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

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

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 $$I(\frac{1}{2}; \frac{3}{2})$$