Blog

adsense

Solution of Exercise 19: Equation of a line (Connection)
——————

Solve problem 7.1 page 34 Math textbook 10 Connecting knowledge volume 2

In the coordinate plane for \(\overrightarrow{n}(2;1), \overrightarrow{v}(3; 2), A(1; 3), B(-2; 1)\)

a) Write the general equation of the line \(\Delta _{1}\) passing through A and having the normal vector \(\overrightarrow{n}\).

b) Make a parametric equation of the line \(\Delta _{2}\) passing through B and having direction vector \(\overrightarrow{v}\).

c) Make a parametric equation for the line AB.

Solve problem 7.1 page 34 Math textbook 10 Connecting knowledge volume 2

In the coordinate plane for \(\overrightarrow{n}(2;1), \overrightarrow{v}(3; 2), A(1; 3), B(-2; 1)\)

a) Write the general equation of the line \(\Delta _{1}\) passing through A and having the normal vector \(\overrightarrow{n}\).

b) Make a parametric equation of the line \(\Delta _{2}\) passing through B and having direction vector \(\overrightarrow{v}\).

c) Make a parametric equation for the line AB.

Solution method

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

– 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 general equation of the line \(\Delta _{1}\) passing through A and having the normal vector $\overrightarrow{n}$:

2(x – 1) + 1.(y – 3) = 0 or 2x + y -5 = 0.

b) Parametric equation of the line \(\Delta _{2}\) passing through B and having direction vector $\overrightarrow{v}$.

\(\left\{\begin{matrix}x=-2+3t\\ y=1+2t\end{matrix}\right.\)

c) The line AB has direction vector: \(\overrightarrow{AB}(-3; -2)\).

=> Select direction vector: \(\overrightarrow{u}(3; 2)\).

Parametric equation of line AB:

\(\left\{\begin{matrix}x=-2+3t\\ y=1+2t\end{matrix}\right.\)

Solve problem 7.2 page 34 Math textbook 10 Connecting knowledge volume 2

Set up the general equation of the coordinate axes

Solve problem 7.2 page 34 Math textbook 10 Connecting knowledge volume 2

Set up the general equation of the coordinate axes

Solution method

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

Detailed explanation

+) Ox axis: there is a normal vector \(\overrightarrow{n}(0; 1)\), passing through the point O(0; 0).

General equation of the line containing the axis Ox: y = 0

+ Oy axis: there is a normal vector \(\overrightarrow{n}(1; 0)\), passing through the point O(0; 0).

General equation of the line containing the axis Oy: x = 0

Solve problem 7.3, page 34, Math textbook 10, Connecting knowledge volume 2

Given two lines \(\Delta _{1}:\left\{\begin{matrix}x=1+2t\\ y=3+5t\end{matrix}\right.\) and 2x + 3y – 5 = 0.

a) Write the general equation for \(\Delta _{1}\)

b) Make a parametric equation for \(\Delta _{2}\)

Solve problem 7.3, page 34, Math textbook 10, Connecting knowledge volume 2

Given two lines \(\Delta _{1}:\left\{\begin{matrix}x=1+2t\\ y=3+5t\end{matrix}\right.\) and 2x + 3y – 5 = 0.

a) Write the general equation for \(\Delta _{1}\)

b) Make a parametric equation for \(\Delta _{2}\)

Solution method

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

– 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) \(\Delta _{1}\) has direction vector \(\overrightarrow{u}(2;5)\)

=> \(\Delta _{1}\) has a normal vector \(\overrightarrow{n}(5;-2)\)

The general equation: 5(x – 1) – 2(y – 3) = 0, or 5x – 2y +1 = 0.

b) \(\Delta _{2}\) has a normal vector \(\overrightarrow{n}(2;3)\)

=> \(\Delta _{2}\) has direction vector \(\overrightarrow{n}(3;-2)\)

\(\Delta _{2}\) passes through the point with coordinates: (1; 1)

Parametric equation: \(\left\{\begin{matrix}x=1+3t\\ y=1-2t\end{matrix}\right.\)

Solve problem 7.4 page 34 Math textbook 10 Connecting knowledge volume 2

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

a) Find the equation of the altitude drawn from A.

b) Make an equation for the median line drawn from B.

adsense

Solve problem 7.4 page 34 Math textbook 10 Connecting knowledge volume 2

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

a) Find the equation of the altitude drawn from A.

b) Make an equation for the median line drawn from B.

Solution method

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

– 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 equation of altitude drawn from A of triangle ABC takes the vector \(\overrightarrow{BC}(-5; -1)\) as the selection vector.

=> The equation of the altitude through A and having the selection vector \(\overrightarrow{BC}(-5; -1)\) is:

-5(x – 1) – 1.(y – 2) = 0 Or 5x + y – 7 = 0.

b) Let M(x; y) be the midpoint of AC. Infer the coordinates of point M as: \(\left\{\begin{matrix}x=\frac{1-2}{2}=\frac{-1}{2}\\ y=\frac{2- 1}{2}=\frac{1}{2}\end{matrix}\right.\Rightarrow M\left ( \frac{-1}{2};\frac{1}{2} \right )\ )

+ the equation of the median drawn from B has the direction vector \(\overrightarrow{BM}(-3.5; 0.5)\)

=> Choose a direction vector of the line: \(\overrightarrow{u}(-7; 1)\)

The parametric equation of the line through B has the direction vector \(\overrightarrow{u}(-7; 1)\):

\(\left\{\begin{matrix}x=3-7t\\ y=t\end{matrix}\right.\)

Solve problem 7.5 page 34 Math textbook 10 Connecting knowledge volume 2

(Equation of intercept of the line)

Prove that the line passing through two points A(a; 0) and B(0; b) with ab \(\neq \) 0 has the equation \(\frac{x}{a}+\frac{ y}{b}=1\)

Solve problem 7.5 page 34 Math textbook 10 Connecting knowledge volume 2

(Equation of intercept of the line)

Prove that the line passing through two points A(a; 0) and B(0; b) with ab \(\neq \) 0 has the equation \(\frac{x}{a}+\frac{ y}{b}=1\)

Solution method

– Find the direction vector of the line AB => Normal vector.

=> General equation of the straight line.

Detailed explanation

Line AB has direction vector \(\overrightarrow{AB}(-a; b)\).

=> The line with the normal vector is: \(\overrightarrow{n}(b; a)\).

=> The general equation of the line is: b.(x – a) + a.(y – 0) = 0 or bx + a. y – ab = 0 (1)

Dividing both sides of (1) by ab \(\neq \) 0 we get: \(\frac{x}{a}+\frac{y}{b}=1\).

So the line passing through two points A(a; 0) and B(0; b) with ab \(\neq \) 0 has the equation \(\frac{x}{a}+\frac{y}{ b}=1\)

Solve problem 7.6 page 34 Math textbook 10 Connecting knowledge volume 2

According to Google Maps, Noi Bai airport has a latitude of 21.2^o North, longitude 105.8^o East, Da Nang airport has a latitude of 16.1^o North, longitude 108.2^o Winter. An airplane, flying from Noi Bai to Da Nang airport. At time t hours, counting from the start, the plane is at a position of latitude x^o North, longitude y^o East is calculated by the formula

\(\left\{\begin{matrix}x=21,2-\frac{153}{40}t\\ y=105.8+\frac{9}{5}t\end{matrix}\right .\)

a) How long does the flight from Hanoi to Da Nang take?

b) At 1 hour from take-off, the aircraft has flown over the 17th parallel (17 .)^o North) yet?

Solve problem 7.6 page 34 Math textbook 10 Connecting knowledge volume 2

According to Google Maps, Noi Bai airport has a latitude of 21.2^o North, longitude 105.8^o East, Da Nang airport has a latitude of 16.1^o North, longitude 108.2^o Winter. An airplane, flying from Noi Bai to Da Nang airport. At time t hours, counting from the start, the plane is at a position of latitude x^o North, longitude y^o East is calculated by the formula

\(\left\{\begin{matrix}x=21,2-\frac{153}{40}t\\ y=105.8+\frac{9}{5}t\end{matrix}\right .\)

a) How long does the flight from Hanoi to Da Nang take?

b) At 1 hour from take-off, the aircraft has flown over the 17th parallel (17 .)^o North) yet?

Solution method

a) If the plane arrives at Da Nang, then x = 16.1 and y = 108.2. Substituting into the given equation, find t

b) At 1 hour, t = 1. Substitute the given equation to find the result

Detailed explanation

a) If the plane arrives at Da Nang, then x = 16.1 and y = 108.2.

We have: \(\left\{\begin{matrix}16,1=21,2-\frac{153}{40}t\\ 108.2=105.8+\frac{9}{5}t \end{matrix}\right.\)

\(\Leftrightarrow t=\frac{4}{3}\)

So the flight from Hanoi to Da Nang takes nearly 1.33 hours.

b) At 1 hour then t = 1 substitute the equation with:

\(\left\{\begin{matrix}x=21,2-\frac{153}{40}.1=17,375\\ y=105.8+\frac{9}{5}.1=107, 6\end{matrix}\right.\)

So at 1 o’clock, the plane passed the 17th parallel.