CAT 2025 Lesson : Trigonometry & Coordinate - Other Line Properties

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

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: $2$

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: $1.5$