Question

What is the equation of the line that is perpendicular to y, =, 1 over 3 x, 4. And contains the point negative 2, negative 5,?.

Answer 1

The equation of the line is, $\rm y = 3x + 1$.

Given that,

The equation of the line is,

$\rm y = \dfrac{1}{3} x+4$

And contains the point (-2, -5).

We have to determine,

What is the equation of the line that is perpendicular to the given line?

According to the question,

The equation of the line is,

$\rm = \dfrac{1}{3}x + 4$

On comparing with the standard equation of the line y = mx +c.

The slope of the line $m_1$ is 1/3.

When two lines are perpendicular the relation between these slopes is,

$\rm m_1\times m_1 = {-1}\\\\\dfrac{-1}{3} \times m_2 = -1\\\\-1 \times m_2 = -1 \times 3\\\\-m_2 = -3\\\\m_2 = 3$

And line contains the point (-2, -5).

Then,

$\rm y = mx +c \\\\-5 = 3 (-2) + c\\\\-5 = -6+c \\\\c = 6-5 \\\\c = 1$

Therefore,

The equation of the line that is perpendicular to the given line and contains the point (-2, -5) is,

$\rm y = mx +c \\\\y = 3x +1$

Hence, The required equation of the line is, $\rm y = 3x + 1$.

For more details refer to the link given below.

brainly.com/question/14388443