Exercise 1

Find the symmetric point A', of the point A = (3, 2), with the line of symmetry: r \equiv 2x + y - 12 = 0.

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Exercise 2

Identify the type of triangle formed by the points: A = (4, -3), \quad B = (3, 0), \quad and \quad C = (0, 1).

Exercise 3

Calculate the equation of the line that passes through the point P = (-3, 2) and is perpendicular to the line r \equiv 8x - y - 1 = 0.

Exercise 4

The line r \equiv x + 2y - 9 = 0 is the perpendicular bisector of the line segment AB whose endpoint A has the coordinates (2, 1). Find the coordinates of the other endpoint.

Exercise 5

Calculate the angle between the lines whose equations are:

1. { r }_{ 1 } \equiv 3x + 4y - 12 = 0; \qquad { r }_{ 2 } \equiv 6x + 8y + 1 = 0

2. { s }_{ 1 } \equiv 2x + 3y - 5 = 0; \qquad { s }_{ 2 } \equiv 3x - 2y + 10 = 0

Exercise 6

A straight line is parallel to the line r \equiv 5x + 8y - 12 = 0, and it is 6 units from the origin. What is the equation of this line?

Exercise 7

Determine the equations of the angle bisectors formed by the lines:

r \equiv 24x - 7y - 2 = 0; \qquad s \equiv 3x + 4y -4 = 0

Exercise 8

The vertices of a parallelogram are A = (3, 0), \quad B = (1, 4), \quad C = (-3, 2) \quad and \quad D = (-1, -2). Calculate the area.

Exercise 9

Given the triangle formed by the points A = (-1, -1), \quad B = (7, 5) \quad and \quad C = (2, 7), calculate the equations of the heights and determine the orthocenter of the triangle.

Exercise 10

A line is perpendicular to the line r \equiv 5x - 7y + 12 = 0 and it is 4 units away from the origin. Determine the equation of this line.

 

 

Solution of exercise 1

Find the symmetric point A', of the point A = (3, 2), with the line of symmetry: r \equiv 2x + y - 12 = 0.

2x + y - 12 = 0

y= -2x + 12

Comparing the above equation with y = mx + c

m = -2, \quad c = 12

 

{ m }_{ 1 } \times { m }_{ 2 } = -1

{ m }_{ 1 } = -2

-2 \times { m }_{ 2 } = -1

{ m }_{ 2 } = \frac { -1 }{ -2 }

{ m }_{ 2 } = \frac { 1 }{ 2 }

 

Using the co-ordinates of A(3,2) to construct the line equation of AA'

y - { y }_{ 1 } = m(x - { x }_{ 1 })

{ r }_{ AA' } \equiv y - 2 = \frac { 1 }{ 2 }(x - 3)

y - 2 = \frac { 1 }{ 2 }(x - 3)

2(y - 2) = x - 3

2y - 4 = x - 3

2y = x - 3 +4

2y = x + 1

 

We have equations of both linear lines and now we will solve simultaneously to find the middle point:

\begin{cases} 2y=x+1 \\ y=-2x+12 \end{cases}\begin{matrix} eq\quad i \\ eq\quad ii \end{matrix}

Multiplying equation ii by 2:

2y = -4x + 24   equation iii

Since the unknown y is same in equation iii and equation i that means we can equate both sides of equation:

-4x + 24 = x + 1

24 - 1 = x + 4x

23 = 5x

x = \frac { 23 }{ 5 }

 

Substituting the value of x in the equation i:

2y = x + 1

2y = \frac { 23 }{ 5 } + 1

2y = \frac { 23 + 5 }{ 5 }

2y = \frac { 28 }{ 5 }

y = \frac { 28 }{ 10 }

y = \frac { 14 }{ 5 }

Hence the mid-point is M( \frac { 23 }{ 5 }, \frac { 14 }{ 5 })

Since M is the mid-point of AA' and we know the coordiantes of A that means we can calculate the coordinates of A' with the help of mid-point formula:

Mid point formula = ( \frac { { x }_{ 1 } + { x }_{ 2 } }{ 2 }, \frac { { y }_{ 1 } + { y }_{ 2 } }{ 2 } )

A = (3, 2), \quad M=( \frac { 23 }{ 5 }, \frac { 14 }{ 5 }), \quad A' = (x, y)

( \frac { 23 }{ 5 }, \frac { 14 }{ 5 }) = ( \frac { 3 + x }{ 2 }, \frac { 2 + y }{ 2 })

Equating each side of coordinates:

\frac { 23 }{ 5 } = \frac { 3 + x }{ 2 }

23 \times 2 = (3 + x) \times 5

46 = 15 + 5x

46 - 15 = 5x

31 = 5x

x = \frac { 31 }{ 5 }

 

\frac { 14 }{ 5 } = \frac { 2 + y }{ 2 }

14 \times 2 = (2 + y) \times 5

28 = 10 + 5y

28 - 10 = 5y

18 = 5y

y = \frac { 18 }{ 5 }

Hence, A' = ( \frac { 31 }{ 5 } , \frac { 18 }{ 5 })

Solution of exercise 2

Identify the type of triangle formed by the points: A = (4, -3), \quad B = (3, 0), \quad and \quad C = (0, 1).

The distance of each side of the triangle will indicate the type of triangle. To calculate the distance, we will be using the distance formula.

d = \sqrt { { ({ x }_{ 1 } - { x }_{ 2 }) }^{ 2 } + { ({ y }_{ 1 } - { y }_{ 2 }) }^{ 2 } }

d(\overline { AB } ) = \sqrt { { (3 - 4) }^{ 2 } + { (0 - 3) }^{ 2 } } = \sqrt { 1 + 9 } = \sqrt { 10 }

d(\overline { BC } ) = \sqrt { { (0 - 3) }^{ 2 } + { (1 - 0) }^{ 2 } } = \sqrt { 9 + 1 } = \sqrt { 10 }

d(\overline { AC } ) = \sqrt { { (0 - 4) }^{ 2 } + { (1 - 3) }^{ 2 } } = \sqrt { 16 + 16 } = 4 \sqrt { 2 }

Since d(\overline { AB } ) = d(\overline { BC } ) \neq d(\overline { AC } ), hence we can conclude that the triangle is an Isosceles.

 

{[d(\overline { AC } )]}^{ 2 } = 32

{[d(\overline { AB } )]}^{ 2 } = 10

{[d(\overline { BC } )]}^{ 2 } = 10

{[d(\overline { AC } )]}^{ 2 } > {[d(\overline { AB } )]}^{ 2 } + {[d(\overline { BC } )]}^{ 2 }, hence we can conclude that the triangle is an obtuse triangle.

Solution of exercise 3

Calculate the equation of the line that passes through the point P = (-3, 2) and is perpendicular to the line r \equiv 8x - y - 1 = 0.

 

y = 8x - 1

y = mx + c

Comparing both equations to find gradient and y-intercept:

m = 8, \quad c = -1

 

{ m }_{ 1 } \times { m }_{ 2 } = -1

{ m }_{ 1 } = 8

8 \times { m }_{ 2 } = -1

{ m }_{ 2 } = \frac { -1 }{ 8 }

{ m }_{ 2 } = - \frac { 1 }{ 8 }

 

y - { y }_{ 1 } = m(x - { x }_{ 1 })

{ r }_{ AA' } \equiv y - 2 = -\frac { 1 }{ 8 }(x + 3)

8(y - 2) = -x - 3

8y - 16 = -x - 3

8y = -x - 3 + 16

8y = -x + 13

Solution of exercise 4

The line r \equiv x + 2y - 9 = 0 is the perpendicular bisector of the line segment AB whose endpoint A has the coordinates (2, 1). Find the coordinates of the other endpoint.

r \equiv x + 2y - 9 = 0

2y = -x + 9

y = \frac { -1 }{ 2 }x + \frac { 9 }{ 2 }

y = mx + c

Comparing both equations:

m = \frac { -1 }{ 2 }, \quad c = \frac { 9 }{ 2 }

 

{ m }_{ 1 } \times { m }_{ 2 } = -1

{ m }_{ 1 } = \frac { -1 }{ 2 }

\frac { -1 }{ 2 } \times { m }_{ 2 } = -1

{ m }_{ 2 } = \frac { -1 }{ \frac { -1 }{ 2 } }

{ m }_{ 2 } = 2

 

y - { y }_{ 1 } = m(x - { x }_{ 1 })

y - 1 = 2(x - 2)

y - 1 = 2x - 4

y = 2x - 4 + 1

y = 2x - 3

 

\begin{cases} y = \frac { -1 }{ 2 }x + \frac { 9 }{ 2 } \\ y = 2x - 3 \end{cases}

Solving both equations simultaneously will give us the cooridnates of mid-point because of the perpendicular bisector:

2x - 3 = -\frac { 1 }{ 2 }x + \frac { 9 }{ 2 }

2x + \frac { 1 }{ 2 }x = \frac { 9 }{ 2 } + 3

\frac { x + 4x }{ 2 } = \frac { 9 + 6 }{ 2 }

\frac { 5x }{ 2 } = \frac { 15 }{ 2 }

5x \times 2 = 15 \times 2

10x = 30

x = \frac { 30 }{ 10 }

x = 3

 

Substituting the value of x in the second equation:

y = 2x - 3

y = 2(3) - 3

y = 6 - 3

y = 3

 

Mid-point = (3,3), A = (2, 1), B = (x,y)

Mid point formula = ( \frac { { x }_{ 1 } + { x }_{ 2 } }{ 2 }, \frac { { y }_{ 1 } + { y }_{ 2 } }{ 2 } )

(3,3) = ( \frac { x + 2 }{ 2 }, \frac { y + 1 }{ 2 } )

Equating each side of coordinates:

3 = \frac { x + 2 }{ 2 }

3 \times 2 = x + 2

6 = x + 2

6 - 2 = x

x = 4

 

3 = \frac { y + 1 }{ 2 }

3 \times 2 = y + 1

6 = y + 1

6 - 1 = y

y = 5

 

Hence the coordinates of B = (4,5)

Solution of exercise 5

Calculate the angle between the lines whose equations are:

1. { r }_{ 1 } \equiv 3x + 4y - 12 = 0; \qquad { r }_{ 2 } \equiv 6x + 8y + 1 = 0

3x + 4y - 12 = 0

4y = -3x + 12

y = \frac { -3 }{ 4 }x + 3

y = mx + c

m = -\frac { 3 }{ 4 }, \quad c = 3

 

6x + 8y + 1 = 0

8y = -6x - 1

y = \frac { -6 }{ 8 }x - \frac { 1 }{ 8 }

y = \frac { -3 }{ 4 }x - \frac { 1 }{ 8 }

m = \frac { -3 }{ 4 }, \quad c = - \frac { 1 }{ 8 }

 

\tan { \alpha } = \frac { -\frac { 3 }{ 4 }-(-\frac { 3 }{ 4 }) }{ 1 + (-\frac { 3 }{ 4 }) \times -\frac { 3 }{ 4 }) } = 0

\alpha = \tan ^{ -1 }{ 0 }

\alpha = 0

 

2. { s }_{ 1 } \equiv 2x + 3y - 5 = 0; \qquad { s }_{ 2 } \equiv 3x - 2y + 10 = 0

2x + 3y - 5 = 0

3y = -2x + 5

y = -\frac { 2 }{ 3 }x + \frac { 5 }{ 3 }

y = mx + c

m = -\frac { 2 }{ 3 }, \quad c = \frac { 5 }{ 3 }

 

3x - 2y + 10 = 0

-2y = -3x - 10

2y = 3x + 10

y = \frac { 3 }{ 2 } + \frac { 10 }{ 2 }

y = \frac { 3 }{ 2 } + 5

y = mx + c

m = \frac { 3 }{ 2 }, \quad c = 5

 

\tan { \alpha } = \frac { -\frac { 3 }{ 2 }-(-\frac { 2 }{ 3 }) }{ 1 + (-\frac { 2 }{ 3 }) \times -\frac { 3 }{ 2 }) } = \frac { \frac { 13 }{ 6 } }{ 0 }

\alpha = 90

Solution of exercise 6

A straight line is parallel to the line r \equiv 5x + 8y - 12 = 0, and it is 6 units from the origin. What is the equation of this line?

Since the line is parallel but deviated from the origin, the only difference between both lines will be the constant.

s \equiv 5x + 8y + D = 0

d(O,s) = 6

\frac { D }{ \sqrt { { 5 }^{ 2 } + { 8 }^{ 2 } } } = 6

\frac { D }{ \pm \sqrt { 89 } } = 6

D = \pm 6 \sqrt { 89 }

 

5x + 8y \pm 6 \sqrt { 89 } = 0

 

Solution of exercise 7

Determine the equations of the angle bisectors formed by the lines:

r \equiv 24x - 7y - 2 = 0; \qquad s \equiv 3x + 4y -4 = 0

 

\frac { 24x - 7y - 2}{ \sqrt { 576 + 49 } } = \pm \frac { 3x + 4y - 4}{ \sqrt { 9 + 16 } }

\frac { 24x - 7y - 2}{ 25 } = \pm \frac { 3x + 4y - 4}{ 5 }

\begin{cases} x - 3y + 2\\ 39x + 13y - 22 \end{cases} \quad \begin{matrix} = 0\\ = 0 \end{matrix}

Solution of exercise 8

The vertices of a parallelogram are A = (3, 0), \quad B = (1, 4), \quad C = (-3, 2) \quad and \quad D = (-1, -2). Calculate the area.

Area = d(A,B) \times d(C, { r }_{ AB })

d=\sqrt { { ({ x }_{ 1 }-{ x }_{ 2 }) }^{ 2 }+{ ({ y }_{ 1 }-{ y }_{ 2 }) }^{ 2 } }

A = (3, 0), \quad B = (1, 4), \quad C = (-3, 2) \quad and \quad D = (-1, -2)

d(A,B) = \sqrt { { (3-1) }^{ 2 }+{ (0-4) }^{ 2 } }

d(A,B) = \sqrt { 4+16 }

d(A,B) = \sqrt { 20 }

d(A,B) = 2 \sqrt { 5 }

 

\frac { x - 3 }{ 1 - 3 } = \frac { y - 0}{ 4 - 0}

{ r }_{ AB } \equvi 2x + y -6 = 0

d(C , { r }_{ AB }) = \frac { \left| 2(-3) + 2 - 6 \right| }{ \sqrt { 5 } } = \frac { 10 }{ \sqrt { 5 } }

A = 2 \times \sqrt { 5 } \times \frac { 10 }{ \sqrt { 5 } } = 20 \quad { units }^{ 2 }

Solution of exercise 9

Given the triangle formed by the points A = (-1, -1), \quad B = (7, 5) \quad and \quad C = (2, 7), calculate the equations of the heights and determine the orthocenter of the triangle.

 

{ h }_{ A } is perpendicular to BC

{ m }_{ BC } = \frac { 7 - 5 }{ 2 - 7} = \frac { 2 }{ -5 }

{ m }_{ 1 } \times { m }_{ 2 } = -1

{ m }_{ { h }_{ A } } = \frac { 5 }{ 2 } \qquad A(-1, -1)

y + 1 = \frac { 5 }{ 2 } (x+1)

{ h }_{ A } \equiv 2y = 5x + 3

 

{ h }_{ B } is perpendicular to AC

{ m }_{ AC } = \frac { 7 + 1 }{ 2 + 1 } = \frac { 8 }{ 3 }

{ m }_{ 1 } \times { m }_{ 2 } = -1

{ m }_{ { h }_{ B } } = - \frac { 3 }{ 8 } \qquad B(7, 5)

y - 5 = - \frac { 3 }{ 8 } (x - 7)

{ h }_{ B } \equiv 8y = 61 - 3x

 

{ h }_{ C } is perpendicular to AB

{ m }_{ AB } = \frac { 5 + 1 }{ 7 + 1 } = \frac { 6 }{ 8 } = \frac { 3 }{ 4 }

{ m }_{ 1 } \times { m }_{ 2 } = -1

{ m }_{ { h }_{ C } } = - \frac { 4 }{ 3 } \qquad B(2, 7)

y - 7 = - \frac { 4 }{ 3 } (x - 2)

{ h }_{ C } \equiv 3y = 29 - 4x

 

\begin{cases} 2y = 5x + 3 \\ 8y = 61 - 3x \end{cases}

O( \frac { 49 }{ 23 }, \frac { 157 }{ 23 } )

Solution of exercise 10

A line is perpendicular to the line r \equiv 5x - 7y + 12 = 0 and it is 4 units away from the origin. Determine the equation of this line.

 

Since the line is perpendicular but deviated from the origin, the only difference between both lines will be the constant, and of course, the axis will switch.

s \equiv 5x - 7y + D = 0

d(O, s) = 4 \qquad \frac { D }{ \sqrt { { 7 }^{ 2 } + { 5 }^{ 2 } } } = 4

\frac { D }{ \pm \sqrt { 74 } } = 4

D = \pm \sqrt { 74 }

7x + 5y \pm \sqrt { 74 } = 0

 
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Emma

I am passionate about travelling and currently live and work in Paris. I like to spend my time reading, gardening, running, learning languages and exploring new places.