Problems in the Cartesian coordinate plane
On this webpage you can find the math homework with answers for geometry of the 9th grade:
 – the problems 23  27 show the examples of solutions and answers for the theme "Formula for calculating the coordinates of a vector from the coordinates of its beginning and end";
 – in the problems 28  33, it is considered how to solve the geometry, if it is necessary to determine the coordinates of the middle of the segment, the length of the vector having used its coordinates in the xyplane;
 – the solution of the problems in the Cartesian coordinate system is explained in the math tests 34  38;
 – the theme "Points, triangles, quadrangles in the xy coordinate plane" is presented in the problems 39  42 of the textbook;
 – in the examples of problems 4347 of the workbook, it is shown how to find various solutions for the theme "Equation of a circle";
 – the question of how to calculate and write the parametric equation of a straight line is considered in the tests 48  53.
The simplest problems in the xy coordinate plane
Definition:
The radius vector of an arbitrary point M is a vector drawn from the origin of the coordinate plane to the point M.
Theorem:
The coordinates of any point of the plane are equal to the corresponding coordinates of its radiusvector.
Given:
M(x;y) is any point
Is the radiusvector of the point M
Prove: {x; y}
Proof:
Case 1 (Figure 1). If x>0
By the parallelogram rule , where х=OM_{1}
_{ }= OM_{1} = x
_{ }= OM_{2} = y
So,
Therefore {x; y}
Case 2 (Figure 2). If x<0
Formula for calculating the coordinates of a vector having used the coordinates of its origin and its end.
Theorem:
Each coordinate of the vector is equal to the difference of the corresponding coordinates of its end and its beginning.
Given:
is an arbitrary vector
A (a_{1};a_{2}) are the coordinates of the origin of the vector AB
B (b_{1};b_{2}) are the coordinates of the end of the vector AB
Prove: {b_{1 }– a_{1}; b_{2 }– a_{2}}
Proof:
By the triangle rule, the vector ==
= {b_{1 }– a_{1}; b_{2 }– a_{2}}, where {b_{1};b_{2}} and {a_{1};a_{2}}, since they are the radiusvectors.
***
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:
The point C lies on the ordinate axis, i.e. yaxis: СO_{y}, then the coordinates of the point C (0;h)
Since CO is the height in the given isosceles triangle, then CO is the median too.
Therefore, AO=OB=2a : 2 = a
Then the coordinates of the point A(–a;0), the coordinates of the point B(a;0), since A and B O_{x} (i.e. they lie on the xaxis).
***
Problem 24.
Find: the coordinates of the vector , if given are the coordinates of the beginning and the end of the vector
1) A(2;7) and B(–2;7); 2) A(–5;1) and B(–5;27)
Solution:
1) = B–A={–2;7}–{2;7} = {–2–2;7–7}={–4;0}
2) = {–5;27}– {–5;1} = {–5+5;27–1}={0;26}
***
Problem 25.
Find: the coordinates of the vector , if given are the coordinates of the beginning and the end of the vector
1) A(–3;0) and B(0;4); 2) A(0;3) and B(–4;0)
Solution:
1) = B–A=
2) = B–A=
***
Problem 26.
Find: the coordinates of the vector , and the coordinates of the beginning and the end of the vector
Solution:
1) If A(0;0), B(1;1), then = B–A=
2) If A(x;–3), B(2;–7), {5;y}, then = B – A =
5 = 2 – x x= –3
y = –4
3) If A(a;b), {c;d}, then = B – A
B=
4) If A(1;2), {0;0}, then = B – A
B=
***
Problem 27.
Given:
MNPQ is a square
P (–3;3) are the coordinates of the point
NQ ∩ PM = O(0;0)
Find: the coordinates of the points M,N,Q
Solution:
Since the diagonals in the square are divided at the intersection point in half, then PO=OM , since PO and OM are collinear vectors, but they are ↑↓ multidirectional vectors.
Then {3;–3} M(3;–3)
Draw a perpendicular joining the point P in the direction to the yaxis to the intersection with the diagonal NQ.
Then the point N is the vertex of the square. Since the vertex of the square is located in the first (I) coordinate quarter, then N(3;3);
Since the diagonals at the intersection point are divided in half, then NO=OQ
, since these vectors are collinear, but differently directed ↑↓  i.e. they are collinear and oppositely directed vectors
Then {–3;–3} Q(–3;–3)
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 middle of a segment
Each coordinate of the middle of the segment is equal to the halfsum of the corresponding coordinates of its end and its origin.
Problem 28.
Given:
The segment AB, A(x_{1}; y_{1}), B(x_{2}; y_{2})
CAB, AC=CB
Prove: C
Proof:
Since the vector OC is equal to half the sum of two other vectors OA and OB emanating from the same point O, then
Since and are the radiusvectors of points A and B, then {x_{1}; y_{1}} and {x_{2}; y_{2}}.
Then
C, since is the radiusvector of the point C.
Formula, how to find the coordinates of the middle of a segment
C
Example 1.
Given:
The points A and B are the ends of the segment AB, the point M is the midpoint of the segment AB
a) If given are the coordinates of points A(2; –3); B(–3; 1), then how to find the coordinates of the middle of the segment AB
Using the formula to find the coordinate of the middle of the segment, we get M MM
b) If given are the 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 length of the vector having used its coordinates
Given: the coordinates of the vector
{a_{1}; a_{2}}
Prove: The length of the vector or the magnitude of a vector is equal to the square root of the sum of squares of its coordinates
Proof:
From the point О (0;0) we draw the vector =
But is a radius vector of the point А. Then A(a_{1}; a_{2})
Having used the triangle ΔOAA_{1} by the Pythagorean theorem:
a_{1}=OA_{1}; a_{2}=OA_{2}=AA_{1}
OA^{2}=
OA=
The formula how to find the length of a vector having used the coordinates of this vector
Example 2.
{11;–11}, then =
=
{10;0}, then = 10
Problem 30.
Given: the coordinates of the point
M_{1}(x_{1}; y_{1}), M_{2}(x_{2}; y_{2})
Find: distance between two points
d= M_{1}M_{2} = ?
Solution:
{ x_{2}– x_{1}; y_{2}– y_{1}},
, = M_{1}M_{2 }= d
Then –
Formula how to find the distance between two points
***
Problem 31.
Given: the coordinates of the beginning of the segment, the coordinates of the end of the segment, the coordinates of the middle of the segment
1  2  3  4  
A  (2;–3)  ?  (0;0)  (0;1) 
B  (–3;1)  (4;7)  (–3;7)  ? 
M  ?  (–3;–2)  ?  (3;–5) 
Find: the coordinates of the beginning, the end and the midpoint of the segment AB.
Solution:
Using the formula to find the coordinate of the middle of the segment, we obtain M
1) MM(–0,5 ; –1)
2) If B (4;7) and M (–3;–2), then we have to find the coordinates of the beginning of the segment A(x;y) – ?
–3= –6=4+x x=–10
 –2= –2•2=7+y y=– 11 
3) MM(–1,5 ; 3,5)
4) If A (0;1) and M (3;–5), then we have to find the coordinates of the end of the segment B(x;y) – ?
3= 6=x+0 x=6
 –5= 2•(–5)=y+1 y=– 11 
5  6  7  8  
A  (c;d)  (3;5)  (1;3)  (3t+5;7) 
B  ?  (3;8)  ?  (t+7;–7) 
M  (a;b)  ?  (0;0)  ? 
5) If A (c;d) and M (a;b), then we have to find the coordinates of the end of the segment B(x;y) – ?
a= 2a=c+x x=2a–c
 b= 2b=d+y y=2b–d 
6) MM(3 ; 6,5)
7) If A (1;3) and M (0;0), then we have to find the coordinates of the end of the segment B(x;y) – ?
0= 0=1+x x=–1
 0= 0=3+y y=–3 
8) MM(2t+6 ; 0)
***
Problem 32.
Given: the coordinates of the vector
{5;9}; {–3;4}; {–10;–10}; {10;17}
Find: the length of vectors ; ; ; – ?
Solution:
Using the formula to calculate the length of a vector having used its coordinates,
a) ====
b) =====5
c) ====10
d) == =
***
Problem 33.
Given: the coordinates of points A and B
1  2  3  4  
A  (2;7)  (–5;1)  (–3;0)  (0;3) 
B  (–2;7)  (–5;–7)  (0;4)  (–4;0) 
d  ?  ?  ?  ? 
Find: the distance between two points – ?
Solution:
Using the formula to calculate the distance between two points
, we obtain
1) d = = = 4
2) d = = = 8
3) d = ===5
4) d = ===5
***
Solving problems on a twodimensional Cartesian plane
Problem 34.
Given:
twodimensional coordinate system,
triangle ΔABC,
AM is the median
coordinates of the angles of a triangle
A(0;1), B(1;–4), C(5;2)
Find: the length of the median AM– ?
Solution:
Using the formula to find the coordinate of the middle of the segment, we get M MM(3 ; –1), since the point M is the middle of BC
Using the formula to calculate the distance between two points
, we obtain
AM ===
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:
Since the point O is the origin of the xy coordinate plane, then the coordinates of the point О (0;0).
Since OA=a and the point A lies on the abscissa axis O_{x}, i.e. xaxis, then the coordinates of the point A (a;0).
By the rule of the parallelogram and since OC is the vector, then
=
So, the vector has coordinates {a+b;c}
Since is the radius vector of the point C, then the coordinates of the point C (a+b;c).
Using the formula to calculate the distance between two points
, we obtain
AC ==
CO = ==
Answer: C (a+b;c) ; AC=; CO =
***
Problem 36.
Given:
the twodimensional coordinate system,
a triangle ,
the coordinates of the angles 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=, then triangle ΔABC  is isosceles.
2) Find: S_{Δ}_{ABC} – the area of the triangle ΔABC – ?
Solution:
We draw the height AM to the base BC.
Using the formula for calculating the area of a triangle, we obtain
S_{Δ}_{ABC}= AM•BC
Using the formula to find the coordinate of the middle of the segment, we get M MM(3 ; –1), since the point M is the midpoint of BC
Using the formula to calculate the distance between two points
, we obtain
AM ===
BC ===2
Then the area of the triangle S_{Δ}_{ABC} =••2==13
Answer: S_{Δ}_{ABC} =13.
***
Problem 37.
Given:
the 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_{Δ}_{MNP} – ?
Solution:
Using the formula to calculate the distance between two points
, we obtain
MN====
NP===
PM===
Using the formula to calculate the perimeter of a triangle, we get P_{Δ}_{MNP} =MN+NP+PM=++
Answer: P_{Δ}_{MNP} =++
***
Problem 38.
Given:
MNPQ is a quadrilateral
The coordinates of the vertices or the 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=, then MNPQ is a parallelogram
2) Find: the length of MP, NQ – ?
Solution:
Using the formula to calculate the distance between two points
, we obtain
MP === ==3
NQ === = 5
Answer: MP= 3, NQ = 5
***
Points, triangles, quadrangles on the coordinate plane
Problem 39.
Given:
the coordinates of the points C(4;–3), D(8;1)
the point A lies on the yaxis, i.e. O_{y}
AC=AD
Find: the coordinates of the point A – ?
Solution:
1) Since the point A lies on the yaxis, i.e. the ordinate axis O_{y}, then its coordinates are (0;y).
2) Using the formula to calculate the distance between two points , we obtain
AD==
3) AC==
4) Since AC=AD, then =
64 + (y – 1)^{2}= 16 + (3 + y)^{2}
64 + 1 – 2y+y^{2}=16 + 9 + 6y + y^{2}
–2y – 6y= 25 – 65
–8y = –40
y = 5
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:
1) Since the points A and B lie on the xaxis, i.e. the abscissa axis O_{x}_{,}
AO=OB=80 : 2 = 40 (cm)
The coordinates of the points B(40;0) and A(–40;0)
2) Since OC = 160 cm and the point C lies on the yaxis, i.e. O_{y}
The coordinates of the point C(0;160)
3) Since K is the midpoint of BC, then, using the formula to find the coordinate of the middle of the segment, we obtain K
KK(20 ; 80)
4) Then, using the formula to calculate the distance between two points , we obtain
AK= = ==100 (cm)
5) Since N is the midpoint of the segment AC, then, using the formula to find the coordinate of the middle of the segment, we obtain
N NN(–20 ; 80)
6) Using the formula to calculate the distance between two points , we obtain
NB = = ==100 (cm)
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 middle of the side AB
BC=a, AC=b
Prove: the length of the height in a rightangled triangle is half the length of the hypotenuse.
Proof:
Since BC=a, AC=b, then
the point C (0;0) – since the point C is the origin of the coordinate plane
Then 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 the squares of the sides of the parallelogram is equal to the sum of squares of the segments connecting the opposite vertices of the parallelogram.
AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}
Proof:
Using the formula to calculate the distance between two points , we obtain
AB= AB^{2} = b^{2}+c^{2}
AD=AD^{2} = a^{2}
AC=AC^{2} = (a+b)^{2} + c^{2}
BD=AC^{2} = (a–b)^{2} + c^{2}
Since AB=CD, AD=BC, then
AB^{2} + AD^{2} + BC^{2} + CD^{2} = 2(AB^{2} + AD^{2}) = 2(b^{2}+c^{2} + a^{2})
AC^{2} + BD^{2} = (a+b)^{2} + c^{2} + (a–b)^{2} + c^{2} = a^{2} + 2ab + b^{2} + c^{2} + a^{2} – 2ab + b^{2} + c^{2} = 2(a^{2} + b^{2} + c^{2})
Т.е. AB^{2} + BC^{2} + CD^{2} + AD^{2} = AC^{2} + BD^{2}
***
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:
In a rectangular or Cartesian coordinate system, the equation of the circle of the center at the point C(x_{0}; y_{0}) and the radius r has the form
(x – x_{0})^{2} + (y – y_{0})^{2} = r ^{2}
Given:
The circle (C; r) of the center at the point C,
the coordinates of the point C(x_{0}; y_{0})
Prove: the equation of the circle (x – x_{0})^{2} + (y – y_{0})^{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 CM^{2} = r^{2}
r^{2} = (x – x_{0})^{2} + (y – y_{0})^{2} (1)
The coordinates of the point M satisfy the equation (1).
If N(x_{1};y_{1}) does not lie on the circle (C;r), then
Since NC ≠ r , then the coordinates of the point N do not satisfy equation (1).
Hence, the equation of the circle (x – x_{0})^{2} + (y – y_{0})^{2} = r ^{2}
If the center of the circle, i.e. the point C has the coordinates C(x_{0}; y_{0}) = O(0;0) the equation of the circle is x^{2} + y^{2} = r ^{2}
***
Problem 43.
Given: the equation of a circle with the center at the point A passing through 2 points
a) x^{2} + y^{2} = 9
b) (x – 1)^{2} + (y + 2)^{2} = 4
c) (x + 5)^{2} + (y – 3)^{2} = 25
d) (x – 1)^{2} + y^{2} = 4
e) x^{2} + (y + 2)^{2} = 2
Find: the center of the circle and its radius r
Solution:
a) Let there be a circle of the center A and the radius r or briefly:
circle (A;r),
and a point B with coordinates (x;y), that lies on a given circle or briefly:
B(x;y) circle.
But AB = r, therefore, AB^{2}=r^{2} r^{2}=9 r^{2}=3^{2}, i.e. r = 3.
Since x^{2} + y^{2} = 9, then x_{0} =0 and y_{0}=0, then the center of the circle has the coordinates A(0;0). The graph of the circle is shown in Figure а).
б) Since (x – 1)^{2} + (y + 2)^{2} = 4, then by the equation of the circle
x_{0} = 1 and y_{0}= – 2.
Then A(1;–2), where A is the center of the circle.
Since r^{2} = 4, then r = 2. The graph of the circle is shown in Figure b)
в) Since (x + 5)^{2} + (y – 3)^{2} = 25, then by the equation of the circle
x_{0} = –5 and y_{0}= 3.
Then A(–5;3), where A is the center of the circle.
Since r^{2} = 25, then r = 5.
Draw this circle. The graph of the circle is shown in Figure c)
г) Since (x – 1)^{2} + y^{2} = 4, then by the equation of the circle
x_{0} = 1 and y_{0}= 0.
Then A(1;0), where A is the center of the circle.
Since r^{2} = 4, then r = 2.
Draw this circle. The graph of the circle is shown in Figure d)
д) Since x^{2} + (y + 2)^{2} = 2, then by the equation of the circle
x_{0} = 0 and y_{0}= –2.
Then A(0;–2), where A is the center of the circle.
Since r^{2} = 2, then r ≈ 1,4.
Draw this circle. The graph of the circle is shown in Figure e)
***
Problem 44.
Given: the circle is given by equation
a) x^{2} + y^{2} = 25
b) (x – 1)^{2} + (y + 3)^{2} = 9
Determine: which of the points A, B, C, D, E belong to the circle (A; r),
if given are the coordinates of the points A(3;–4); B(1;0); C(0;5); D(0;0); E(0;1)
Solution:
a) If A(3;–4), where x=3 and y=–4, then 3^{2} + (–4)^{2} = 25
9+16=25
25=25
The point A belongs to the circle (A; r) or the point A circle (A;r)
If B(1;0), where x=3 and y=0, then 1^{2} + 0^{2} = 25
1≠25
Therefore, the point B does not belong to the circle (A; r) or the point B circle (A;r)
If C(0;5), where x=0 and y=5, then 0^{2} + 5^{2} = 25
25=25
This means that the point C belongs to the circle (A; r) or the point Ccircle (A;r)
If D(0;0), where x=0 and y=0, then 0^{2} + 0^{2} = 25
0≠25
This means that the point D does not belong to the circle (A; r) or the point Dcircle (A;r)
If E(0;1), where x=0 and y=1, then 0^{2} + 1^{2} = 25
1≠25
Hence, the point E does not belong to the circle (A; r) or the point Ecircle (A;r)
b) If A(3;–4), where x=3 and y=–4, then (3 – 1)^{2} + (–4 + 3)^{2} = 9
4+1=9
5≠9
The point A does not belong to a circle (A; r) or the point Acircle (A;r)
If B(1;0), where x=1 and y=0, then (1 – 1)^{2} + (0 + 3)^{2} = 9
9=9
The point B belongs to the circle or the point Bcircle (A;r)
If C(0;5), where x=0 and y=5, then (0 – 1)^{2} + (5 + 3)^{2} = 9
1+64≠9
65≠9
The point C does not belong to the circle (A; r) or the point Ccircle (A;r)
If D(0;0), where x=0 and y=0, then (0 – 1)^{2} + (0 + 3)^{2} = 9
10≠9
The point D does not belong to the circle (A; r) or the point Dcircle (A;r)
If E(0;1), where x=0 and y=1, then (0 – 1)^{2} + (1 + 3)^{2} = 9
17≠9
The point E does not belong to a circle (A; r) or the point Ecircle (A;r)
***
Problem 45.
Given: the circle given by equation
The equation of a circle through 2 points (x+5)^{2} + (y – 1)^{2} = 16
The circle (C;r), where r=4, C(–5;1)
The points A(–2;4), B(–5;–3)
Determine: which of the points A or B belong to the circle (C;r)
Solution:
Using the formula to calculate the distance between two points , we obtain
CA===3,
3 >4, then the point A is outside the circle (C;r)
CB===4,
4=4, then the point B lies on the circle (C;r)
***
Problem 46.
Given:
the circle (C;r), where the diameter of the circle d=MN
the coordinates of the points M(–3;5), N(7;–3)
Write: the equation of a circle of the center C passing through 2 points M and N
Solution:
Using the formula to calculate the distance between two points , we obtain
CA====2,
Since d=2r, then r =• d = =
Using the formula to find the coordinate of the middle of the segment, we obtain the coordinates of the center of the circle C CC(2;1), since the point C is the middle of MN.
Using the equation of the circle equation (x – x_{0})^{2} + (y – y_{0})^{2} = r ^{2}, therefore, (x – 2)^{2} + (y – 1)^{2} =
(x – 2)^{2} + (y – 1)^{2} = 41
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 the center at the point C passing through 2 points
Solution:
Using the equation of the circle equation (x – x_{0})^{2} + (y – y_{0})^{2} = r ^{2}, therefore, x^{2} + (y – y_{0})^{2} = r ^{2}
Since the point B lies on the circle and its coordinates B(x=0;y=9).
Then the point B lies on the ordinate axis O_{y}, i.e. the yaxis, then CB=r.
x^{2} + (y – y_{0})^{2} = r ^{2} 0^{2}+ (9 – y_{0})^{ 2} = r ^{2}
Since the point A lies on the circle and its coordinates
A(x=–3;y=0).
Then the point A lies on the abscissa axis O_{x}, i.e. the xaxis, then CB=r.
(–3)^{2} + (0 – y_{0})^{2} = r ^{2} 9 + y_{0}^{2} = r ^{2}
Then (9 – y_{0})^{ 2} = 9 + y_{0}^{2}
y_{0}^{ 2} – 18 y_{0 }+ 81 = 9 + y_{0}^{2}
– 18 y_{0 } = – 72
y_{0 } = 4
Therefore, C(0;4).
We obtain the equation of a circle of the form: x^{2} + (y – 4)^{2} = r ^{2}
Let us find the radius r.
Since the points A and B belong to the circle, then
(–3)^{2} + (0 – 4)^{2} = r^{2} 9+16 = r^{2} r^{2} = 25  or  0^{2} + (9 – 4)^{2} = r^{2} r^{2} = 25

So we get the equation of the circle x^{2} + (y – 4)^{2} = 25
Answer: x^{2} + (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) l_{1}: y = y_{0} is the equation of the straight line passing through the point M_{0}, perpendicular to the straight line y, the yaxis, the ordinate axis O_{y}, parallel to the straight line x, the xaxis, the axis of abscissas O_{x}  
3) l_{2}: x = x_{0 } is the equation of a straight line passing through the point M_{0}, parallel to the straight axis of ordinates O_{y}, perpendicular to the xaxis, i.e. the axis of abscissas O_{x}  
4) y = 0 the equation of the straight line, the xaxis, i.e. the axis of abscissas O_{x}, passing through the origin point of the coordinate plane 5) x = 0 the equation of the yaxis, i.e. the ordinate axis O_{y}, 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 yaxis O_{y}
b) (x –2)^{2} + (y +5)^{2} = 3, parallel to the xaxis O_{x}
Solution:
a) (x +3)^{2} + (y – 2)^{2} = 25 the center of the circle (3; 2), where r=5
The straight line passes parallel to the axis O_{y} , then its equation is x=x_{0}.
Therefore, x= –3.
b) (x – 2)^{2} + (y +5)^{2} = 3 the center of the circle (2;–5), where r=
The straight line passes parallel to the axis O_{x} , then its equation is y=y_{0}.
Therefore, y= –5.
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Problem 49.
Given: the equations of two straight lines
b_{1}: 4x + 3y – 6=0
the straight line b_{2} is given by the equation: 2x + y – 4=0
Find: the coordinates A(x;y), where A is the intersection point of the straight lines b_{1} and b_{2}
Solution:
We will compose and solve the system of equations of the straight lines
The sum of equations (1) and (2) –2x = –6
Then x=3.
We substitute x=3 in the equation 4x+3y=6.
Then 12+3y= –6 y= –2
Answer: A(3; –2)
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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:
The general equation of the straight line has the form AB=ax+by+c=0, where having used the coordinates, it is necessary to find a=? b=? c=?
Since the points A and B lie on the straight line AB (or briefly: A and B AB),
then their coordinates satisfy the equation of the straight line AB
–a=–3c a=3c
Then we substitute a=3c in the equation a–b=–c. Therefore,
3c–b=–c
–b=–4c
b=4c
Then the equation of the line AB through two points: ax+by+c=0
3c+4cy+c=0
c(3x+4y+1)=0
3x+4y+1=0
Answer: the equation of a straight line having used two given points is 3x+4y+1=0
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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:
The general equation of the line AB: ax+by+c=0, where it is necessary to find a=? b=? c=?
Since the points A and B lie on the line AB, then their coordinates satisfy the equation of the line AB
–10,5b=1,5c
c=–7b
Then in the equation 2a+5b=–c we substitute
c=–7b, then, 2a+5b=7b
2a=7b–5b
a=b
Then the equation of the line AB should be written: ax+by+c=0
bx+by+(–7b)=0
b(x + y – 7)=0
x + y – 7=0
Answer: for these two points, the equation of the line AB is x+y–7=0
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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:
Since the points A and B lie on the line AB, therefore their coordinates satisfy the equation of the straight line AB.
Since the line AB intersects with the coordinate axes, then
coordinate I – (x;0)
coordinate II – (0;y)
Therefore, 3x–4•0+12=0
3x=–12
x=–4 A(–4;0)
3•0–4y+12=0
–4y=–12
y=3 B(0;3)
Next, we construct a straight line AB in the coordinate plane.
To construct a straight line, we first mark the points in the coordinate system: the abscissa of the point A is 4, its ordinate is zero; the abscissa of the point B is zero, its ordinate is 3. Draw a straight line through the marked two points.
Answer: A(–4;0), B(0;3)
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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:
Let us find the coordinate of the point M.
Using the formula to find the coordinate of the middle of the segment, we obtain M M
M(0;3), since the point M is the midpoint of AB.
The canonical equation of the straight line MC passing through the point has the form: ax+by+c=0, where it is necessary to find a=? b=? c=?
Since the points M and C lie on the line MC (or briefly: M,CMC), then their coordinates satisfy the equation of the straight line MC.
–3a=–7c
Having substituted
in the equation –a – 4b = –c, we obtain
b=
Knowing a and b, we find the coordinate equation of the straight line MC: ax+by+c=0
7x–y+3=0
Answer: the linear equation of the straight line MC: 7x–y+3=0
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