mmaymarie18
contestada

What is an equation of the line that passes through the point
(
5
,
−
8
)
(5,−8) and is perpendicular to the line
5
x
−
4
y
=
16
5x−4y=16?

Respuesta :

MikeMart77

Answer:

[tex]4x + 5y = - 20[/tex]

Step-by-step explanation:

1. Understand that a straight line that is perpendicular to another will have an opposite slope to the same.

2. Convert the equation to slope-intercept form, to manipulate it more easily:

[tex]5x - 4y = 16 \\ y = \frac{5}{4} x - 4[/tex]

3. Take the slope, and get the opposite. Then replace the variables with the given points.

[tex] - 8 = - \frac{4}{5} (5) - 4 \\ - 8 = - 4 - 4 = \\ - 8 = - 8[/tex]

Therefore, your finished equation would be:

[tex]y = - \frac{4}{5} x - 4[/tex]

or

[tex] 4 x + 5y = - 20[/tex]

Lanuel

An equation of the line that passes through points (5,−8) and is perpendicular to the line is equal to [tex]y =\frac{-5}{4} x+\frac{7}{4}[/tex]

Given the following data:

  • Points (x, y) = (5, -8)
  • [tex]5x-4y=16[/tex]

First of all, we would determine the slope of the given equation:

[tex]5x-4y=16\\\\4y=5x+16\\\\y=\frac{4}{5}x + 4[/tex]

Slope, [tex]m_1 = \frac{4}{5}[/tex]

In Mathematics, the slopes of two lines are said to be perpendicular when the product of these slopes is equal to negative one (-1).

Mathematically, this is given by;

[tex]m_1 \times m_2 = -1[/tex]  

Substituting the value of [tex]m_1[/tex], we have:

[tex]m_1 \times m_2 = -1\\\\\frac{4}{5} \times m_2 = -1\\\\m_2 = \frac{-5}{4}[/tex]

The standard form of an equation of line is given by the formula;

[tex]y -y_1 =m(x-x_1)[/tex]

Where:

  • x and y are the points.
  • m is the slope.

Substituting the points into the formula, we have;

[tex]y-y_1=m(x-x_1)\\\\y-[-8]=\frac{-5}{4} (x-5)\\\\y+8=\frac{-5}{4} x + \frac{25}{4} \\\\y =\frac{-5}{4} x + \frac{25}{4}-8\\\\y =\frac{-5}{4} x-\frac{-7}{4}\\\\y =\frac{-5}{4} x+\frac{7}{4}[/tex]

Read more: https://brainly.com/question/14096662

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