Line on a Plane, Various Equations.

There are such forms of writing the equation of a line:

1) $y=kx+b,$ where $k$ is the slope coefficient, and $b$ is the segment the line cuts off on the $OY$ axis.

2) $y-y_0=k(x-x_0)$ - the equation of the line passing through a given point $P(x_0,y_0)$ at a given angle $\alpha$ to the $OX$ axis ($k=\tan \alpha$).

3) $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$ - the equation of the line passing through two points $M(x_1,y_1)$ and $N(x_2,y_2)$.

4) $\frac{x}{a}+\frac{y}{b}=1$ - the line equation in intercept form on the axes, where $a$ and $b$ are the lengths of the segments that the line cuts off on the coordinate axes.

5) $\frac{x-x_0}{l}=\frac{y-y_0}{m}$ - the canonical equation of the line, where $\bar{S}=(l,m)$ is the direction vector of the line, i.e., a vector parallel to the line ($\bar{S}\parallel L$), with point $P(x_0,y_0)\in L$.

6) $A(x-x_0)+B(y-y_0)=0$ - the equation of the line $L$, passing through the point $M(x_0,y_0)$ perpendicular to the vector $\bar{N}=(A,B)$. The vector $\bar{N}$ is called the normal vector of the line.

$Ax+By+C=0$ - the general equation of the line $L$, where $\bar{N}=(A,B)$ is the normal vector of the line $L$.

$x\cos \alpha +y\cos \beta -p=0$ - the normal form of the line equation, where $\cos \alpha$ and $\cos \beta$ are the directional cosines of the normal vector $n$, directed from the origin towards the line, and $p>0$ is the distance from the origin to the line.

The general equation of the line is converted to the normal form by multiplying by the normalizing factor $\mu =-\frac{\text{sgn }C}{\sqrt{A^2+B^2}}.$

The distance from the point $P(x_0,y_0)$ to the line $L:Ax+By+C=0$ is calculated using the formula$$d=\left|\frac{A{x}_0+B{y}_0+C}{\sqrt{A^2+B^2}}\right|.$$

The positioning of two lines on the plane.

Conditions for the parallelism of two lines:

1) Let $L_1:k_{1}x+b_1,$ $k_1=\tan {\alpha }_1;$

$L_2:k_{2}x+b_2,$ $k_2=\tan {\alpha }_2.$

Lines $L_1$ and $L_2$ are parallel if and only if $k_1=k_2.$

2) Let $L_1:\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1},$ ${\bar{S}}_1=(l_1,m_1);$

$L_2:\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2},$ ${\bar{S}}_2=(l_2,m_2).$

Lines $L_1$ and $L_2$ are parallel if and only if ${\bar{S}}_1\parallel {\bar{S}}_2\iff \frac{l_1}{l_2}=\frac{m_1}{m_2}.$

3)Let $L_1:A_{1}x+B_{1}y+C_1=0,$ ${\bar{N}}_1=(A_1,B_1);$

$L_2:A_{2}x+B_{2}y+C_2=0,$ ${\bar{N}}_2=(A_2,B_2).$

Lines $L_1$ and $L_2$ are parallel if and only if ${\bar{N}}_1\parallel {\bar{N}}_2\iff \frac{A_1}{A_2}=\frac{B_1}{B_2}.$

Conditions for the perpendicularity of two lines:

1) Let $L_1:k_{1}x+b_1,$ $k_1=\tan {\alpha }_1;$

$L_2:k_{2}x+b_2,$ $k_2=\tan {\alpha }_2.$

$L_1\perp L_2\iff k_1\cdot k_2=-1.$

2) Let $L_1:\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1},$ ${\bar{S}}_1=(l_1,m_1);$

$L_2:\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2},$ ${\bar{S}}_2=(l_2,m_2).$

$L_1\perp L_2\iff {\bar{S}}_1\perp {\bar{S}}_2\iff l_1\cdot l_2+m_1\cdot m_2=0.$

3) Let $L_1:A_{1}x+B_{1}y+C_1=0,$ ${\bar{N}}_1=(A_1,B_1);$

$L_2:A_{2}x+B_{2}y+C_2=0,$ ${\bar{N}}_2=(A_2,B_2).$

$L_1\perp L_2\iff {\bar{N}}_1\perp {\bar{N}}_2\iff A_1\cdot A_2+B_1\cdot B_2=0.$

The angle between two lines:

1) Let $L_1:k_{1}x+b_1,$ $k_1=\tan {\alpha }_1;$

$L_2:k_{2}x+b_2,$ $k_2=\tan {\alpha }_2.$

$\tan (\widehat{(L_1,L_2)})=\frac{k_1-k_2}{1+k_1\cdot k_2}.$

2) Let $L_1:\frac{x-x_1}{l_1}=\frac{y-y_1}{m_1},$ ${\bar{S}}_1=(l_1,m_1);$

$L_2:\frac{x-x_2}{l_2}=\frac{y-y_2}{m_2},$ ${\bar{S}}_2=(l_2,m_2).$

$\cos (\widehat{(L_1,L_2)})=\frac{l_1\cdot l_2+m_1\cdot m_2}{\sqrt{l_1^2+m_1^2}\cdot \sqrt{l_2^2+m_2^2}}.$

3) Let $L_1:A_{1}x+B_{1}y+C_1=0,$ ${\bar{N}}_1=(A_1,B_1);$

$L_2:A_{2}x+B_{2}y+C_2=0,$ ${\bar{N}}_2=(A_2,B_2).$

$\cos (\widehat{(L_1,L_2)})=\frac{A_1\cdot A_2+B_1\cdot B_2}{\sqrt{A_1^2+B_1^2}\cdot \sqrt{A_2^2+B_2^2}}.$

Examples:

2.141.

Solution:

a) The line $L$ is defined by the point $M_0(-1;2)\in L$ and the normal vector $\bar{N}(2;2)$. Required: 1) Write the equation of the line, bring it to general form, and construct the line; 2) Convert the general equation to the normal form and indicate the distance from the origin to the line.

By substituting into formula 6) for the line equation ($A(x-x_0)+B(y-y_0)=0$) the coordinates of the point $(x_0;y_0)=M_0(-1;2)$ and the vector $(A;B)=\bar{N}(2;2)$, we get:

$2(x+1)+2(y-2)=0.$ Now, let's bring this equation to the general form:

$2x+2+2y-4=0\Rightarrow$

$2x+2y-2=0\Rightarrow$

$x+y-1=0.$

The normal form of the line equation looks like $x\cos \alpha +y\cos \beta -p=0$, where $\cos \alpha$ and $\cos \beta$ are the directional cosines of the normal vector $n$, directed from the origin towards the line, and $p>0$ is the distance from the origin to the line.

The general equation of the line is converted to the normal form by multiplying by the normalizing factor $\mu =-\frac{\text{sgn }C}{\sqrt{A^2+B^2}}.$

For our line, we have $A=1;B=1;C=-1\Rightarrow \text{sgn }C=-1.$ Thus, $\mu =-\frac{-1}{\sqrt{1+1}}=\frac{1}{\sqrt{2}}.$

Write the normal equation of the line:

$x+y-1=0\cdot \frac{1}{\sqrt{2}}\Rightarrow$

$\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y-\frac{1}{\sqrt{2}}=0.$

The distance from the origin is $p=\frac{1}{\sqrt{2}}.$

Answer: $2(x+1)+2(y-2)=0;$ general equation $x+y-1=0;$ normal equation of the line $\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y-\frac{1}{\sqrt{2}}=0;$ $p=\frac{1}{\sqrt{2}}.$

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2.142

a) The line $L$ is defined by the point $M_0(-1;2)\in L$ and the direction vector $\bar{S}(3;-1)$. Required: 1) Write the equation of the line, bring it to the general form, and construct the line; 2) Convert the general equation to the normal form and indicate the distance from the origin to the line.

Solution:

Substitute into formula 5) for line equations ($\frac{x-x_0}{l}=\frac{y-y_0}{m}$) the coordinates of the point $(x_0;y_0)=M_0(-1;2)$ and the vector $(l;m)=\bar{S}(3;-1):$ $\frac{x+1}{3}=\frac{y-2}{-1}$

Now, let's bring this equation to the general form:

$-1(x+1)=3(y-2)\Rightarrow$

$-x-1-3y+6=0\Rightarrow$

$x+3y-5=0.$

The general equation of the line is converted to the normal form by multiplying by the normalizing factor $\mu =-\frac{\text{sgn }C}{\sqrt{A^2+B^2}}.$

For our line, we have $A=1;B=3;C=-5\Rightarrow \text{sgn }C=-1.$ Thus, $\mu =-\frac{-1}{\sqrt{1+9}}=\frac{1}{\sqrt{10}}.$

Write the normal equation of the line:

$x+3y-5=0|\cdot \frac{1}{\sqrt{10}}\Rightarrow$

$\frac{1}{\sqrt{10}}x+\frac{3}{\sqrt{10}}y-\frac{5}{\sqrt{10}}=0.$

The distance from the origin is $p=\frac{5}{\sqrt{10}}.$

Answer: $\frac{x+1}{3}=\frac{y-2}{-1};$ general equation $x+3y-5=0;$ normal equation of the line $\frac{1}{\sqrt{10}}x+\frac{3}{\sqrt{10}}y-\frac{5}{\sqrt{10}}=0;$ $p=\frac{5}{\sqrt{10}}.$

2.143.

a) The line $L$ is given by its two points $M_1(1;2)\in L$ and $M_2(-1;0)\in L$. Required: 1) Write the equation of the line, bring it to the general form and construct the line; 2) Convert the general equation to the normal form and indicate the distance from the origin to the line.

Solution.

Substitute into formula 3) for the line equations ($\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$) the coordinates of the points $M_1(1;2)=(x_1;y_1)$ and $M_2(-1;0)=(x_2;y_2):$ $\frac{x-1}{-1-1}=\frac{y-2}{0-2}\Rightarrow \frac{x-1}{-2}=\frac{y-2}{-2}.$

Next, let's bring this equation to the general form:

$-2(x-1)=-2(y-2)\Rightarrow$

$x-1=y-2\Rightarrow$

$x-y+1=0.$

The general equation of the line is converted to the normal form by multiplying by the normalizing factor $\mu =-\frac{\text{sgn }C}{\sqrt{A^2+B^2}}.$

For our line, we have $A=1;B=-1;C=1\Rightarrow \text{sgn }C=1.$ Thus, $\mu =-\frac{1}{\sqrt{1+1}}=-\frac{1}{\sqrt{2}}.$

Write the normal equation of the line:

$x-y+1=0|\cdot -\frac{1}{\sqrt{2}}\Rightarrow$

$-\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y-\frac{1}{\sqrt{2}}=0.$

The distance from the origin is $p=\frac{1}{\sqrt{2}}.$

Answer: $\frac{x-1}{-2}=\frac{y-2}{-2};$ general equation $x-y+1=0;$ normal equation of the line $-\frac{1}{\sqrt{2}}x+\frac{1}{\sqrt{2}}y-\frac{1}{\sqrt{2}}=0;$ $p=\frac{1}{\sqrt{2}}.$

2.150.

For triangle $ABC$ given by the vertices $A(1;2),B(2;-2),C(6;1)$:

Find the equation of side $AB;$

Find the equation of altitude $CD$ and calculate its length $h=|CD|;$

Find the angle between the altitude $CD$ and the median $BM.$

Solution:

Let's create a diagram:

The equation of the line $AB$ can be found using the formula for a line passing through two points $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}.$

In our case, we have $(x_1;y_1)=A(1;2)$ and $(x_2;y_2)=B(2;-2)$.

We substitute the points' coordinates into the line equation. We obtain $$\frac{x-1}{2-1}=\frac{y-2}{-2-2}\Rightarrow x-1=\frac{y-2}{-4}.$$ Now, let's write down the general equation of the line $AB$:

$-4(x-1)=y-2\Rightarrow -4x+4=y-2\Rightarrow 4x+y-6=0.$

2) The equation of line $CD$ can be found using equation (6): $A(x-x_0)+B(y-y_0)=0$ - the equation of the line $L$ that passes through point $M(x_0,y_0)$ perpendicular to vector $\bar{N}=(A,B)$.

In our case, the altitude $CD$ is the line that passes through point $C$ perpendicular to the vector $AB$.

Thus, we have, $$(x_0;y_0)=C=(6;1);\bar{N}=\overline{AB}=(2-1;-2-2)=(1;-4).$$

We substitute these coordinates into the line equation:

$1(x-6)-4(y-1)=0\Rightarrow x-6-4y+4=0\Rightarrow x-4y-2=0.$

Therefore, the equation of the line $CD:$ $x-4y-2=0.$

To find the length of altitude $h=|CD|$, we need to find the coordinates of point $D$, the intersection of lines $CD$ and $AB$:

$\{\begin{array}{l}x-4y-2=0\\ 4x+y-6=0.\end{array}$

Solve the system of equations using the method of elimination:

$\{\begin{array}{l}x-4y-2=0\\ 4x+y-6=0.\end{array}\Rightarrow$ $\{\begin{array}{l}x=4y+2\\ 4x+y-6=0.\end{array}\Rightarrow$

$\{\begin{array}{l}x=4y+2\\ 4(4y+2)+y-6=0.\end{array}\Rightarrow$ $\{\begin{array}{l}x=4y+2\\ 16y+8+y-6=0.\end{array}\Rightarrow$

$\{\begin{array}{l}x=-8/17+2=26/17\\ y=-2/17.\end{array}.$

Therefore, we have $D(\frac{26}{17},-\frac{2}{17})$. Now we can find the length of the altitude $CD$:

$h=|CD|=\sqrt{(x_d-x_c{)}^2+(y_d-y_c{)}^2}=\sqrt{(26/17-6{)}^2+(-2/17-1{)}^2}=$ $\sqrt{{\left(\frac{26-102}{17}\right)}^2+{\left(\frac{-2-17}{17}\right)}^2}=\sqrt{\frac{{76}^2+{19}^2}{{17}^2}}=\sqrt{\frac{6137}{{17}^2}}=\frac{19}{17}\sqrt{17}=\frac{19}{\sqrt{17}}.$

3) The equation of the altitude $CD$ has already been found in step 2). Let's find the equation of the median $BM$. We will find it using the formula for the equation of a line passing through two points.

The coordinates of point $B=(2,-2);$ the coordinates of point $M$ can be found as the midpoint of side $AC:$ $x_M=\frac{x_A+x_C}{2};y_M=\frac{y_A+y_C}{2}.$

$x_M=\frac{1+6}{2}=3.5;$ $y_M=\frac{2+1}{2}=1.5.$

Substitute the coordinates of points $B(2;-2)$ and $M(3.5;1.5)$ into the equation of the line.

$\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}:$

$\frac{x-2}{3.5-2}=\frac{y-(-2)}{1.5-(-2)}\Rightarrow \frac{x-2}{1.5}=\frac{y+2}{3.5}\Rightarrow$

$3.5(x-2)=1.5(y+2)\Rightarrow 3.5x-7=1.5y+3\Rightarrow 3.5x-1.5y-10=0.$

Next, knowing the general equations of the two lines $CD:x-4y-2=0$ and $BM:3.5x-1.5y-10=0$, we can find the angle between them using the formula

$\cos \widehat{(L_1,L_2)}=$ $\frac{A_1\cdot A_2+B_1\cdot B_2}{\sqrt{A_1^2+B_1^2}\cdot \sqrt{A_2^2+B_2^2}},$

where $L_1:A_{1}x+B_{1}y+C_1=0,$ ${\bar{N}}_1=(A_1,B_1);$

$L_2:A_{2}x+B_{2}y+C_2=0,$ ${\bar{N}}_2=(A_2,B_2).$

For our lines, we have: $(A_1,B_1)=(1,-4)$ and $(A_2,B_2)=(3.5,-1.5)$.

From here,

$\cos \widehat{(CD,BM)}=$ $\frac{1\cdot 3.5+(-4)\cdot (-1.5)}{\sqrt{1^2+(-4{)}^2}\cdot \sqrt{{3.5}^2+(-1.5{)}^2}}=\frac{9.5}{\sqrt{17}\cdot \sqrt{14.5}}=\frac{19}{\sqrt{986}}.$

Answer: 1) $AB:4x+y-6=0.$

2) $CD:$ $x-4y-2=0;$ $h=|CD|=\frac{19}{\sqrt{17}};$

3) $\cos \widehat{(CD,BM)}=$ $\frac{19}{\sqrt{986}}.$

2.160. In the isosceles triangle $ABC$, the vertex $C$ is given as $(4,3)$, the equation of the base $AC$ is $2x-y-5=0$, and the equation of the side $AB$ is $x-y=0$. Find the equation of the side $BC$.

Solution.

Let's find the coordinates of the vertex $A$ as the intersection point of the lines $AB$ and $AC$:

$$\{\begin{array}{l}x-y=0\\ 2x-y-5=0\end{array}\Rightarrow \{\begin{array}{l}x=y\\ 2y-y-5=0\end{array}\Rightarrow \{\begin{array}{l}x=y\\ y=5\end{array}\Rightarrow \{\begin{array}{l}x=5\\ y=5\end{array}$$

Thus, we have the coordinates of the vertices at the base of the isosceles triangle as $A(5,5)$ and $C(4,3)$. Let's find the coordinates of the vertex $B(x,y)$. We know that this point belongs to the line $AB:x-y=0$ and that $AB=BC$. Let's write down the formulas for the lengths of the sides $AB$ and $BC$:

$|AB|=\sqrt{(x-5{)}^2+(y-5{)}^2};$

$|BC|=\sqrt{(x-4{)}^2+(y-3{)}^2}.$

Next, to find the coordinates of point $B$, let's solve the system of equations:

$$\{\begin{array}{l}x-y=0\\ \sqrt{(x-5{)}^2+(y-5{)}^2}=\sqrt{(x-4{)}^2+(y-3{)}^2}\end{array}\Rightarrow$$ $$\{\begin{array}{l}x-y=0\\ x^2-10x+25+y^2-10y+25=x^2-8x+16+y^2-6y+9\end{array}\Rightarrow$$

$$\{\begin{array}{l}x-y=0\\ -2x-4y+25=0\end{array}\Rightarrow \{\begin{array}{l}x=y\\ -2y-4y+25=0\end{array}\Rightarrow$$

$$\Rightarrow \{\begin{array}{l}x=y\\ y=\frac{25}{6}.\end{array}$$ We found the coordinates of point$B(\frac{25}{6},\frac{25}{6}).$

Knowing the coordinates of points $B$ and $C$, we can write the equation of line $BC$ as a line passing through two points $(\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}):$

$$\frac{x-4}{\frac{25}{6}-4}=\frac{y-3}{\frac{25}{6}-3}\Rightarrow \frac{x-4}{\frac{25-24}{6}}=\frac{y-3}{\frac{25-18}{6}}\Rightarrow$$

$$\Rightarrow \frac{x-4}{1}=\frac{y-3}{7}\Rightarrow 7x-28=y-3\Rightarrow 7x-y-25=0.$$

Answer: $7x-y-25=0.$

2.165. Given two opposite vertices of a square $A(1;3)$ and $C(-1;1)$. Find the coordinates of the other two vertices and write down the equations of its sides.

Solution:

Let's find the equation of the diagonal $AC$:

$\frac{x-1}{-1-1}=\frac{y-3}{1-3}\Rightarrow \frac{x-1}{-2}=\frac{y-3}{-2}\Rightarrow x-y+2=0.$

Let's find its midpoint: $$x_O=\frac{x_A+x_C}{2};y_O=\frac{y_A+y_C}{2}.$$

$$x_O=\frac{1-1}{2}=0;y_O=\frac{3+1}{2}=2\Rightarrow O=(0;2).$$

Next, let's find the equation of the second diagonal of the square - the line passing through point $O$ perpendicular to the line $AC.$ For the line $AC$, the normal vector has coordinates $\bar{N}=(1;-1).$ The line perpendicular to the line $AC$ is parallel to the normal vector $\bar{N}$. Therefore, the equation of the line $BD$ is written using formula 5) $(\frac{x-x_0}{l}=\frac{y-y_0}{m}),$ where $(x_0,y_0)=O(0;2),$ $(l,m)=\bar{N}=(1,-1):$

$$\frac{x}{1}=\frac{y-2}{-1}\Rightarrow x=-y+2\Rightarrow x+y-2=0.$$

It is clear that $AO=CO=BO=DO.$ Let's find the length of segment $AO:$ $AO=\sqrt{(0-1{)}^2+(2-3{)}^2}=\sqrt{2}.$

Next, we will look for the coordinates of points $B$ and $D$ that lie on the line $BD$ and satisfy $BO=DO=AO.$

$$\{\begin{array}{l}x+y-2=0\\ \sqrt{(0-x{)}^2+(2-y{)}^2}=\sqrt{2}\end{array}\Rightarrow$$

$$\{\begin{array}{l}x=2-y\\ \sqrt{(y-2{)}^2+(2-y{)}^2}=\sqrt{2}\end{array}\Rightarrow$$

$$\{\begin{array}{l}x=2-y\\ |2-y|\sqrt{2}=\sqrt{2}\end{array}\Rightarrow \{\begin{array}{l}x=2-y\\ |2-y|=1\end{array}\Rightarrow \{\begin{array}{l}x=2-y\\ [\begin{array}{l}2-y_1=1\\ 2-y_2=-1\end{array}\end{array}\Rightarrow$$

$\Rightarrow \{\begin{array}{l}x=2-y\\ [\begin{array}{l}{y}_1=1\\ y_2=3\end{array}\end{array}\Rightarrow$ $\{\begin{array}{l}[\begin{array}{l}{x}_1=1\\ x_2=-1\end{array}\\ [\begin{array}{l}{y}_1=1\\ y_2=3\end{array}\end{array}$

Thus, we have found the coordinates of vertices $B(1;1)$ and $D(-1;3).$ Knowing the coordinates of the square's vertices, let's write the equations of its sides using formula (3) - $\frac{x-x_1}{x_2-x_1}=\frac{y-y_1}{y_2-y_1}$ - the equation of a line passing through two points $M(x_1,y_1)$ and $N(x_2,y_2).$

$A(1;3),$ $B(1;1),$ $C(-1;1),$ $D(-1;3).$

$AB:$ $\frac{x-1}{1-1}=\frac{y-3}{1-3}\Rightarrow \frac{x-1}{0}=\frac{y-3}{-2}\Rightarrow$ $-2(x-1)=0\Rightarrow x=1.$

$BC:$ $\frac{x-1}{-1-1}=\frac{y-1}{1-1}\Rightarrow \frac{x-1}{-1}=\frac{y-1}{0}\Rightarrow$ $0(x-1)=-1(y-1)\Rightarrow y=1.$

$CD:$ $\frac{x+1}{-1+1}=\frac{y-1}{3-1}\Rightarrow \frac{x+1}{0}=\frac{y-1}{2}\Rightarrow$ $2(x+1)=0(y-1)\Rightarrow x=-1.$

$DA:$ $\frac{x-1}{-1-1}=\frac{y-3}{3-3}\Rightarrow \frac{x-1}{-2}=\frac{y-3}{0}\Rightarrow$ $0(x-1)=-2(y-3)\Rightarrow y=3.$

Answer:$A(1;3),$ $B(1;1),$ $C(-1;1),$ $D(-1;3);$ $AB:$ $x=1;$ $BC:$ $y=1;$ $CD:$ $x=-1;$ $DA:$ $y=3.$

Tags: Line on a Plane, Various Equations, algebraic lines, geometry