Basic Maths on the Cartesian Coordinate Plane

The simplest problems on the x-y coordinate plane

Definition:

The position vector of an arbitrary point M is a vector connects the origin of the coordinate plane and the point M.

 

Theorem:

The coordinates of any point on the coordinate plane are equal to the corresponding components of the position vector of this point.

Given:

M(x;y) is any point

is the position vector of the point M

Prove: {x; y}

Proof:

Case 1 (Figure 1). If x>0

By the parallelogram rule , where х=OM₁

= OM₁ = x

= OM₂ = y

So,

Therefore {x; y}

 

Case 2 (Figure 2). If x<0

 

 

 

 

Formula for calculating the vector components from coordinates of the initial and terminal points of this vector.

Theorem:

Each vector component is equal to the difference of the corresponding coordinates of the initial and terminal points of this vector.

Given:

is an arbitrary vector

A (a₁;a₂) are the coordinates of the initial point of the vector AB

B (b₁;b₂) are the coordinates of the terminal point of the vector AB

 

Prove: {b₁ – a₁; b₂ – a₂}

Proof:

By the triangle rule, the vector ==

= {b₁ – a₁; b₂ – a₂}, where {b₁;b₂} and {a₁;a₂}, since they are the position vectors.

***

 

Problem 23.

Given:

ΔABC is an isosceles triangle

AB=2a, CO=h

Find: the coordinates of the point A, the point B and the point C

Solution:

Answer: M(3;–3), Q(–3;–3), N(3;3) and

M(3;–3), N(–3;–3), Q(3;3)

***

 

Formula how to find the coordinates of the midpoint of a segment

Each coordinate of the midpoint of a segment is equal to the half-sum of the corresponding coordinates of the initial and terminal points of the segment.

 

Problem 28.

Given:

The segment AB, A(x₁; y₁), B(x₂; y₂)

CAB, AC=CB

 

Prove: C

Proof:

Since the vector OC is equal to half the sum of two vectors OA and OB constructed at the same point O, we see that

Since and are the position vectors of points A and B, we see that {x₁; y₁} and {x₂; y₂}.

Then

C, since is the radius-vector of the point C.

Formula how to find the coordinates of the middle of a segment

C

 

Example.

Given:

The points A and B are endpoints of the segment AB, the point M is the midpoint of the segment AB

a) Given coordinates of points A(2; –3); B(–3; 1) how to find the coordinates of the midpoint of the segment AB

Using the formula to find the coordinate of the midpoint of the segment, we get M MM

b) Given coordinates of points B(4; 7); M(–3; –2), then how to find the coordinates of the point A(x;y).

and

–6 = 4+x and –4 = 7+y

x= –10 and y= –11

Answer: M; A (– 10; – 11)

***

 

Problem 29. How to find the vector magnitude or length from x and y components

Given: vector components

{a₁; a₂}

 

Prove that the magnitude or length of a vector is equal to the square root of sum of squares of x and y components

Proof:

Let us draw a vector at the origin O (0;0) =

But is a position vector of the point A. Then A(a₁; a₂)

Having used the triangle ΔOAA₁ by the Pythagorean theorem:

a₁=OA₁; a₂=OA₂=AA₁

OA²=

OA=

Formula how to find the magnitude of a vector from x and y components

 

Example.

{11;–11}, then =

=

{10;0}, then = 10

 

Problem 30.

Given: the coordinates of the point

M₁(x₁; y₁), M₂(x₂; y₂)

 

Find: distance between two points

d= M₁M₂ = ?

Solution:

Answer: AM =

***

 

Problem 35.

Given:

OACB is a parallelogram

The length of side OA=a

The coordinates of the point B (b;c)

 

Find: the coordinates of the point C(x;y),

the length of sides AC, CO – ?

Solution:

Answer: C (a+b;c) ; AC=; CO =

***

 

Problem 36.

Given:

a two-dimensional coordinate plane,

a triangle ,

the coordinates of the vertices of the triangle A(0;1), B(1;–4), C(5;2)

 

1) Prove: triangle ΔABC is isosceles

Proof:

Using the formula to calculate the distance between two points

, we obtain

AC===

AB===

Since AC=AB=, we see that triangle ΔABC is isosceles.

 

2) Find: S_ΔABC – the area of the triangle ΔABC – ?

Solution:

Answer: S_ΔABC =13.

***

 

Problem 37.

Given:

the Cartesian coordinate plane,

a triangle ΔMNP,

the coordinates of vertices of a triangle

M(4;0), N(12;–2), P(5;–9)

 

Find: the perimeter of the triangle P_ΔMNP – ?

Solution:

Answer: P_ΔMNP =++

***

 

Problem 38.

Given:

MNPQ is a quadrilateral

coordinates of vertices or angles of a quadrilateral are as follows

M(1;1), N(6;1), P(7;4), Q(2;4)

 

1) Prove: MNPQ is a parallelogram

Proof:

Using the formula to calculate the distance between two points

, we obtain

MN=== 5

PQ === 5

NP ===

QM ===

Since MN=PQ=5 and NP=QM=, we see that MNPQ is a parallelogram

 

2) Find: the length of MP, NQ – ?

Solution:

Answer: MP= 3, NQ = 5

***

 

Points, triangles, quadrilaterals on the coordinate plane

Problem 39.

Given:

the coordinates of the points C(4;–3), D(8;1)

the point A lies on the y-axis

AC=AD

 

Find: the coordinates of the point A – ?

Solution:

Answer: A (0; 5).

***

 

Problem 40.

Given:

O(0;0)

ΔABC is isosceles

the median OC=160cm

the base AB=80cm

 

Find: AK; NB

Solution:

Answer: AK = NB = 100 cm

***

 

Problem 41.

Given:

a triangle ΔABC

the right angle C = 90°

The point M lies on the side AB

The point M is the midpoint of the side AB

BC=a, AC=b

 

Prove: the length of the height in a right angled triangle is half the length of the hypotenuse.

Proof:

Since BC=a, AC=b, we see that

the point C (0;0) – since the point C is the origin of the coordinate plane

we have B(a;0), A(0;b).

Using the formula to find the coordinate of the middle of the segment, we obtain M MM(;), since the point M is the midpoint of AB.

Using the formula to calculate the distance between two points , we obtain

MC== =

AM==

Therefore, AM=MB=MC.

***

Problem 42.

Given:

ABCD is a parallelogram

AD=BC=a

the coordinates of points

B(b;c), D(a;0), C(a+b;c)

 

Prove:

The sum of squares of the sides of the parallelogram is equal to the sum of squares of diagonals of a parallelogram (where diagonals are the segments connecting the opposite vertices of the parallelogram).

AB² + BC² + CD² + AD² = AC² + BD²

Proof:

Using the formula to calculate the distance between two points , we obtain

AB= AB² = b²+c²

AD=AD² = a²

AC=AC² = (a+b)² + c²

BD=AC² = (a–b)² + c²

Since AB=CD, AD=BC, we see that

AB² + AD² + BC² + CD² = 2(AB² + AD²) = 2(b²+c² + a²)

AC² + BD² = (a+b)² + c² + (a–b)² + c² = a² + 2ab + b² + c² + a² – 2ab + b² + c² = 2(a² + b² + c²)

We see that AB² + BC² + CD² + AD² = AC² + BD²

 

***

 

The equation of a line and the equation of a circle

Definition: An equation with two variables x and y is called an equation of the line L if the coordinates of any point belonging to the line L satisfy this equation, and the coordinates of any point not lying on this line does not satisfy this equation.

Theorem - the formula of the equation of a circle or the equation of the circle radius:

On a rectangular or Cartesian coordinate system, the equation of the circle of center at the point C(x₀; y₀) and the radius r has the form

(x – x₀)² + (y – y₀)² = r ²

 

Given:

The circle (C; r) of the center at the point C,

the coordinates of the point C(x₀; y₀)

 

Prove: the equation of the circle (x – x₀)² + (y – y₀)² = r ²

Proof:

Consider a point M(x;y), lying on the circle (C;r)

Using the formula to calculate the distance between two points , we obtain

CM=, but CM=r

Then we get CM² = r²

r² = (x – x₀)² + (y – y₀)² (1)

The coordinates of the point M satisfy equation (1).

If N(x₁;y₁) does not lie on the circle (C;r), then

Since NC ≠ r , we see that the coordinates of the point N do not satisfy equation (1).

Hence, the equation of the circle (x – x₀)² + (y – y₀)² = r ²

If the center of the circle, i.e. the point C has the coordinates C(x₀; y₀) = O(0;0) the equation of the circle is x² + y² = r ²

***

 

Problem 43.

Given: the equation of a circle with the center at the point A passing through 2 points

a) x² + y² = 9

b) (x – 1)² + (y + 2)² = 4

c) (x + 5)² + (y – 3)² = 25

d) (x – 1)² + y² = 4

e) x² + (y + 2)² = 2

 

Find: the center of the circle and its radius r

Solution:

Answer: (x – 2)² + (y – 1)² = 41

***

 

Problem 47.

Given:

the circle (C;r), where the coordinates of the center of the circle C(0;y)

The points A and B lie on a circle

The coordinates of the points A(–3;0), B(0;9)

 

Write: the equation of a circle of center at the point C passing through 2 points

Solution:

Answer: x² + (y – 4)² = 25

***

 

The equation of the straight line

The graph of the equation of a straight line passing through a point. Types of equations of the line.

1) the equation of the straight line l: ax + by + c = 0, where a,b,c are the coefficients of the equation of the straight line

2) l1: y = y0 is the equation of the straight line passing through the point M0, perpendicular to the straight line y, the y-axis, the ordinate axis, parallel to the straight line x, the x-axis, the abscissa axis

3) l2: x = x0 is the equation of a straight line passing through the point M0, parallel to the straight axis of ordinates Oy, perpendicular to the x-axis, i.e. the axis of abscissas

4) y = 0 the equation of the straight line, the x-axis, i.e., the axis of abscissas, passing through the origin point of the coordinate plane 5) x = 0 the equation of the y-axis, i.e. the ordinate axis that passes through the origin point of the coordinate plane

 

 

Problem 48.

Write: the equation of a straight line passing through one point, i.e. the center of a circle.

a) (x +3)² + (y – 2)² = 25, parallel to the y-axis

b) (x –2)² + (y +5)² = 3, parallel to the x-axis

Solution:

Answer: A(3; –2)

***

Problem 50.

Given: the coordinates of two points A(1;–1), B(–3;2)

Find the equation of the straight line AB having used the given points

Solution:

Answer: the equation of a straight line having used two given points is 3x+4y+1=0

***

Problem 51.

Given: the coordinates of two points C(2;5), D(5;2)

Solve: find the equation of the line AB passing through 2 points

Solution:

Answer: for these two points, the equation of the line AB is x+y–7=0

***

Problem 52.

Given:

the line AB is given by the equation 3x–4y+12=0

Find: the coordinates of two points A and B that are the points of intersection with the coordinate axes

Solution:

Answer: A(–4;0), B(0;3)

***

 

Problem 53.

Given:

Triangle ΔABC

The coordinates of the vertices of the triangle A(4;6), B(–4;0), C(–1;–4)

The point M lies on the segment AB

Solve: find the equation of a straight line containing the median MC, having used the coordinates of the points

Solution:

Answer: the linear equation of the straight line MC: 7x–y+3=0

***