Equation of the line:

resolved exercises
Applied Mathematics
Hernando Arias de Saavedra Higher Studies Institute (ITEC)
10 pages.
 SOLVED PROBLEMS OF THE EQUATION OF THE
STRAIGHT

1) Find the slope and the angle of inclination of the line that passes through the
A(-3, 2)
Solution
y 2 y1
m
x2 x1
 3 2 5 1
m  
7 3 10 2
and how tan
then
1
tan 
2
153.4
2) The segments that a line determines on the X and Y axes are
respectively 2 and -3. Find its equation.
Solution
Using the segment equation we have
x y 1
2 3
 3x 2 y 6
3x 2 y  6 0
A line passes through the point A(7, 8) and is parallel to the line that passes through the
points C(-2, 2) and D(3, -4). Find its equation
Solution
Since the line must be parallel to the one that passes through the two given points,
then they must have the same slope.
 4 2 6
mCD
3 2 5
Using this slope, point A, and the point-slope equation, we have
6
y 8  (x 7)
5
5y 40 6 x 42
6x 5y 82 0
4) Prove that the points A(-5, 2), B(1, 4), and C(4, 5) are collinear.
Solution
Being collinear means they are located on the same line.
The slope between A and B must be the same as the one between B and C.
m AB mBC
4 2 5 4
1  5 4 1
2 1
6 3
Therefore, they are collinear.
5) Find the equation of the bisector of the segment formed by the coordinate axes.
determinant on the line 5x + 3y - 15 = 0
Solution
We will begin by determining the intersections of the line with the axes.
coordinates, to then determine the midpoint of the segment, right away
we calculate the slope of the given line to obtain that of the perpendicular
to her in order to finally determine the bisector line.
X axis intersection
5x 15 0
x 3
Y-axis intersection
3y 15 0
3y 15
y 5
We now calculate the midpoint of the segment that goes from (3, 0) to (0, 5)
3 0 0 5 3 5
PM , ,
2 2 2 2
We obtain the slope of the given line.
5x 3y 15 0
3y 5  x 15
5
y  x 5
3
5 3
The slope is then  , and the slope of the perpendicular is
3 5
3 3 5
We determine the equation of the line withm and what happens by ,
5 5 2
5 3 3
y x
2 5 5
5 3 9
y x
2 5 25
50y 125 30x 18
30x 50 years 107 0
6) Find the equation of the line with a slope of -4 that passes through the point of
intersection of the lines 2x + y - 8 = 0 and 3x - 2y + 9 = 0.
Solution
We find the point of intersection of the lines by solving the system of
two equations
2x y 8
3x 2 y 9
Clearing in the first and replacing in the second
y = 8 - 2x
3x - 2(8 - 2x) = -9
3x - 16 + 4x = -9
7x = 7
x=1
y=6
Now we write the equation of the line that passes through (1, 6) with slope
equal to -4
y - 6 = -4(x - 1)
y - 6 = -4x + 4
4x + y = 10
7) Find the area of the triangle formed by the coordinate axes and the line of
equation 5x + 4y + 20 = 0
Solution
The intersections of the line with the coordinate axes are determined.
4 y  20 0
4 y 2 0
y 5 
5x 20 0
5x 2  0
x 4
((4,0) and (0,5 )
Then the area is 20
Another way to solve this problem is by writing the equation in the form
from the equation of segments
5x 4 y e a r s  20 0
5x 4 y 2  0
x  y 1
 4 5
8) The following problems refer to a triangle with vertices A(-2, 1), B(4,
7) and C(6, -3)
8.1.- Find the equations of its sides
Solution
In this case, we use the point-point equation for each pair of vertices.
i) equation by AB
y 1x 2
7 1 4 2
 ii) equation by AC
solution
y 1x 2
 3 1 6 2
y 1x 2
4 8
2 y  2  x 2
x  2y 0
iii) equation by BC
y 7x 4
 3 7 6 4
y 7x 4
 10 10
y 7  x 4
x y 11 0

8.2 Find the equation of the line that passes through A and is parallel to side BC
Solution
We determine the slope of the line through BC and then use point A.
we obtain the requested line
x y 11 It has a slope of -1, then
y 1 1  (x 2)
y 1  x 2
x y 1 0

9) Find an expression to determine the distance from a point to a line.

Solution
Be the pointP(x0,y0and the line Ax + By + C = 0.
i) We determine the slope of the line
m  A
B
ii) We obtain the slope of the line perpendicular to it.
m B
A
iii) Now we calculate the equation of the line that passes through P and with
pendingm
y y0 B  x x 
0
A
Oh Oh0 Bx Bx0
Bx Oh Ay0 Bx0 0
iv) We find the intersection of the two lines
Ax By C 
Bx Oh Bx0 Oh0
Multiplying the first by A and the second by B
 A2x  B 2x B 2x0 Bay0 AC
B 2x0 Bay0 AC
x
A2 B 2
Now multiplying the first by B and the second by -A, we have
ABx B 2y  BC
 ABx A2y A2y0  ABx0
Adding
B 2y A2x A2y0 ABx0 BC
A2y0 ABx0 BC
y
A2 B 2
Then the point of intersection is
B 2x0Bay AC0 A2y ABx BC
, 0 0

AB
2 2
AB
2 2

Now we calculate the distance between point P and the point of intersection of
the two lines.
B 2x0 BAy0 AC A 2y0 ABx0 BC
d (x 0 ) 2
 ( y 0 )2
AB
2 2
AB
2 2

A 2x0 B 2x0 B 2x0 Bay0 AC A 2y0 B 2y0 A 2y0 ABx0 BC

( ) 2 ( )2
A 2 B 2 A 2 B 2
2 2
A 2x0 Bay AC0 B 2y 0ABx BC0

A 2 B 2 A 2 B 2

A 2 Ax0 By0 C   B 2  By0 Ax0 C 

2 2

( A 2 B 2) 2
 A  B   Ax  By  C
2 2
0 0  2

 AB  2 2 2

Ax0 By0 C
A 2 B 2
Therefore, the distance from a point to a line can be calculated as the
quotient between the equation of the line by replacing the coordinates of the
given point and the square root of the sum of the squares of the coefficients
from x to y and from y.

10) The coordinates of a point P are (2, 6) and the equation of a line L is
4x + 3y = 12. Find the distance from point P to line L.
Solution
Applying the previous result, it is found that
4(2) 18 12 8 18 12 14
d
4 2 3 2 25 5
11) Find the distance from the line 3x –4y + 12 = 0 to the point (4, -1).
Solution
3 4  4   1   12 12 4 12 28
d
9 16 25 5

12) Determine the values of A and B in the equation Ax –By + 4 = 0 if

passes through the points C(-3, 1) and D(1, 6)
Solution
If each of the points belongs to the line, it means they satisfy the
equation of the line, then
 3A B 4 0 y A  6B 4 0
Forming the system of two equations
 3A B 4
A 6B 4
Isolating A in the second
A 6B 4

and replacing it in the first one

 3  6B 4 B 4
 18B 12 B 4
 19B 1 6
16
B
19
And replacing
16 96 76 20
A 6 4
19 19 19
13) Find the value of k so that the line kx + (k –1)y–18 = 0 is parallel to
the line 4x + 3y + 7 = 0
Solution
Two lines are parallel if they have the same slope, then
i) pending from the first
 k
k 1
pending of the second
4

3
iii) making them equal
k 4
k 1 3
3k 4k 1
k 1

14) Demonstrate that the line passing through the points (4, -1) and (7, 2) bisects the
segment with endpoints (8, -3) and (-4, -3)
Solution
 ii) we find the equation of the line that passes through the two points
y 1x 4
2 1 7 4

Or well
y 1x 4
3 3
y 1 x 4
y x 5
iii) and finally we must prove that the midpoint found
it belongs to the line, for that we substitute it in the equation of the
recta.
 3 2 5
 3 3
15) Reduce the line with the equation 5x –7y -11 = 0 to standard form.
Solution
We multiply the equation by R
5Rx 7Ry 11R 0
cos 5R
sin 7 R
Squaring both equalities
you2 25R2
sin 2 49R2
Adding member to member
cos2  sin 2 74R2
1 74R2
1
R2
74
1
R
74
Since C is negative, we take the positive sign for the radical.
And the equation in its normal form is
5 7 11
x y 0
74 74 74

16) Find the distance from the origin to the line 2x –3y + 9 = 0
Solution 1 using the distance from a point to a line
2 0  3  0  9 9
d
4 9 13
Solution 2: knowing that when writing a line in its normal form the term
The constant indicates the distance of the line from the origin.
 2Rx 3Ry 9R 0
cos 2R
sin 3 R
cos2  sine 2 13R2
1 13R2
1
R
13
Since C is positive, the negative sign is taken for the radical.
2 3 9
 x y 0
13 13 13
9
It follows that the distance is
13
17) Determine the value of k so that the distance from the origin to the line x + ky
-7 = 0 sea 2.
Solution
Using the formula for the distance from a point to a line we have
Ax0 By0 C
d
A 2 B 2
0  k(0) 7
2
1 k2
7
2
1 k2
7
1 k2  2
2
49
1 k2
4
45
k2
4
3
k 5
2
18) Find the distance between the two parallel lines 3x + 5y - 11 = 0, and the
line 6x + 10y - 5 = 0.
Solution 1 (Alamos method)
We take a point from the first line, let it be (2, 1)
We calculate the slope of the second line.
3
m 
5
We obtain the slope of its perpendicular
5
m
3
We determine the equation of the line that passes through (2,1) with slope.m
5
y 1  x 2 
5x 3y 7
6x 10 years 5
Multiplying the first by 10 and the second by 3
50x 30 years 70
18x 30y 15
Adding
68x=85
85
x
68
Multiplying the first by 6 and the second by -5
30x 1 8 years 42
 30x 50 years 2 5
Adding
 68 years 17
17
y 
68
5 1
Finally, we calculate the distance between (2, 1) and ,
4 4
2 2 2 2
5 1 3 5 34 34
d 2  1 
4 4 4 4 16 4
Solution 2
using the procedure of the distance from a point to a line
Taking the same chosen point (2,1) we calculate the distance from that point.
to the second given line
6 2  10  1  5 12 10 5 17 17 136 136 34
d
36 100 136 136 136 8 4
Solution 3
writing the two equations in standard form and keeping in mind the fact
that the constant term of it marks the distance from the line to the origin and
finally subtracting the two distances we will also have the distance between the
two parallel lines.
Writing the first line in standard form
3 5 11
x y 0
34 34 34
Writing the second line in standard form
6 10 5
x  0
136 136 136
11
Then the first line is located at units of the origin and the second
34
5
a
136
Finally subtracting the lesser from the greater
11 5 22 5 17 34

19) The vertices of a triangle are A(-2, 3), B(5, 5) and C(4, -1). Find the
equation of the angle bisector of interior angle ACB
Solution
i) we found the equations of sides AC and BC
equation by AC
y 3x 2
 1  3 4 2
y 3x 2
4 6
6 y  18 4 x 8
4x 6 y  10 0
2x 3y 5 0

Equation by side BC
y 5x 5
 1  5 4 5
y 5x 5
6 1
 y 5 6 x 30
6x y 25 0
ii) We consider a pointP(x,y)about the bisector, this point must
meet the condition that it must be equidistant from the sides AC and BC
6x y 25 2x 3y 5

36 1 4 9
6x y 25 2x 3y 5

37 13
6 13x 13y 25 13 2 37x 3 37y 5 37
6  
13 2 37 x 3 37 13 
y 5 37 25 13 0
The negative sign is because they are on opposite sides of the origin.