Relative position of two lines. Angles and distances | Grade 10

Two roads on a map either cross, run side by side, or are the same road drawn twice. A surveyor also wants to know at what angle they cross and how far a house is from the road. All of this follows from the normal vectors of the lines.

Relative position of two lines

Let $d_1: a_1x+b_1y+c_1=0$ and $d_2: a_2x+b_2y+c_2=0$ with normal vectors $\vec{n}_1=(a_1, b_1)$ and $\vec{n}_2=(a_2, b_2)$ .

Example Three cases

Decide the position of each pair. a) $x+y-3=0$ and $2x-y=0$ . b) $2x-4y+6=0$ and $x-2y-5=0$ . c) $2x-4y+6=0$ and $x-2y+3=0$ .

Solution.

a) $1\cdot(-1)-2\cdot 1=-3\neq 0$ : they intersect. From $y=2x$ : $x+2x-3=0$ , so $x=1$ , $y=2$ . The point is $(1, 2)$ .
b) $2\cdot(-2)-1\cdot(-4)=0$ , so the normals are collinear. Since $\frac{2}{1}=\frac{-4}{-2}=2\neq\frac{6}{-5}$ , the lines are parallel.
c) $\frac{2}{1}=\frac{-4}{-2}=\frac{6}{3}=2$ : the equations are proportional, so the lines coincide.

Angle between two lines

The angle $\varphi$ between two lines is between $0^\circ$ and $90^\circ$ and $\cos\varphi=\frac{|\vec{n}_1\cdot\vec{n}_2|}{|\vec{n}_1|\,|\vec{n}_2|}=\frac{|a_1a_2+b_1b_2|}{\sqrt{a_1^2+b_1^2}\sqrt{a_2^2+b_2^2}}.$ The lines are perpendicular exactly when $a_1a_2+b_1b_2=0$ .

Two lines crossing at

45^\circ

Distance from a point to a line

For two parallel lines, pick any point on one and measure to the other. With $ax+by+c_1=0$ and $ax+by+c_2=0$ the distance is $\frac{|c_1-c_2|}{\sqrt{a^2+b^2}}$ .

Heights and areas of a triangle

Triangle

ABC

Example Height and area

Triangle $ABC$ has $A(1, 1)$ , $B(5, 1)$ and $C(3, 5)$ . Find the height from $A$ and the area.

Solution. $\overrightarrow{BC}=(-2, 4)$ , so a normal vector of $BC$ is $(2, 1)$ (or $(4, 2)$ ). The line $BC$ : $2(x-5)+(y-1)=0$ , i.e. $2x+y-11=0$ . The height from $A$ is $h_A=\frac{|2+1-11|}{\sqrt5}=\frac{8}{\sqrt5}.$ Also $BC=\sqrt{4+16}=2\sqrt5$ , so the area is $\frac12\cdot 2\sqrt5\cdot\frac{8}{\sqrt5}=8$ . Check with the base $AB=4$ and the height $4$ (the $y$ -coordinate of $C$ minus $1$ ): $\frac12\cdot 4\cdot 4=8$ ✓.

Key terms

English	Tiếng Việt
relative position	vị trí tương đối
intersect	cắt nhau
coincide	trùng nhau
parallel	song song
angle between two lines	góc giữa hai đường thẳng
distance from a point to a line	khoảng cách từ một điểm đến đường thẳng
height	đường cao
area	diện tích

Summary

Compare $\vec{n}_1,\vec{n}_2$ : not collinear means the lines intersect; collinear means parallel or coincident (check $c$ ).
$\cos\varphi=\frac{|\vec{n}_1\cdot\vec{n}_2|}{|\vec{n}_1||\vec{n}_2|}$ , and perpendicular means $a_1a_2+b_1b_2=0$ .
$d(M_0, d)=\frac{|ax_0+by_0+c|}{\sqrt{a^2+b^2}}$ .
Heights and areas of triangles follow from distance and length.

Practice

Decide the position of $2x-y+1=0$ and $4x-2y+3=0$ .
Find the intersection of $3x+2y-6=0$ and $x-y+2=0$ .
Find the angle between $x+y=0$ and $x+3y-2=0$ (to $0.01^\circ$ ).
Find $m$ so that $(m+1)x+2y-3=0$ is perpendicular to $x-y+4=0$ .
Find the distance from the origin to $5x+12y-26=0$ .
Find the distance from $A(2, -1)$ to $3x-4y+5=0$ .
Find the distance between $2x-y+1=0$ and $2x-y-4=0$ .
For $A(1, 1)$ , $B(5, 4)$ , $C(2, 5)$ find the line $AB$ , the height from $C$ and the area of $ABC$ .
Challenge. Find the points of the $x$ -axis whose distance to $3x+4y-5=0$ is $3$ .
Challenge. Find the distance from $A(1, 1)$ to the line through $B(2, -1)$ with direction vector $\vec{u}=(3, 4)$ .

Show solutions

$\frac{2}{4}=\frac{-1}{-2}=\frac12\neq\frac13$ : parallel (distinct).
From $x=y-2$ : $3(y-2)+2y-6=0$ , so $5y=12$ , $y=\frac{12}{5}$ and $x=\frac25$ . The point is $\left(\frac25, \frac{12}{5}\right)$ .
$\cos\varphi=\frac{|1+3|}{\sqrt2\sqrt{10}}=\frac{4}{\sqrt{20}}=\frac{2}{\sqrt5}$ , so $\varphi\approx 26.57^\circ$ .
$(m+1)\cdot 1+2\cdot(-1)=0$ , so $m=1$ .
$\frac{|{-26}|}{\sqrt{25+144}}=\frac{26}{13}=2$ .
$\frac{|6+4+5|}{\sqrt{9+16}}=\frac{15}{5}=3$ .
$\frac{|1-(-4)|}{\sqrt{4+1}}=\frac{5}{\sqrt5}=\sqrt5$ .
$\overrightarrow{AB}=(4, 3)$ , $\vec{n}=(3, -4)$ : $3(x-1)-4(y-1)=0$ , i.e. $3x-4y+1=0$ . Height from $C$ : $\frac{|6-20+1|}{5}=\frac{13}{5}=2.6$ . $AB=5$ , so the area is $\frac12\cdot 5\cdot 2.6=6.5$ .
$M(t, 0)$ : $\frac{|3t-5|}{5}=3$ , so $3t-5=\pm15$ , giving $t=\frac{20}{3}$ or $t=-\frac{10}{3}$ . The points are $\left(\frac{20}{3}, 0\right)$ and $\left(-\frac{10}{3}, 0\right)$ .
$\vec{n}=(4, -3)$ : $4(x-2)-3(y+1)=0$ , i.e. $4x-3y-11=0$ . Distance from $A$ : $\frac{|4-3-11|}{5}=\frac{10}{5}=2$ .