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Analytic geometry on a plane

Example 6.5. Assume, there is a line 6x+8y +11 = 0 and a point M 0 (2; −1).
Let’s find the distance from the point to the line. ,

Solution. Let’s convert the given equation to the normal one. Following the formula (6.15), µ =
1 1
− √ = − (a sign minus is taken because C = 11 > 0). Having multiplied the given equation
2
6 +8 2 10
3
4
on digit µ, we will get a normal equation of the line: x + y + 11 = 0. Let’s find a sought distance using
5 5 10
4
formula (6.18): d = 3 · 2 + · (−1) + 11 = 1, 5.
5 5 10
6.7. The angle between two lines on a plane

Let’s consider two lines (l 1 ) and (l 2 ), that are given by equations with angle coefficients: y =
k 1 x + b 1 and y = k 2 x + b 2 , where k 1 = tg α 1 , k 2 = tg α 2 . Let φ be the angle between lines (l 1 )
and (l 2 ), 0 ≤ φ ≤ π (fig. 6.16). As it is known from elementary geometry, tg φ = tg(α 2 −α 1 ) =
tg α 2 −tg α 1
or
1+tg α 1 ·tg α 2
k 2 − k 1
tg φ = . (6.19)
1 + k 1 · k 2
Formula (6.19) determines one of angles between lines. The second one equals to π − φ. Con-
ditions of parallelity and perpendicularity of lines follow O from formula (6.19).

y
l 2

φ
l 1

α 2
. . . . . . . . . α 1
0 x

Figure 6.16 – Angle between two lines

1. If (l 1 ) ∥ (l 2 ), then φ = 0, tgφ = 0. In this case a numerator of the fraction in
(6.19) equals to zero. So,
k 2 = k 1 (6.20)

is the condition of parallelity of lines.
π
2. If (l 1 ) ⊥ (l 2 ), then φ = , tg φ is not defined. In this case a denominator of
2
1
the fraction in (6.19) equals to zero. So, 1 + k 1 · k 2 = 0, and k 2 = − (6.18)
k 1
is the condition of perpendicularity of lines.
Example 6.6. Assume, there are two lines: y = 2x − 1 and y = −3x + 2.
Let’s find the angle between them. ,

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