Question

Write the standard form of the equation of the line passing through the point (1, -1) and perpendicular to the line 3x - 4y - = 20.

A. - 4x – 3y = 1
B. 3x + 4y = 1
C. 3x - 4y = 1
D. 4x + 3y = 1​

Answer 1

The standard form of the equation of the line passing through the point (1, -1) and perpendicular to the line 3x - 4y - = 20 is 3y + 4x  = 1

Perpendicular lines:

Equation of lines are perpendicular to each other if the product of their slope is equals to -1.

Therefore,

• m₁ m₂ = -1

Using slope intercept equation:

• y = mx + c

where

m = slope

c = y-intercept

Standard form equation is represented as follows:

• Ax + By = C

The line is perpendicular to 3x - 4y = 20. Lets find the slope.

3x - 20 = 4y

y = 3/4 x - 5

The slope of this equation is 3/4 .

Let's find the slope of the equation we are looking for.

3/4 m₁ = - 1

m₁ = - 4 / 3

Therefore, the equation also passes through (1, -1).

-1 = -4/3(1) + b

b = 4 / 3 - 1

b = 4 - 3/ 3

b = 1 / 3

Therefore, let's convert to standard form

y = - 4 /3 x + 1 / 3

y + 4 / 3 x = 1 / 3

multiply through by 3

The equation is 3y + 4x  = 1

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