines (the line which bisects the angle between the twolines) is the locus of a point which is equidistant (having equal perpendiculardistance) from the two lines.

Say, we have two lines

L1 : A1x + B1y + C1 = 0L2 : A2x + B2y + C2 = 0

If point R(p, q) lies on the bisector, then length of perpendicular from P tobe both lines should be equal.


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i.e.

generalizing for any point (x, y) the equation of the angle bisector isobtained as:

A1x+B1y+C1/A12 + B1

2 = + A2x+B2y+C2/A22 + B2

2

Note:

1. This equation gives two bisectors: one-acute angle bisector and the otherobtuse bisector.

2. Rule to decide the particular bisector

To determine a bisector which lies in the same relative position with respectto the lines as a given point S(x3, y3) does, make the signs of theexpressions A2x3 + B1y3 + C1 and A2x3 + B2y3 + C2 identical. (say positive)then A1x+B1y+C1/A1

2 + B12 = + A2x+B2y+C2/A2

2 + B22 gives the bisector

towards this point. If the signs are different multiply one of the equationswith 1 throughout, so that positive sign is obtained. Then above equationwith changed equations of lines will given the required bisector.

3. If (x3, y3) (0, 0) and A2A1 + B2B1 > 0 then the bisector towards the

origin is the obtuse angle bisector.

4. Alternative method: to determine whether the bisector is of the obtuse oracute angle, determines both the bisectors and calculate angle between oneof them and the initial line. The bisector for which |tan | > 1 is the obtuseangle bisector.

Length of the Perpendicular from a Point on a Line

The distance of a point from a line is the length of the perpendicular drawn from the point on theline. Given the equation of the line are different forms, the length of the perpendicular can beobtained in different forms.

First form:

The normal equation helps us in finding the distance of a point from a straight line. Suppose wehave to find the distance of the point P(x1, y1) from the line l1 whose equation is x cos + y sin


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= p. Let l2 be the line through P parallel to the line l1. Let d be the distance of P from l1. Then, thenormal from O to l2 is of length p + d. Hence the equation of l2 is

x cos + y sin = p + d.

Since P(x1, y1) lies on it. x1 cos + y1 sin = p + d

d = x1 cos + y1 sin p.

Note:

1. Rule to find the perpendicular distance of a given point from a given line in normal form.

In the left side of the equation (right side being zero), substitute the coordinates of the point.

The result gives the perpendicular distance.

2. Complete distance formula. If the point P and the origin O, instead of lying on the oppositesides of l as in figure given above, lie on the same side of line l1 it may be proved by proceedingexactly in the same manner that


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d = (x1 cos + y1 sin p)

Hence, the complete distance formula is

d = + (x1 cos + y1 sin p)

The distance of a point from a line is the length of the perpendicular drawn from the point to theline. Let L : Ax + By + C = 0 be a line, whose distance from the point P(x 1, y1) is d. Draw aperpendicular PM from the point P to the line L. If the line meets the x and y-axes at the points Qand R respectively, then coordinates of the points are Q(C/A, 0) and R(0, C/B). Thus the areaof the triangle PQR is given by

area(?PQR) = 1/2 PM QR, which gives PM = 2 are (PQR)/QR (1)also, area (?PQR) = 1/2 |x1 (0 + C/B) + (C/A) (C/B y1) + 0(y1 0)|

= 1/2 |x1 C/B + y1 C/A + C2/AB|

Or 2 area (?PQR) = |C/AB| |Ax1 + By1 + C| and

QR = (0 + C/A)2 + (C/B 0)2 = |C/AB| A2 + B2

Substituting the values of area (?PQR) and QR in (1), we get

PM = |Ax1+By1+C|/A2+B2.

Or d = |Ax1+By1+C|/A2+B2.

Thus the perpendicular distance (d) of a line Ax + By + C = 0 from a point

(x1, y1) is given by d = |Ax1+By1+C|/A2+B2.


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Family of lines

The general equation of the family of lines through the point of intersection of two given lines isL + L = 0, where L = 0 and L = 0 are the two given lines, and is a parameter.

Illustration:

A variable line through the point of intersection of the linesx/a + y/b = 1 and x/b + y/a = 1 meets the coordinate axes in A and B. Show that the locus of themidpoint of AB is the curve 2xy(a + b) = ab(x + y).

Solution:

Let (h, k) be the midpoint of the variable line AB.The equation of the variable line AB is

(bx + ay ab) + (ax + by ab) = 0

Coordinates of A are (ab(1+)b+a).

Coordinate of B are (0, ab(1+)b+a).

Mid point of AB is (ab(1+)2(b+a), ab(1+)2(a+b))


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h = ab(1+)2(b+a); k = ab(1+)2(a+b) 1/2h = b+a/ab(1+); 1/2k =a+b/ab(1+)

1/2h + 1/2k = a+b/ab (h + k)ab = 2hk (a + b).

Hence the locus of the midpoint of AB is (x + y) ab = 2xy (a + b).

To find the equation to the straight lines which pass through a given point (x1, y1) and makeequal angles with the given straight line y = m1x + c.

If m is the slope of the required line and is the angle which this line makes with the given line,then tana = + m1m/1+m1m.

(i) The above expression for tan, gives two values of m, say mA and mB.

(ii) The required equations of the lines through the point (x1, y1) and making equal angles with the given line are y y1 = mA (x x1), y y1 = mB(x x1).

Illustration:

Find the equations to the sides of an isosceles right-angled triangle, the equation of whosehypotenuse is 3x + 4y = 4 and the opposite vertex is the point (2, 2).

Solution:

The problem can be restated as:

Find the equations to the straight lines passing through the given point (2, 2) and makingequal angles of 45o with the given straight line 3x + 4y 4 = 0


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Slope of the line 3x + 4y 4 = 0 is m1 = 3/4

tan 45o = + m1m/1+m1m, i.e., 1 = + m+3/4/13/4m

So that mA = 1/7, and mB = 7.

Hence the required equations of the two lines are

y 2 = mA (x 2) and y 2 = mB(x 2).

7y x 12 = 0 and 7x + y = 16.

Illustration:

The straight lines 3x + 4y = 5 and 4x 3y = 15 intersect at the point A. On these lines points Band C are chosen so that AB = AC. Find the possible equations of the line BC passing through(1, 2).

Solution:

The two given straight lines are at right angles.Since AB = AC, the triangle is an isosceles right angled triangle.

The required equation is of the form y 2 = m(x 1) (1)

with tan 45o = + m+3/4/13m/4 = + m4/3/1+4m/3


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1 = + m+3/4/13m/4 = + m4/3/1+4m/3 m = 7, 1/7.

Substitute the value of m in (1). We get the required equations.

Illustration:

Find the equation of the straight line passing through (2, 7) and having intercept of length 3units between the straight lines4x + 3y = 12 and 4x + 3y = 3.

Solution:

Distance AC between the two given parallel lines

= |c1c2/a

2

+b

2

| = 123/16+9 = 9/5.

Let AB be the intercept of length 3 units.

BC = 12/5. If is the angle between BC and AB, then tan = 9/12 = 3/4.

Slope of the parallel lines = 4/3 = m2.

If m1 is the slope of the required line, then tan = m1m/1+m1m 3/4 = + m14/3/1+4/3m1

i.e. m1 + 4/3 = 3/4 (1 4/3 m1) and m1 + 4/3 = 3/4 (1 4/3 m1).

The slopes are

(i) m1 = 7/24

(ii) m1 =


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(the line is parallel to the y-axis).

The required equations of the lines are 7x + 24y + 182 = 0 andx + 2 = 0.

Alternative solution:

Equation of the line, through P(2, 7) and making angle with the x-axis, is x+2/cos = y+7/sin = r.

If this line intersects the given lines at A and B, with AB = 3, the points A and B are A(2 + r1cos , 7 + r1 sin ) and B (2 + (r1 + 3) cos , 7 + (r1 + 3) sin ).

Since A and B lie on the lines 4x + 3y = 3 and 4x + 3y = 12, we have

4r1 cos + 3r1 sin = 32 and4r1 cos + 3r1 sin + 12 cos + 9 sin = 41, so that

12 cos + 9 sin = 0 or 4 cos + 3 sin = 3.

Solving this equation we find that = /2 and tan = 7/24.

Hence the required lines are x + 2 = 0

and y + 7 = 7/24 (x + 2) i.e. 7x + 24y + 182 = 0.

Concurrency of Straight Lines

The condition for 3 lines a1x + b1y + c1 = 0, a2x + b2y + c2 = 0,a3x + b3y + c3 = 0 to be concurrent is

(i) = 0.

(ii) There exist 3 constants l, m, n (not all zero at the same time) such that IL1 + mL2 + nL3 = 0,where L1 = 0, L2 = 0 and L3 = 0 are the three given straight lines.

(iii) The three lines are concurrent if any one of the lines passes through the point ofintersection of the other two lines.


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Illustration:

Check if lines

a1 x + b1 y + c1 = 0 (1)a2 x + b2 y + c2 = 0 (2)

(2a1 3a2)x + (2b1 3b2)y + (2c1 3c2) = 0 (3)

are concurrent?Solution:

We can try to find , and by observation as follow:

L3 2L1 + 3L2 = 0

Enquiry: many lines can pass through the intersection of two lines. Can we find a general

equation of these lines?

If L1 = 0 and L2 = 0 are two lines then equation of family of lines passing through theirintersection is given by

L1 + L2 = 0 (A)

Where is any parameter. (Equation A is satisfied by the point of intersection of L1 = 0 and L2= 0)

Note:

To determine a particular line one more condition is required so as to determine or eliminate .

Illustration:

If x (2q + p) + y(3q + p) = 0

(x + y 1) + q/p (2x + 3y 1) = 0, p 0Solution:This equation represents the family of lines passing through the intersection of lines x + y 1 = 0and 2x + 3y 1 = which is fixed point i.e. (2, 1).

If p = 0 then equation becomes

q(2x + 3y 1) = 0

this also represents a line which passes through fixed points (2, 1).


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Hence the given equation represents family of lines passing through a fixed point (2, 1) forvariable p, q.

Illustration:

Find the equation of a line, through the intersection of 2x + 3y 7 = 0 and x + 3y 5 = 0 andhaving distance from origin as large as possible.

Point of intersection of two lines is A(2, 1)

Now, with OA as radius and O itself as centre draw a circle.

There will be infinitely many lines through A and each except one of them produces a chordof circle and hence their distance from origin i.e. centre of circle is less than OA i.e. radius ofcircle.

But the exceptional one which infact is a tangent to circle at A will be at a distance OAfrom O.

Thus, tangent to circle at A will be the line through A and is farthest from origin.

Now, OA tangent at A.

(slope of OA) (slope of tangent at A) = 1

Or, 10/20 (slope of tangent at A) = 2

equation of required line is


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(y 1) = 2(x 2)

Or 2x + y 5 = 0

Illustration:Find the point of concurrency of the altitudes drawn from the vertices (at1t2, a(t1 + r2)), (at2t3, a2t2+ t3)) and (at3t1, a(t3 + t1)) respectively of a triangle ABC.

Solution:

Slope of AD = t3.

Equation of AD is y a(t1 + t2) = t3(x + at1t2). (1)

Equation of CF is y a(t3 + t1) = t2(x at3t1). (2)

Subtracting (1) from (2), we get

x = a y = a(t1 + t2 + t1t2t3).

Hence the point of concurrency of the altitudes is

(a, a(t1 + t2 + t3 + t1t2t3)).

Position of two points with respect to a given line

Let the line be ax + by + c = 0 and P(x1, y1), Q(x2, y2) be two points.


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3 + 5 1 =7 > 0 sin + cos 1 > 0

sin(/4 + ) > 1/2

/4 < /4 + < 3/4 0 < < /2.

Illustration:

Find a, if (, 2) lies inside the triangle having sides along the lines

2x + 3y = 1, x + 2y 3 = 0, 6y = 5x 1.

Solution:

Let A, B, C be vertices of the triangle.

A (7, 5), B (5/4, 7/8),

C (1/3, 1/9).

Sign of A w.r.t. BC is ve.

If p lies in-side the ABC, then sign of P will be the same as sign of a w.r.t. the line BC

5 62 1 < 0. (1)


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Similarly 2 + 32 1 > 0. (2)

And, + 22 3 < 0. (3)

Solving, (1), (2) and (3) for and then taking intersection,

We get ? (1/2, 1) (3/2, 1).

Illustration:

The equations of the perpendicular bisectors of the sides AB and AC of a triangle ABC arerespectively x y + 5 = 0 and x + 2y = 0. If the co-ordinates of A are (1, 2), find the equation ofBC.

Solution:From the figure,

E (x1+1/2, y12/2),

F (x2+1/2, y22/2).

Alt text : equations of the perpendicular bisectors of sides of triangle

Since E and F lie on OE and OF respectively,

x1 y1 + 13 = 0 (1)

and x2 + 2y2 3 = 0 (2)

Also, slope of AB = 1 and slope of AC is 2, so that

x1 + y1 + 1 = 0. (3)

And 2x2 y2 4 = 0 (4)


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Solving these equations, we get the co-ordinates of B and C as

B (7, 6) and C (11/5, 2/5)

Equation of BC is 14x + 23y 40 = 0.

Illustration:

Two fixed points A and B are taken on the co-ordinate axes such that OA = a and OB = b. Twovariable points A and B are taken on the same axes such that OA + OB = OA + OB. Find thelocus of the point of intersection of AB and AB.

Solution:

Let A (a, 0), B (0, b), A (a, 0), B (0, b).Equation of AB is x/a' + y/b' = 1. . (1)

and equation of AB is x/a + y/b' = 1. . (2)

Subtracting (1) from (2), we get, x (1/a 1/a') + y(1/b' 1/b) = 0.

x(a'a)/aa' + y(bb')/bb' = 0. [Using a a = b b]

x/a(bb'+a) + y/bb', 0 b = a(a+b)y/aybx. .. (3)

From (2) bx + ay = (4) we get x + y = a + b

which is the required locus.Angle Bisectors

To find the equations of the bisectors of the angle between the lines

a1x + b1y+ c1 = 0 and a2x + b2y + c2 = 0.

A bisector is the locus of a point, which moves such that the perpendiculars drawn from it to thetwo given lines, are equal.

The equations of the bisectors are

a1x+b1y+c1/a12+b1

2 = + a2x+b2y+c2/a22+b2

2.


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AP is the bisector of an acute angle if,

Tan (PAN) = tan (/2) is such that |tan /2| < 1.

AP is an obtuse angle bisector if,

Tan (PAN) = tan (/2) is such that |tan /2| > 1.

Notes :

When both c1 and c2 are of the same sign, evaluate a1a2 + b1b2. If negative, then acute anglebisector is a1x+b1y+c1/a1

2+b12 = + a2x+b2y+c2/a2

2+b22.

When both c1 and c2 are of the same sign, the equation of the bisector of the angle whichcontains the origin is a1x+b1y+c1/a1

2+b12 = + a2x+b2y+c2/a2

2+b22.

Bisectors of the angle containing the point (, ) is a1x+b1y+c1/a12+b1

2 = +a2x+b2y+c2/a2

2+b22 if a1 + b1 + c1 and a2 + b2 + c2 have the same sign.

Bisectors of the angle containing the point (, ) is a1x+b1y+c1/a12+b1

2 = +a2x+b2y+c2/a2

2+b22 if a1 + b1 + c1 and a2 + b2 + c2 have the opposite sign.

Illustration:

For the straight lines 4x + 3y 6 = 0 and 5x + 12y + 9 = 0 , find the equation of the

(i) bisector of the obtuse angle between them,


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(ii) bisector of the acute angle between them,

(iii) bisector of the angle which contains (1, 2)

Solution:

Equations of bisectors of the angles between the given lines are

4x+3y6/42+32 = + 5x+12y+9/52+122

9x 7y 41 = 0 and 7x + 9y 3 = 0.

If is the angle between the line 4x + 3y 6 = 0 and the bisector 9x 7y 41 = 0, then tan => 1.

Hence

(i) The bisector of the obtuse angle is 9x 7y 41 = 0.

(ii) The bisector of the acute angle is 7x + 9y 3 = 0.

For the point (1, 2)

4x + 3y 6 = 4 1 + 3 2 6 > 0.

5x + 12y + 9 = 12 12 + 9 > 0.

Hence equation of the bisector of the angle containing the point (1, 2) is 4x+3y6/5 =5x+12y+9/13 9x 7y 41 = 0.

straight line nie

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