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**Solution for Exercise 20: Relative position between two lines. Angle and Distance (Connection)**

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### Solve problem 7.7 page 41 Math textbook 10 Connecting knowledge volume 2

Consider the relative positions between the following pairs of lines:

a) \(\Delta _{1}:3\sqrt{2}x+\sqrt{2}y-\sqrt{3}=0\) and \(\Delta _{2}: 6x+2y-\sqrt {6}=0\)

b) \(d _{1}: x-\sqrt{3}y+2=0\) and \(d _{2}: \sqrt{3}x-3y+2=0\)

c) \(m _{1}: x-2y+1=0\) and \(m _{2}: 3x+y-2=0\)

Solve problem 7.7 page 41 Math textbook 10 Connecting knowledge volume 2

Consider the relative positions between the following pairs of lines:

a) \(\Delta _{1}:3\sqrt{2}x+\sqrt{2}y-\sqrt{3}=0\) and \(\Delta _{2}: 6x+2y-\sqrt {6}=0\)

b) \(d _{1}: x-\sqrt{3}y+2=0\) and \(d _{2}: \sqrt{3}x-3y+2=0\)

c) \(m _{1}: x-2y+1=0\) and \(m _{2}: 3x+y-2=0\)

**Solution method**

Based on direction vectors \(\overrightarrow {{u_1}} ,\overrightarrow {{u_2}} \) or normal vectors \(\overrightarrow {{n_1}} ,\overrightarrow {{n_2}} \) of \(\overrightarrow {{\Delta _1}} ,\overrightarrow {{\Delta _2}} \) we have:

+ \({{\Delta _1}}\) and \({{\Delta _2}}\) parallel or identical ⇔ \(\overrightarrow {{u_1}} \) and \(\overrightarrow {{u_2} } \) in the same direction ⇔ \(\overrightarrow {{n_1}} \) and \(\overrightarrow {{n_2}} \) in the same direction.

+ \({{\Delta _1}}\) and \({{\Delta _2}}\) intersect ⇔ \(\overrightarrow {{u_1}} \) and \(\overrightarrow {{u_2}} \) not in the same direction ⇔ \(\overrightarrow {{n_1}} \) and \(\overrightarrow {{n_2}} \) are not in the same direction.

**Detailed explanation**

a) \(\Delta _{1}\) has a selection vector: \(\overrightarrow{n_{1}}(3\sqrt{2};\sqrt{2})\)

\(\Delta _{2}\) has a selection vector: \(\overrightarrow{n_{2}}(6; 2)\)

We have \(\overrightarrow{n_{1}}\) and \(\overrightarrow{n_{2}}\) in the same direction, so \(\Delta _{1}\) and \(\Delta _{2} \) parallel or overlap.

We have: \(3\sqrt{2}x+\sqrt{2}y-\sqrt{3}=0\) \(\Leftrightarrow \) \(3\sqrt{2}x+\sqrt{2}y- \sqrt{3}=0\)

So \(\Delta _{1}\) and \(\Delta _{2}\) coincide.

b) We have: \(x-\sqrt{3}y+2=0\) \(\Leftrightarrow \) \(\sqrt{3}x-3y+2\sqrt{3}=0\)

Which \(\sqrt{3}x-3y+2\sqrt{3} \neq \sqrt{3}x-3y+2\) so \(d _{1}\) and \(d _{2} \) parallel.

c) \(m _{1}\) has a normal vector: \(\overrightarrow{n_{1}}(1;-2)\)

\(m _{2}\) has a normal vector: \(\overrightarrow{n_{2}}(3;1)\)

We have \(\overrightarrow{n_{1}}\) and \(\overrightarrow{n_{2}}\) not in the same direction, so \(d _{1}\) and \(d _{2}\ ) intersecting.

### Solve lesson 7.8, page 41, Math 10 Textbook, Connecting knowledge volume 2

Calculate the angle between the following pairs of lines:

a) \(\Delta _{1}:\sqrt{3}x+y-4=0\) and \(\Delta _{2}: x+\sqrt{3}y+3=0\)

b) \(d_{1}:\left\{\begin{matrix}x=-1+2t\\ y=3+4t\end{matrix}\right.\) and \(d_{2}:\ left\{\begin{matrix}x=3+s\\ y=1-3s\end{matrix}\right.\) (t, s are parameters)

Solve lesson 7.8, page 41, Math 10 Textbook, Connecting knowledge volume 2

Calculate the angle between the following pairs of lines:

a) \(\Delta _{1}:\sqrt{3}x+y-4=0\) and \(\Delta _{2}: x+\sqrt{3}y+3=0\)

b) \(d_{1}:\left\{\begin{matrix}x=-1+2t\\ y=3+4t\end{matrix}\right.\) and \(d_{2}:\ left\{\begin{matrix}x=3+s\\ y=1-3s\end{matrix}\right.\) (t, s are parameters)

**Solution method**

Given two straight lines

\({\Delta _1}:{a_1}x + {b_1}y + {c_1} = 0\) and \({\Delta _2}:{a_2}x + {b_2}y + {c_2} = 0\ ).

For normal vectors \(\overrightarrow {{n_1}} \left( {{a_1};{b_1}} \right)\) and \(\overrightarrow {{n_2}} \left( {{a_2};{ b_2}} \right)\) in the application. Then, the angle \(\varphi \) between those two lines is determined through the formula

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

**Detailed explanation**

a)

\(\Delta _{1}\) has a normal vector \(\overrightarrow{n_{1}}(\sqrt{3}; 1)\)

\(\Delta _{2}\) has a normal vector \(\overrightarrow{n_{2}}(1; \sqrt{3})\)

Let \(\varphi \) be the angle between the two lines \(\Delta _{1}\) and \(\Delta _{2}\), we have:

\(cos\varphi =\left | cos(\overrightarrow{n_{1}},\overrightarrow{n_{2}})\right |=\frac{|\sqrt{3}.1+1.\sqrt{ 3}|}{\sqrt{1^{2}+3}.\sqrt{3+1^{2}}}=\frac{\sqrt{3}}{2}\)

Hence the angle between \(\Delta _{1}\) and \(\Delta _{2}\) is \(\varphi =30^{o}\).

b)

\(d _{1}\) has direction vector \(\overrightarrow{u_{1}}(2; 4)\)

\(d _{2}\) has direction vector \(\overrightarrow{u_{2}}(1; -3)\)

Let \(\varphi \) be the angle between the two lines \(d _{1}\) and \(d _{2}\), we have:

\(cos\varphi =\left | cos(\overrightarrow{u_{1}},\overrightarrow{u_{2}})\right |=\frac{|2.1-3.4|}{\sqrt{2^{2 }+1^{2}}.\sqrt{4^{2}+3^{2}}}=\frac{2\sqrt{5}}{5}\)

Therefore the angle between \(\Delta _{1}\) and \(\Delta _{2}\) is \(\varphi \approx 26.6^{o}\).

### Solve problem 7.9 page 41 Math textbook 10 Connecting knowledge volume 2

In the coordinate plane Oxy, give the point A(-2; 0) and the line \(\Delta \): x + y – 4 = 0.

a) Calculate the distance from point A to the line \(\Delta \)

b) Write an equation for the line a that passes through the point M(-1; 0) and is parallel to \(\Delta \).

c) Write an equation for the line b that passes through the point N(3; 0) and is perpendicular to \(\Delta \).

Solve problem 7.9 page 41 Math textbook 10 Connecting knowledge volume 2

In the coordinate plane Oxy, give the point A(-2; 0) and the line \(\Delta \): x + y – 4 = 0.

a) Calculate the distance from point A to the line \(\Delta \)

b) Write an equation for the line a that passes through the point M(-1; 0) and is parallel to \(\Delta \).

c) Write an equation for the line b that passes through the point N(3; 0) and is perpendicular to \(\Delta \).

**Solution method**

a) The distance from point M to the line \(\Delta \), denoted by \(d\left( {M,\Delta } \right)\), is calculated by the formula

\(d\left( {M,\Delta } \right) = \frac{{\left| {a{x_0} + b{y_0} + c} \right|}}{{\sqrt {{a^2 } + {b^2}} }}\)

b) The general equation of the form ax + by + c =0, takes \(\overrightarrow n \left( {a;b} \right)\) as a normal vector.

c) The line \(\Delta \) passes through the point \(A\left( {{x_0};{y_0}} \right)\) and has direction vector \(\overrightarrow u \left( {a;b } \right)\). The parametric equation of the line \(\Delta \) is \(\left\{ \begin{array}{l}

x = {x_0} + at\\

y = {y_0} + bt

\end{array} \right.\;\;\;\;\;\;\;\;\)

**Detailed explanation**

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a) The distance from point A to the line \(\Delta\) is: \(d_{(A;\Delta )}=\frac{|0-2+4|}{\sqrt{1^{2} +1^{2}}}=\sqrt{2}\)

b) line a is parallel to \(\Delta\) so line a has the form: x + y + c = 0.

Since a passes through M, so: -1 + 0 + c = 0, so c = 1.

So the equation of the line a: x + y + 1 = 0.

c) The line b is perpendicular to \(\Delta\) so the line b has the direction vector which is the normal vector of the line b: \(\overrightarrow{u}(1; 1)\)

The parametric equation of line b is: \(\left\{\begin{matrix}x=t\\ y=3+t\end{matrix}\right.\)

### Solve lesson 7.10, page 41, Math textbook 10, Connecting knowledge volume 2

In the coordinate plane, let triangle ABC have A(1; 0), B(3; 2) and C(-2; 1).

a) Find the length of the altitude drawn from the vertex A of triangle ABC.

b) Calculate the area of triangle ABC.

Solve lesson 7.10, page 41, Math textbook 10, Connecting knowledge volume 2

In the coordinate plane, let triangle ABC have A(1; 0), B(3; 2) and C(-2; 1).

a) Find the length of the altitude drawn from the vertex A of triangle ABC.

b) Calculate the area of triangle ABC.

**Solution method**

a)

+ Write the equation of the line BC

+ Calculate the length of the altitude drawn from A of triangle ABC is the distance from A to the line BC.

b)

Calculate the length of segment BC

+ Calculate the area of triangle ABC as: \(S_{ABC}=\frac{1}{2}d_{(A;BC)}.BC\)

**Detailed explanation**

a)

+ Write the equation of the line BC: whose direction vector is \(\overrightarrow{BC}(-5;-3)\) and passes through B(3; 2).

=> The line BC has a normal vector: \(\overrightarrow{n}(3; -5)\)

The equation of the line BC is: 3(x – 3) – 5(y – 2) = 0, Or 3x – 5y +1 = 0

+ The length of the altitude drawn from A of triangle ABC is the distance from A to the line BC.

Applying the distance formula has: \(d_{(A; BC)}=\frac{|3.1-5.0+1|}{\sqrt{3^{2}+5^{2}}}=\frac {2\sqrt{34}}{17}\)

b)

+ The length of segment BC is: \(BC = \sqrt{3^{2}+5^{2}}=\sqrt{34}\)

+ Area of triangle ABC is: \(S_{ABC}=\frac{1}{2}d_{(A;BC)}.BC=\frac{1}{2}.\frac{2\sqrt{ 34}}{17}.\sqrt{34}=2\)

### Solve problem 7.11 page 41 Math textbook 10 Connecting knowledge volume 2

Prove that the two lines d: y = ax + b (a \(\neq \) 0) and d’: y = a’x + b’ (a’ \(\neq \) 0) are perpendicular to each other. if and only if aa’ = -1.

Solve problem 7.11 page 41 Math textbook 10 Connecting knowledge volume 2

Prove that the two lines d: y = ax + b (a \(\neq \) 0) and d’: y = a’x + b’ (a’ \(\neq \) 0) are perpendicular to each other. if and only if aa’ = -1.

**Solution method**

+) Assuming the lines d and d’ are perpendicular to each other, we prove aa’ = -1.

+) Assuming a.a’ = -1, we prove that the lines d and d’ are perpendicular to each other.

**Detailed explanation**

+) Assuming the lines d and d’ are perpendicular to each other, we prove aa’ = -1. Indeed,

Line d has a normal vector: \(\overrightarrow{n}(a; -1)\)

The line d’ has a normal vector: \(\overrightarrow{n’}(a’; -1)\)

Since the lines d and d’ are perpendicular, \(\overrightarrow{n}.\overrightarrow{n’}=0\)

Infer: a.a’ + (-1).(-1) = 0, or a.a’ = -1.

+) Assuming a.a’ = -1, we prove that the lines d and d’ are perpendicular to each other. Indeed,

Consider the dot product: \(\overrightarrow{n}.\overrightarrow{n’}= a.a’ + (-1).(-1) = -1 + 1 = 0\)

=> \(\overrightarrow{n}\perp \overrightarrow{n’}\)

So the lines d and d’ are perpendicular to each other.

### Solve lesson 7.12, page 41, Math 10 Textbook, Connecting knowledge, volume 2

In the coordinate plane, an audio signal is transmitted from one location and received by three signal recorders at three positions O(0; 0), A(1; 0), B(1; 3) be at the same time. Please determine the location of the audio signal.

Solve lesson 7.12, page 41, Math 10 Textbook, Connecting knowledge, volume 2

In the coordinate plane, an audio signal is transmitted from one location and received by three signal recorders at three positions O(0; 0), A(1; 0), B(1; 3) be at the same time. Please determine the location of the audio signal.

**Solution method**

We have: \(IO=\sqrt{(x-0)^{2}+(y-0)^{2}}\),

\(IA= \sqrt{(x-1)^{2}+(y-0)^{2}}\),

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

Since IO = IA = IB, we have a system of equations: \(\left\{ {\begin{array}{*{20}{c}}

{{{(x – 0)}^2} + {{(y – 0)}^2} = {{(x – 1)}^2} + {{(y – 0)}^2}}\ \

{{{(x – 1)}^2} + {{(y – 0)}^2} = {{(x – 1)}^2} + {{(y – 3)}^2}}

\end{array}} \right.\)

Solve the system to find the value x, y is the point to find

**Detailed explanation**

Call the signal generating point I(x; y).

Because position I is all received by three signal recorders at O, A, B at the same time, so: IO = IA = IB.

We have: \(IO=\sqrt{(x-0)^{2}+(y-0)^{2}}\),

\(IA= \sqrt{(x-1)^{2}+(y-0)^{2}}\),

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

Since IO = IA = IB, we have a system of equations:

\(\left\{\begin{matrix}(x-0)^{2}+(y-0)^{2}=(x-1)^{2}+(y-0)^{2} \\ (x-1)^{2}+(y-0)^{2}=(x-1)^{2}+(y-3)^{2}\end{matrix}\right.\ Leftrightarrow \left\{\begin{matrix}-2x+1=0\\ -6y +9 =0\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix}x=\frac{1} {2}\\ y=\frac{3}{2}\end{matrix}\right.\)

So the point to find is \(I(\frac{1}{2}; \frac{3}{2})\)