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Trigonometry & Coordinate

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CAT 2025 Lesson : Trigonometry & Coordinate - Other Line Properties

4. Other types

4.1 Other Properties of Points & Lines

The following are certain key properties/formulae that you need to remember.

If	then
$m_1$ and $m_2$ are the slopes of $2$ parallel lines,	$m_1 = m_2$
$m_1$ and $m_2$ are the slopes of $2$ perpendicular lines,	$m_1 \times m_2 = -1$
$m_1$ and $m_2$ are the slopes of $2$ lines and θ is the angle formed between the lines	tan θ = $\lvert \dfrac{m_1 - m_2}{1 + m_1 m_2} \rvert$
the equation of a line is $ax + by + c = 0$ and another point ( $x_1, y_1$ ) lies on the plane, such that $l$ is the length of the perpendicular drawn from this point to the line	$l = \dfrac{\lvert ax_1 + by_1 + c \rvert}{\sqrt{a^2 +b^2}}$
$ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ are two parallel lines (as slopes are equal), and if $l$ is the perpendicular distance between the two lines,	$l = \dfrac{\lvert c_1 - c_2 \rvert}{\sqrt{a^2 + b^2}}$
a set of collinear points P ( $x_1, y_1$ ), Q ( $x_2, y_2$ ), R ( $x_3, y_3$ ), etc. are provided, and PQ, QR and RP are the distances between the respective points,	(i) slopes of lines connecting any two of these point will be equal; (ii) the largest of PQ, QR and RP should be equal to the sum of the other two distances.

Example 14

A and B are two lines and A passes through the points (4, 8) and (-2, -1). Find the equation of B, if

(i) A and B are parallel lines and B passes through the point (3, -2)
(ii) A and B are perpendicular lines, which are intersecting at a point on

y

- axis

Solution

(i) Since A and B are parallel lines,

Slope of A = Slope of B

Coordinates of A are (4, 8) and (-2, -1)

x_1 = 4, y_1 = 8, x_2 = -2

and

y_2 = -1

Slope of A =

\dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{(-1) - 8}{(-2) - 4} = \dfrac{-9}{-6} = \dfrac{3}{2}

Equation of the line B,

We know the slope and one coordinate of line B,

Equation =

(y - y_1) = m(x - x_1)

Slope

(m) = \dfrac{3}{2}, x_1 = 3

and

y_1 = -2

⇒

y - (-2) = \dfrac{3}{2} (x - 3)

⇒

2(y + 2) = 3x - 9

⇒

2y + 4 = 3x - 9

⇒

3x - 2y - 13 = 0

⇒

y = \dfrac{3}{2} x - \dfrac{13}{2}

(ii) Since A and B are perpendicular lines,

Slope of A

\times

Slope of B = -1

Slope of A =

\dfrac{3}{2}

, Slope of B =

\dfrac{-2}{3}

Equation of B,

Since the lines are intersecting at

y

-axis, one of the points of line B will be (

0, b

), where

b

is the

y

intercept of line A.

Equation of line B,

y = mx + b

Coordinates of A are (

4, 8

) and (

-2, -1

)

x_1 = 4, y_1 = 8, x_2 = -2

and

y_2 = -1

Equation of line A =

\dfrac{(y - y_1)}{(y_2 - y_1)} = \dfrac{(x - x_1)}{(x_2 - x_1)} = \dfrac{(y - 8)}{((-1) - 8)} = \dfrac{(x - 4)}{((-2) - 6)}

⇒

2(y - 8) = 3(x - 4)

⇒

2y - 16 = 3x - 12

⇒

y = \dfrac{3}{2}x + 2

Therefore, the equation of Line B =

y = - \dfrac{2}{3}x + 2

Answer: (i)

y = \dfrac{3}{2}x - \dfrac{13}{2}

; (ii)

y = - \dfrac{2}{3}x + 2

Example 15

What is the distance (in units) between the point (

3, 3

) and

5x - 12y = 5

Solution

Distance between a point (

3, 3

) and line

5x - 12y - 5 = 0

is

D =

\dfrac{|ax_1 + by_1 + c|}{\sqrt{a^2 + b^2}} = \dfrac{|5 \times 3 + (-12) \times 3 - 5|}{\sqrt{5^2 + (-12)^{2}}} = \dfrac{|-26|}{\sqrt{169}} = 2

Answer:

2

Example 16

What is the distance (in units) between the lines

16x + 30y = 45

and

8x + 15y + 3 = 0

Solution

The two lines can be rewritten as

8x + 15y - 22.5 = 0

and

8x + 15y + 3 = 0

. Now, we can apply the formula as the coefficients of

x

and

y

are the same in both the lines.

Given two parallel lines

ax + by + c_1 = 0

and

ax + by + c_2 = 0

, distance between the lines is

\dfrac{|c_2 - c_1|}{\sqrt{a^2 + b^2}} = \dfrac{|3 - ( - 22.5)|}{\sqrt{8^2 + 15^2}} = \dfrac{25.5}{17} = 1.5

Answer:

1.5

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