Find the distance between y=−56x−1 and the parallel

Ответ:

kitttimothy55

29.01.2022

The distance between the two parallel lines is y = -56·x - 1 and the parallel line that passes through the point (6, -4) is 5.91 units

Step-by-step explanation:

The parameters given are;

Equation of the line y = -56·x - 1

The point the parallel line passes = (6, -4)

Given that the slope and intercept form of a line is y = m·x + c

Where:

m = The slope = (y₂ - y₁)/(x₂ - x₁)

c = y- intercept

Comparing y = m·x + c with y = -56·x - 1 gives;

m = -56, c = -1

The equation of the parallel line is presented as follows

y - (-4) = (x - 6)(-56)

y = -56x +336 - 4 = -56x + 332

The formula for the distance, d, between two parallel lines is given by the following relation;

$d = \dfrac{\left | c_1 - c_2\right |}{\sqrt{{a}^{2}+{b}^{2}} }$

Where:

c₁ and c₂ are the coordinates of the point constant y-intercept in both equations when the line equation is written as follows;

y = -56·x - 1 → y + 56·x + 1 = 0

a and b are the x and y coefficients

Which gives;

$d = \dfrac{\left | 1 - 332\right |}{\sqrt{{56}^{2}+{1}^{2}} }$ which gives $d = \dfrac{331 \cdot \sqrt{3137} }{3137} = 5.91$

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