rankers coordinate geometry presentation

Basic Coordinate Geometry- Session for RankersBatch

Vidyalankar Institute

March 8, 2010

Basic Coordinate Geometry- Session for Rankers Batch

Motivation for Coordinate Geometry

Coordinate geometry is a mixture of Geometry and Algebra,A geometrical point is represented by a algebraic values (x , y). Thisgives us the power of both thinking procedures.For example : A line ax + by + c = 0 is an algebraic equationwhich satisfies all (x , y) which we call its solution set (in Algebra)but same can be interpreted as a point lying on a line

Transformations

Shift of origin

Invariant

Distance between two pointsSlope of a line

Rotation of axes about origin

Invariant

Distance between two pointsDistance between origin and a line

Rotation can be attained using multiplication by e iθ or

orthonormal matrix[

cos θ sin θ− sin θ cos θ

]to result rotation by

angle θ or a point (x , y)

Distance of point from a line

Distance of a point (x0, y0) from a line ax + by + c = 0 alonga direction θ is given by

∣∣∣∣ ax0 + by0 + ca cos θ + b sin θ

∣∣∣∣Shortest distance of (x0, y0) from a line ax + by + c = 0 is

AM =|ax0 + by0 + c |√

a2 + b2

Foot of the perpendicular & Image of a point

Foot of the perpendicular from (x0, y0) on the lineax + by + c = 0

x − x0

y − y0

b= −

(ax0 + by0 + c

a2 + b2

Image of the point (x0, y0) in the line ax + by + c = 0

x − x0

y − y0

b= −2

(ax0 + by0 + c

a2 + b2

Linear Combination of lines u=0,v=0

Linear combination of lines u = 0, v = 0 is

αu + βv = 0 or u + kv = 0

where α, β ∈ R1

Linear combination αu + βv = 0 always passes through theintersection of the constituent lines u = 0 and v = 0

1In linear combination, u & v can represent any geometrical structure.Basic Coordinate Geometry- Session for Rankers Batch

Two lines are parallel

Two lines a1x + b1y + c1 = 0 and a2x + b2y + c2 = 0 areparallel iff

b26= c1

Special case : If these lines are coincident then

ExampleProblem solving : Equation of a line parallel to 3x + 2y + 5 = 0 is3x + 2y + c = 0 where c is identifiable using another condition.

Two lines are non-parallel

If two lines are non-parallel then they intersect in a point, withacute and obtuse angle between them.

tan θ =

∣∣∣∣ m1 −m2

1 + m1m2

∣∣∣∣where θ is acute angle between them

If they are coincident then θ = 0 and hence m1 = m2If they are perpendicular2 then θ = 0 and hence m1m2 = −1

ExampleEquation of line perpendicular to line ax + by + c = 0 isbx − ay + c2 = 0. Just interchange the coefficients a, b and replaceone sign negative

2Slope of line perpendicular to a line of slope m1 is −1/m1

Three lines are concurrent

Three lines aix + biy + ci = 0 are concurrent iff

∣∣∣∣∣∣a1 b1 c1a2 b2 c2a3 b3 c3

∣∣∣∣∣∣ = 0

If they are not concurrent then they form a triangle and areaof this triangle is

C1C2C3

where Ci is the cofactor of ci term

Angle Bisector, Origin, Acute angle bisector

Equation of Angle bisector

a1x + b1y + c1√a21 + b2

= ±a2x + b2y + c2√a22 + b2

Equation of angle bisector containing origin (providedc1, c2 > 0)

a1x + b1y + c1√a21 + b2

=a2x + b2y + c2√

a22 + b2

If a1a2 + b1b2 > 0 ⇔ Obtuse angle contains the originIf a1a2 + b1b2 < 0 ⇔ Acute angle contains the origin

DefinitionLocus is defined as path traced by a variable point moving underconstraints. Mathematically, the path traced is the relationship between thex and y coordinate of any instance taken by the locus point (variable point).

Steps to solve problems of locus1 Let the locus point be (h, k)(we call this incoming variables)

[this gives an instance for the moving point]2 Introduce some new variables (we call them outgoing

variables) so that we proceed in the solution3 Find a relationship (geometrical) between the introduced

variable(s)4 Find relation between the incoming variables (h, k) and

outgoing variables5 Replace (h, k) by (x , y)

Two lines - Angle bisector/Origin

ExampleTwo lines y = x + 1 and y + 1 = 0 are given then

find equation of angle bisector containing the origin

find obtuse angle between them?

where does the origin lie in (acute or obtuse angle)?

find obtuse angle bisector

Three lines

ExampleGiven three lines 2x + y − 3 = 0, x + 4y − 5 = 0 and3x + 5y − 1 = 0. Find the area bounded by these three lines

ExampleFind the value of c such that 2x + y − 3 = 0, x + 4y − 5 = 0 and3x + 5y + c = 0 are concurrent using methods mentioned below

Using determinant approach

Using linear combination of the form u + kv = 0 andαu + βv = 0

Try this scenario using both kinds of linear combination forms2x + y − 3 = 0, x + 4y − 5 = 0 and ax + 14y − 21 = 0

Three lines

Point and a line

ExampleThe equations of the perpendicular bisectors of the sides AB andAC of triangle ABC are x − y + 5 = 0 and x + 2y = 0 respectively.If the point A is (1,−2) find the equation of the line .

Example

If a point P(a2, a) lies in the region corresponding to the obtuseangle between y + 2 = 0 and y = x + 1 then find the intervalwhere a belongs to?

Example

If a point P(a2, a) lies in the region corresponding to the acuteangle between the lines 2y = x and 4y = x then find the intervalwhere a belongs to?

Point and a line

Example

Point and a line

Example

Point & Line continued...

Problem :A ray of light is sent along the line 2x − 3y = 5. After refractingacross the line x + y = 1 it enters the opposite side after turning by15◦away from the line x + y = 1. FInd the equation of the linealong which the refracted ray travels.

Diagram

Problem :A ray of light is sent along the line 2x − 3y = 5. After refractingacross the line x + y = 1 it enters the opposite side after turning by15◦away from the line x + y = 1. FInd the equation of the linealong which the refracted ray travels.

Diagram

Example

Find the point (x , y) on 2x + 3y = 6 which is atShortest distance from (3, 2)

Distance of 2 from (3, 2)

Example

Find the point (x , y) on 2x + 3y = 6 which is atShortest distance from (3, 2)

Distance of 2 from (3, 2)

Problems on Locus

Example

If the coordinates of a variable point P be(

t +1t, t − 1

)where t

is a variable quantity, then find the locus of the point P

ExampleThe variable line x cos θ + y sin θ = 2 cuts the X and Y axis at Aand B respectively. Find the locus of vertex of point P of therectangle OAPB where O is the origin.

ExampleA variable line is at a constant distance p from the origin andmeets the coordinate axes in A and B. Show that locus of thecentroid of ∆OAB is x−2 + y−2 = 9p−2

Problems on Locus

Example

t +1t, t − 1

)where t

Problems on Locus

Example

t +1t, t − 1

)where t

Problems on Locus

Example

P is any pt on the line x − a = 0. If A is the point (a, 0) andPQ,the bisector of the angle OPA, meets the x-axis in Q. Provethat the locus of the foot of the perpendicular from Q on OP is

(x − a)2(x2 + y2) = a2y2

Problems on Locus

Example

P is any pt on the line x − a = 0. If A is the point (a, 0) andPQ,the bisector of the angle OPA, meets the x-axis in Q. Provethat the locus of the foot of the perpendicular from Q on OP is

(x − a)2(x2 + y2) = a2y2

Problems on Locus

Example

A quadratic equation f (x) = ax2 + bx + c = 0 where a, b, c ∈ R isgiven. For a given value of a and c we vary b such that the locus ofvertex is y = g(x) then

1 Is a line parallel to y − axis2 Is parallel to x − axis3 Is a quadratic curve4 Is a point

Maxima-Minima Problems

Example

A man starts from P(−3, 4) and reaches Q(0, 1) touching X − axisat R(α, 0) such that PR + RQ is minimum then α =?

Example

Consider the points A(0, 1) and B(2, 0), P is a point on the linex + y + 1 = 0. Find coordinates of point P such that |PA− PB| is

MaximumMinimum

Also find the max and min value attained by |PA− PB|

Maxima-Minima Problems

Example

A man starts from P(−3, 4) and reaches Q(0, 1) touching X − axisat R(α, 0) such that PR + RQ is minimum then α =?

Example

Consider the points A(0, 1) and B(2, 0), P is a point on the linex + y + 1 = 0. Find coordinates of point P such that |PA− PB| is

MaximumMinimum

Also find the max and min value attained by |PA− PB|

Transformations

ExampleGiven a line 3x + 4y + 1 = 0 in coordinate system, on shifting theorigin to a new location (h, k) if the x − intercept of the line in thenew coordinate system is 2 then what is its y − intercept

ExampleIn the above problem, if instead of shifting the origin to a newlocation the system of axes is rotated about the origin by someangle θ with new x − intercept of 2 then what is the y − interceptmade by the line?

ExampleThrough what angle should the axes be rotated so that the equation9x2 − 2

√3xy + 7y2 = 10 may be changed to 3x2 + 5y2 = 5?

Transformations

Linear Combination

Example

For a family of lines 5x + 3y − 2 + λ(3x − y − 4) = 0 andx − y + 1 + µ(2x − y − 2) = 0 then

Equation of the straight line that belongs to both the families

Locus of the point of intersection of lines from each familyintersecting at right angles to each other. Hence also derivethe equation of a circle given coordinates of the end points ofits diameter.

Example

If a2 + c2 − b2 + 2ac = 0 then the family of straight linesax + by + c = 0 is concurrent at the points?

Linear Combination

Example

Linear Combination

Example

Linear combination continued...

Example

Given the family of lines a(3x + 4y + 6) + b(x + y + 2) = 0. Theline of the family situated at the greatest distance from the pointP(2, 3) has equationA) 4x + 3y + 8 = 0 B) 5x + 3y + 10 = 0 C) 15x + 8y + 30 = 0 D)None

Example

If 6a2 − 3b2 − c2 + 7ab − ac + 4bc = 0 then the family of linesax + by + c = 0 is concurrent atA) (−2,−3) B) (3,−1) C) (2, 3) D) (−3, 1)

Linear combination continued...

Example

Given the family of lines a(3x + 4y + 6) + b(x + y + 2) = 0. Theline of the family situated at the greatest distance from the pointP(2, 3) has equationA) 4x + 3y + 8 = 0 B) 5x + 3y + 10 = 0 C) 15x + 8y + 30 = 0 D)None

Example

If 6a2 − 3b2 − c2 + 7ab − ac + 4bc = 0 then the family of linesax + by + c = 0 is concurrent atA) (−2,−3) B) (3,−1) C) (2, 3) D) (−3, 1)

Miscellaneous Problems

Problem #1 :

If4∑

(x2i + y2

i ) ≤ 2x1x3 + 2x2x4 + 2y2y3 + 2y1y4

then the points (xi , yi ) for i = 1 to 4 areA) vertices of a rectangleB) CollinearC) ConcyclicD) None of these

rankers coordinate geometry presentation

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