Find the value of x

What is the equation of a line that is perpendicular to −x+3y=9 and passes through the point (−3, 2) ?

vekshin1

The equation of line that is perpendicular to the line $-x+3y =9$ and passes through the point $\left({-3,2}\right)$ is $\boxed{{\mathbf{y=-3x-7}}}$ .

Further explanation:

It is given that the equation of line is $-x+3y =9$ and passes through point $\left({-3,2}\right)$ .

Rewrite the given equation $-x+3y =9$ as follows:

$\begin{aligned}-x+3y&=9\\3y&=9+x\\y&=\frac{9}{3}+\frac{1}{3}x\\y&=\frac{1}{3}x+3\\\end{aligned}$

Now, compare the obtained equation of line $y=\frac{1}{3}x+3$ with the standard equation of line $y=mx+b$ .

$\begin{aligned}m&=\frac{1}{3}\\b&=3\\\end{aligned}$

Therefore, the slope is $\frac{1}{3}$ .

It is given that both lines are perpendicular to each other so the product of slope must be equal to $-1$ .

${m_1}\cdot {m_2}=-1$ …… (1)

Substitute $\frac{1}{3}$ for ${m_1}$ in equation (1) to obtain the value of slope ${m_2}$ .

$\begin{aligned}\frac{1}{3}\cdot{m_2}&=-1\\{m_2}&=-3\\\end{aligned}$

Therefore, the slope is $-3$ .

It is given that the line passes through point $\left({-3,2}\right)$ .

The point-slope form of the equation of a line with slope $m$ passes through point $\left({{x_1},{y_1}}\right)$ is represented as follows:

$y-{y_1}=m\left({x-{x_1}}\right)$ …… (2)

Substitute $-3$ for ${x_1}$ , $2$ for ${y_1}$ and $-3$ for $m$ in equation (2) to obtain the equation of line.

$\begin{aligned}y-2&=-3\left({x-\left({-3}\right)}\right)\\y-2&=-3\left({x+3}\right)\\y&=-3x-9+2\\y&=-3x-7\\\end{aligned}$

Therefore, the equation of line is $y=-3x-7$ .

Thus, the equation of line that is perpendicular to the line $-x+3y=9$ and passes through the point $\left({-3,2}\right)$ is $\boxed{{\mathbf{y=-3x-7}}}$ .

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Answer Details:

Grade: Junior High School

Subject: Mathematics

Chapter: Coordinate Geometry

Keywords: Coordinate Geometry, linear equation, system of linear equations in two variables, variables, mathematics, equation of line, line, passes through point.

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A 12 foot ladder is leaning against a building. If the bottom of the ladder is sliding along the pavement directly away from the

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Using Pythagoras theorem, the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.

Let distance from the wall to the foot of the ladder is 'x' feet and the height of the top of the ladder is 'y' feet.

Pythagoras theorem, $x^{2} + y^{2} = (12)^{2}$ --->(1)

Given, $\frac{dx}{dt}= 2feet/second$ at x=3

Put x=3 in Pythagoras theorem equation (1)

$(3)^{2} + y^{2} = 144$

$y^{2} = 144 - 9$

$y^{2}$ = 135

y = 11.61

Derive equation (1) w.r.t to 't'

$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$ ---->(2)

substitute the value of 'x', 'dx/dt' and 'y' in equation (2), we get the fast of the top of the ladder moving down when the foot of the ladder is 3 feet from the wall

$2(3)(2) + 2 (11.61)\frac{dy}{dt} = 0$

12 + 23.22 $\frac{dy}{dt}$ = 0

$\frac{dy}{dt}= \frac{-12}{23.22}$

$\frac{dy}{dt} = -0.518$

Hence, using Pythagoras theorem the top of the ladder moving down when the foot of the ladder is 3 feet from the wall is of -0.518 feet/sec.

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Answer:

I'm guessing 3 or 2

Step-by-step explanation:

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Question 1:
The labeled points are points from the drawing are:
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The intersection of the two lines is a point since the necessary condition for the two lines to intersect is that they should be on the same plane, hence they form a point.
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The name for a given line is given from the end points that forms the line. Hence given that the two points forming the line is B and P, the correct answer is option C]

Question 4:
There are 5 letters and you use 2 to name the line, so the number of choices will be given by:
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Collinear points are three or more points which lie on the same single straight line. In our case the points are:
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Find the value of x ​

Find the value of x