## Solution for Exercise 4: Relative position and angle between two lines. Distance from a point to a line (C7 – Math 10 Kite)

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### Solving exercises Lesson 1, page 86, Math textbook 10 Kite, episode 2

Consider the relative positions of each of the following pairs of lines

a) $${d_1}:3x + 2y–5 = 0$$ and $${d_2}:x – 4y + 1 = 0$$ ;

b) $${d_3}:x – 2y + 3 = 0$$ and $${d_4}: – {\rm{ }}2x + 4y + 10 = 0$$ ;

c) $${d_5}:4x + 2y – 3 = 0$$ and $${d_6}:\left\{ \begin{array}{l}x = – \frac{1}{2} + t\ \y = \frac{5}{2} – 2t\end{array} \right.$$

Solution method

Solving the system of intersection equations:

System of equations with solutions $$\Rightarrow$$ intersecting each other

Parallel system of equations with no solutions $$\Rightarrow$$

System of equations with an infinite number of solutions $$\Rightarrow$$ coincide

Solution guide

a) The coordinates of the intersection of two lines $${d_1},{d_2}$$ is the solution of the system of equations:

$$\left\{ \begin{array}{l}3x + 2y – 5 = 0\\x – 4y + 1 = 0\end{array} \right.\Leftrightarrow \left\{ \begin{array}{ l}x = \frac{9}{7}\\y = \frac{4}{7}\end{array} \right.$$

The system of equations has only one solution, so the two lines intersect.

b) The coordinates of the intersection of two lines $${d_3},{d_4}$$ is the solution of the system of equations:

$$\left\{ \begin{array}{l}x – 2y + 3 = 0\\ – 2x + 4y + 10 = 0\end{array} \right.$$ .

The system of equations has no solution. So 2 lines are parallel to each other

c) The coordinates of the intersection of two lines $${d_5},{d_6}$$ corresponding to t satisfy the equation:

$$4\left( { – \frac{1}{2} + t} \right) + 2\left( {\frac{5}{2} – 2t} \right) – 3 = 0 \Leftrightarrow 0t = 0$$ .

This equation has a solution for all t. Hence $${d_5} \equiv {d_6}$$.

### Solve the exercise Exercise 2 page 86 Math textbook 10 Kite episode 2

Calculate the measure of the angle between two lines $${d_1}:2x–y + 5 = 0$$ and $${d_2}:x – 3y + 3 = 0$$.

Solution method

In the coordinate plane, for two lines $${\Delta _1}$$ and $${\Delta _2}$$ whose direction vectors are $$\overrightarrow {{n_1}} = {\rm{ }}\left( {{a_1};{\rm{ }}{b_1}} \right),{\rm{ }}\overrightarrow {{n_2}} {\rm{ }} = {\rm{ }} \left( {{a_2};{b_2}} \right)$$ we have:

$$\cos \left( {{\Delta _1},{\Delta _2}} \right) = \left| {\cos \left( {\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} } \right)} \right| = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\ sqrt {a_2^2 + b_2^2} }}.$$

Solution guide

The normal vector of the line $${d_1}$$ is: $$\overrightarrow {{n_1}} = \left( {2; – 1} \right)$$

The normal vector of the line $${d_2}$$ is: $$\overrightarrow {{n_2}} = \left( {1; – 3} \right)$$

We have: $$\cos \left( {{d_1},{d_2}} \right) = \left| {\cos \left( {\overrightarrow {{n_1}} ;\overrightarrow {{n_2}} } \ right)} \right| = \frac{{\left| {2.1 + \left( { – 1} \right).\left( { – 3} \right)} \right|}}{{\sqrt {{ {\left( 2 \right)}^2} + {{\left( { – 1} \right)}^2}} .\sqrt {{1^2} + {{\left( { – 3} \ right)}^2}} }} = \frac{{\sqrt 2 }}{2}$$

So $$\left( {{d_1},{d_2}} \right) = {45^o}$$

### Solve exercises Exercise 3 page 86 Math textbook 10 Kite episode 2

Calculate the distance from a point to a line in each of the following cases:

a)$$A\left( {1; – 2} \right){\rm{ }}v\`a {\rm{ }}{\Delta _1}:{\rm{ }}3x – y + { \rm{ }}4{\rm{ }} = {\rm{ }}0$$ ;

b) B(-3; 2) and $${\Delta _2}:\left\{ \begin{array}{l}x = – 2 + t\\y = 1 – 2t\end{array} \right .$$

Solution method

In the coordinate plane Oxy, for the line $$\Delta$$ there is the equation $${\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} ) > 0} \right)$$ and the point $$M\left( {{x_o};{y_0}} \right)$$. The distance from the point M to the line $$\Delta$$, denoted by $$d\left( {M,\Delta } \right)$$ is calculated by the formula: $$d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}$$

Solution guide

a) The distance from point A to $${\Delta _1}$$ is: $$d\left( {A,{\Delta _1}} \right) = \frac{{\left| {3.1 – 1. \left( { – 2} \right) + 4} \right|}}{{\sqrt {{3^2} + {{\left( { – 1} \right)}^2}} }} = \ frac{9}{{\sqrt {10} }}$$

b) The general equation of the line $${\Delta _2}$$ is: $$2x + y + 3 = 0$$

The distance from point B to $${\Delta _2}$$ is: $$d\left( {A,{\Delta _2}} \right) = \frac{{\left| {2.\left( { – 3} \right) + 1.2 + 3} \right|}}{{\sqrt {{2^2} + {1^2}} }} = \frac{1}{{\sqrt 5 }}$$

### Solving exercises Lesson 4, page 86, Math textbook 10 Kite, episode 2

For what value of parameter m are the following two lines perpendicular?

$${\Delta _1}:mx – y + 1 = 0$$ and $${\Delta _2}:2x – y + 3 = 0$$.

Solution method

The two lines $${\Delta _1}$$,$${\Delta _2}$$ are perpendicular if and only $$\overrightarrow {{n_1}} ;\overrightarrow {{n_2}}$$ are square. angle to each other

Solution guide

The normal vector of is: $$\overrightarrow {{n_1}} = \left( {m; – 1} \right)$$

The normal vector of is: $$\overrightarrow {{n_2}} = \left( {2; – 1} \right)$$

So who the lines $${\Delta _1}$$,$${\Delta _2}$$ are perpendicular if and only $$\overrightarrow {{n_1}} ;\overrightarrow {{n_2}}$$ perpendicular i.e. $$\overrightarrow {{n_1}} .\overrightarrow {{n_2}} = 0 \Leftrightarrow 2m + 1 = 0 \Leftrightarrow m = \frac{{ – 1}}{2}$$

### Solve the exercises Exercise 5 page 86 Math textbook 10 Kite episode 2

Given three points A(2;- 1), B(1 ; 2) and C(4;- 2). Calculate the measure of angle BAC and the angle between two lines AB and AC.

Solution method

$$\cos \widehat {BAC} = \cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)$$

$$\cos \left( {AB,AC} \right) = \left| {\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)} \right|$$

Solution guide

We have: $$\overrightarrow {AB} = \left( { – 1;3} \right);\overrightarrow {AC} = \left( {2; – 1} \right)$$

So $$\cos \widehat {BAC} = \cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right) = \frac{{ – 1.2 + 3.\left( { – 1} \ right)}}{{\sqrt {{{\left( { – 1} \right)}^2} + {3^2}} .\sqrt {{2^2} + {{\left( { – 1 } \right)}^2}} }} = \frac{{ – 1}}{{\sqrt 2 }} \Rightarrow \widehat {BAC} = {135^o}$$

So$$\cos \left( {AB,AC} \right) = \left| {\cos \left( {\overrightarrow {AB} ,\overrightarrow {AC} } \right)} \right| = \frac{ {\left| { – 1.2 + 3.\left( { – 1} \right)} \right|}}{{\sqrt {{{\left( { – 1} \right)}^2} + {3 ^2}} .\sqrt {{2^2} + {{\left( { – 1} \right)}^2}} }} = \frac{1}{{\sqrt 2 }} \Rightarrow \widehat {BAC} = {45^o}$$

### Solution of Exercises Lesson 6, page 86, Math Textbook 10, Kite episode 2

Given three points A(2;4), B(-1; 2) and C(3;-1). Write the equation of the line that passes through B and is equidistant from A and C.

Solution method

In the coordinate plane Oxy, for the line $$\Delta$$ there is the equation $${\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} ) > 0} \right)$$ and the point $$M\left( {{x_o};{y_0}} \right)$$. The distance from the point M to the line $$\Delta$$, denoted by $$d\left( {M,\Delta } \right)$$ is calculated by the formula: $$d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}$$

Solution guide

Let $$\Delta$$ be the line that passes through B and whose normal vector is $$\overrightarrow n = \left( {a;b} \right)$$

So the equation $$\Delta$$ is: $$a\left( {x + 1} \right) + b\left( {y – 2} \right) = 0 \Leftrightarrow {\rm{a}}x + by + \left( {a – 2b} \right) = 0$$

We have: $$d\left( {A,\Delta } \right) = d\left( {C,\Delta } \right) \Leftrightarrow \frac{{\left| {3a + 2b} \right|} }{{\sqrt {{a^2} + {b^2}} }} = \frac{{\left| {4a – 3b} \right|}}{{\sqrt {{a^2} + { b^2}} }} \Leftrightarrow \left[\begin{array}{l}3a+2b=4a–3b\\3a+2b=–4a+3b\end{array}\right\Leftrightarrow\left[\begin{array}{l}a=5b\left(1\right)\\7a=b\left(2\right)\end{array}\right$$[\begin{array}{l}3a+2b=4a–3b\\3a+2b= –4a+3b\end{array}\right\Leftrightarrow\left[\begin{array}{l}a=5b\left(1\right)\\7a=b\left(2\right)\end{array}\right\)

From (1) we can choose a normal vector: $$\overrightarrow n = \left( {5;1} \right)$$. So the equation for the line $$\Delta$$ is: $$5x + y + 3 = 0$$

From (2) we can choose a normal vector: $$\overrightarrow n = \left( {1;7} \right)$$. So the equation for the line $$\Delta$$ is: $$x + 7y – 13 = 0$$

### Solve the exercises Lesson 7 page 86 Math textbook 10 Kite episode 2

There are two ships A and B both starting from two berths, moving in a straight line at sea. On the radar display of the control station (referred to as the coordinate plane Oxy with units on the axes in kilometers), after starting t (hours) ($$t \ge 0$$ ), position

of ship A whose coordinates are determined by the formula $$\left\{ \begin{array}{l}x = 3 – 35t\\y = – 4 + 25t\end{array} \right.$$ , The position of ship B has coordinates (4 – 30t; 3 – 40t).

a) Calculate the cosine of the angle between the paths of two trains A and B.

b) How long after departure are the two closest trains?

c) If train A is at rest in the initial position and train B is moving, what is the shortest distance between the two trains?

Solution method

a) In the coordinate plane, for two lines $${\Delta _1}$$ and $${\Delta _2}$$ whose direction vectors are $$\overrightarrow {{u_1}} = {\, respectively. rm{ }}\left( {{a_1};{\rm{ }}{b_1}} \right),{\rm{ }}\overrightarrow {{u_2}} {\rm{ }} = {\rm{ }}\left( {{a_2};{b_2}} \right)$$ we have:

$$\cos \left( {{\Delta _1},{\Delta _2}} \right) = \left| {\cos \left( {\overrightarrow {{u_1}} ;\overrightarrow {{u_2}} } \right)} \right| = \frac{{\left| {{a_1}{a_2} + {b_1}{b_2}} \right|}}{{\sqrt {a_1^2 + b_1^2} .\ sqrt {a_2^2 + b_2^2} }}.$$

b) Step 1: In the coordinate plane Oxy, for the line $$\Delta$$ has the equation $${\rm{a}}x + by + c = 0\left( {{a^2} + {b^2} > 0} \right)$$ and the point $$M\left( {{x_o};{y_0}} \right)$$. The distance from the point M to the line $$\Delta$$, denoted by $$d\left( {M,\Delta } \right)$$ is calculated by the formula: $$d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}$$

Step 2: Evaluate by parameter t

c) In the coordinate plane Oxy, for the line $$\Delta$$ there is the equation $${\rm{a}}x + by + c = 0\left( {{a^2} + {b^) 2} > 0} \right)$$ and the point $$M\left( {{x_o};{y_0}} \right)$$. The distance from the point M to the line $$\Delta$$, denoted by $$d\left( {M,\Delta } \right)$$ is calculated by the formula: $$d\left( {M, \Delta } \right) = \frac{{\left| {{\rm{a}}{x_o} + b{y_o} + c} \right|}}{{\sqrt {{a^2} + { b^2}} }}$$

Solution guide

a) Ship A moves in a vector direction $$\overrightarrow {{u_1}} = \left( { – 35;25} \right)$$

Ship B moves in the vector direction $$\overrightarrow {{u_2}} = \left( { – 30; – 40} \right)$$

Let $$\alpha$$ be the angle between the two paths of the two ships, we have:

$$\cos \alpha = \left| {\cos \left( {\overrightarrow {{u_1}} ;\overrightarrow {{u_2}} } \right)} \right| = \frac{{\left| {\ left( { – 35} \right).\left( { – 30} \right) + 25.\left( { – 40} \right)} \right|}}{{\sqrt {{{\left( { – 35} \right)}^2} + {{25}^2}} .\sqrt {{{\left( { – 30} \right)}^2} + {{\left( { – 40} \ right)}^2}} }} = \frac{1}{{5\sqrt {74} }}.$$

b) After t hours, the position of train A is point M with coordinates: $$M\left( {3 – 35t; – 4 + 25t} \right)$$

After t hours, the position of train B is point N with coordinates: $$N\left( {4 – 30t;3 – 40t} \right)$$

Thus, $$\overrightarrow {MN} = \sqrt {{{\left( {1 + 5t} \right)}^2} + {{\left( {7 – 65t} \right)}^2}} = \sqrt {4250{t^2} – 900t + 50} = \sqrt {4250{{\left( {t – \frac{9}{{85}}} \right)}^2} + \frac{ {40}}{{17}}} \ge \sqrt {\frac{{40}}{{17}}} \approx 1.53\left( {km} \right)$$

The smallest MN is approximately 1.53km when $$t = \frac{9}{{85}}$$

So after $$\frac{9}{{85}}$$ hours from the time of departure, the two trains are closest to each other and are 1.53km apart.

c) The initial position of ship A at $${M_o}$$ corresponds to $$t = 0$$ , then $${M_o}\left( {3; – 4} \right)$$

Train B moves in a straight line with normal vector $$\overrightarrow n = \left( {40; – 30} \right)$$ and passes through the point $$K\left( {4;3} \right)\ ) The general equation of is: \(40\left( {x – 4} \right) – 30\left( {y – 3} \right) = 0 \Leftrightarrow 4x – 3y – 7 = 0$$ $$\Delta$$

We have: $$d\left( {{M_o},\Delta } \right) = \frac{{\left| {4.3 – 3.\left( { – 4} \right) – 7} \right|} }{{\sqrt {{4^2} + {{\left( { – 3} \right)}^2}} }} = \frac{{17}}{5} = 3.4\left( { km} \right)$$

So if train A is at rest at the initial position and train B is moving, the shortest distance between the two trains is 3.4km.