Exercise 1

Write the equation (in all possible forms) of the line that passes through the points A = (1, 2) and B = (2, 5).

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Exercise 2

Identify the type of triangle formed by the points: A = (6, 0), B = (3, 0) and C = (6, 3).

Exercise 3

Determine the slope and y-intercept of the line 3x + 2y - 7 = 0.

Exercise 4

Find the equation of the line r which passes through the point A = (1, 5) and is parallel to the line s \equiv 2x + y + 2 = 0.

Exercise 5

Find the equation of the line that passes through the point (2, -3) and is parallel to the straight line that joins the points (4, 1) and (-2, 2).

Exercise 6

The points A = (-1, 3) and B = (3, -3) are vertices of an isosceles triangle ABC that has its apex C on the line 2x - 4y + 3 = 0. If AC and BC are the equal sides, calculate the coordinates of Point C.

Exercise 7

The line r \equiv 3x + ny - 7 = 0 passes through the point A = (3, 2) and is parallel to the line s \equiv mx + 2y - 13 = 0. Calculate the values of m and n.

Exercise 8

Given triangle ABC with coordinates A = (0, 0), B = (4, 0) and C = (4, 4), calculate the equation of the median that passes through the vertex C.

Exercise 9

A parallelogram has a vertex A = (8, 0), and the point of intersection of its two diagonals is M = (6, 2). If the other vertex is at the origin, calculate:

1 The other two vertices.

2 The equations of the diagonals.

3 The length of the diagonal.

 

 

Solution of exercise 1

Write the equation (in all possible forms) of the line that passes through the points A = (1, 2) and B = (2, 5).

Two-Point Form Equation.

\frac { x - 1 }{ -2 - 1} = \frac { y - 2 }{ 5 - 2 }

General form.

x + y - 3 = 0

Slope–intercept form.

y = - x + 3

Point–slope form.

y - 2 = -1 . (x - 1)

Parametric form.

A (1, 2) \qquad B(-2, 5) \qquad \vec { AB } = (-3, 3)

\left\{\begin{matrix} x = 1 - 3k \\ y = 2 + 3k \end{matrix}\right

 

Solution of exercise 2

Identify the type of triangle formed by the points: A = (6, 0), B = (3, 0) and C = (6, 3).

d(\overline{ AB }) = \sqrt { { (3 - 6) }^{ 2 } + { (0 + 0) }^{ 2 } } = 3

d(\overline{ BC }) = \sqrt { { (6 - 3) }^{ 2 } + { (3 - 0) }^{ 2 } } = 3 \sqrt { 2 }

d(\overline{ AC }) = \sqrt { { (6 - 6) }^{ 2 } + { (3 - 0) }^{ 2 } } = 3

d(\overline{ AB }) = d(\overline{ BC }) \neq d(\overline{ AC }) Isosceles

{ [d(\overline{ BC })] }^{ 2 } = { [d(\overline{ AB })] }^{ 2 } + { [d(\overline{ AC })] }^{ 2 } Right triangle

 

Solution of exercise 3

Determine the slope and y-intercept of the line 3x + 2y - 7 = 0.

3x + 2y - 7 = 0 \qquad y = - \frac { 3 }{ 2 } x + \frac { 7 }{ 2 }

m = - \frac { 3 }{ 2 } \qquad b = \frac { 7 }{ 2 }

Solution of exercise 4

Find the equation of the line r which passes through the point A = (1, 5) and is parallel to the line s \equiv 2x + y + 2 = 0.

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

y - 5 = -2 . (x - 1)

2x + y - 7 = 0

 

Solution of exercise 5

Find the equation of the line that passes through the point (2, -3) and is parallel to the straight line that joins the points (4, 1) and (-2, 2).

r \left | \right | s \qquad { m }_{ r } = { m }_{ s } = \frac { 2 - 1 }{ - 2 - 4 } = - \frac { 1 }{ 6 }

{ m }_{ r } = - \frac { 1 }{ 6 } \qquad A(2, -3)

y + 3 = - \frac { 1 }{ 6 } (x - 2)

x + 6y + 16 = 0

 

Solution of exercise 6

The points A = (-1, 3) and B = (3, -3) are vertices of an isosceles triangle ABC that has its apex C on the line 2x - 4y + 3 = 0. If AC and BC are the equal sides, calculate the coordinates of Point C.

C \in r

2 { x }_{ c } - 4 { y }_{ c } + 3 = 0

d(\overline{ AC }) = d(\overline{ BC })

\sqrt { { ({ x }_{ c } + 1) }^{ 2 } + { ({ y }_{ c } - 3) }^{ 2 } } = \sqrt { { ({ x }_{ c } - 3) }^{ 2 } + { ({ y }_{ c } + 3) }^{ 2 } }

2 { x }_{ c } - 3 { y }_{ c } - 2 = 0

\left\{\begin{matrix} 2 { x }_{ c } - 4 { y }_{ c } + # = 0 \\ 2 { x }_{ c } - 3 { y }_{ c } - 2 = 0 \end{matrix}\right \qquad C( \frac { 17 }{ 2 }, 5)

 

Solution of exercise 7

The line r \equiv 3x + ny - 7 = 0 passes through the point A = (3, 2) and is parallel to the line s \equiv mx + 2y - 13 = 0. Calculate the values of m and n.

A \in r

3 . 3 + n . 2 - 7 = 0 \qquad n = -1

r \left | \right | s

\frac { 3 }{ m } = \frac { -1 }{ 2 } \qquad m = -6

 

Solution of exercise 8

Given triangle ABC with coordinates A = (0, 0), B = (4, 0) and C = (4, 4), calculate the equation of the median that passes through the vertex C.

{ M }_{ AB } (\frac { 0 + 4 }{ 2 }, \frac { 0 + 0 }{ 2 }) \qquad { M }_{ AB } (2, 0) \qquad B(4, 0)

\frac { x - 4 }{ 4 - 2 } = \frac { y - 0 }{ 4 - 0 } \qquad 2x - y - 4 = 0

 

Solution of exercise 9

A parallelogram has a vertex A = (8, 0), and the point of intersection of its two diagonals is M = (6, 2). If the other vertex is at the origin, calculate:

 

1 The other two vertices.

M is the midpoint of \overline{ OB }

(6, 2) = (\frac { 0 + { x }_{ B } }{ 2 }, \frac { 0 + { y }_{ B } }{ 2 } )

\frac { 0 + { x }_{ B } }{ 2 } = 6 \qquad \frac { 0 + { y }_{ B } }{ 2 } = 2

B(12, 4)

 

M is the midpoint of \overline{ AC }

(6, 2) = (\frac { 8 + { x }_{ c } }{ 2 }, \frac { 0 + { y }_{ c } }{ 2 })

\frac { 8 + { x }_{ C } }{ 2 } = 6 \qquad \frac { 0 + { y }_{ C } }{ 2 } = 2

C(4, 4)

 

2 The equations of the diagonals.

Equation of AC

\frac { x - 8 }{ 4 - 8 } = \frac { y - 0 }{ 4 - 0 }

x + y - 8 = 0

Equation of OB

\frac { x - 12 }{ 12 - 0 } = \frac { y - 4 }{ 4 - 0 }

x - 3y = 0

 

3 The length of the diagonal.

AC = \sqrt { { (4 - 8) }^{ 2 } + { (4 - 0) }^{ 2 } } = 4 \sqrt { 2 }

OB = \sqrt { { (12 - 0) }^{ 2 } + { (4 - 0) }^{ 2 } } = 4 \sqrt { 10 }

 

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Hamza

Hi! I am Hamza and I am from Pakistan. My hobbies are reading, writing and playing chess. Currently, I am a student enrolled in the Chemical Engineering Bachelor program.