What answer exercise 7b

What is the equation of the line perpendicular to 3x+y= -8that passes through -3,1? Write your answer in slope-intercept form. S

Gekata [30.6K]

Slope intercept form of a line perpendicular to 3x + y = -8, and passing through (-3,1) is $y=\frac{1}{3} x+2$

<u>Solution:</u>

Need to write equation of line perpendicular to 3x+y = -8 and passes through the point (-3,1).

Generic slope intercept form of a line is given by y = mx + c

where m = slope of the line.

Let's first find slope intercept form of 3x + y = -8

3x + y = -8

=> y = -3x - 8

On comparing above slope intercept form of given equation with generic slope intercept form y = mx + c , we can say that for line 3x + y = -8 , slope m = -3

And as the line passing through (-3,1) and is perpendicular to 3x + y = -8, product of slopes of two line will be -1 as lies are perpendicular.

Let required slope = x

$\begin{array}{l}{=x \times-3=-1} \\\\ {=>x=\frac{-1}{-3}=\frac{1}{3}}\end{array}$

So we need to find the equation of a line whose slope is $\frac{1}{3}$ and passing through (-3,1)

Equation of line passing through $(x_1 , y_1)$ and having lope of m is given by

$\left(y-y_{1}\right)=\mathrm{m}\left(x-x_{1}\right)$

$\text { In our case } x_{1}=-3 \text { and } y_{1}=1 \text { and } \mathrm{m}=\frac{1}{3}$

Substituting the values we get,

$\begin{array}{l}{(\mathrm{y}-1)=\frac{1}{3}(\mathrm{x}-(-3))} \\\\ {=>\mathrm{y}-1=\frac{1}{3} \mathrm{x}+1} \\\\ {=>\mathrm{y}=\frac{1}{3} \mathrm{x}+2}\end{array}$

Hence the required equation of line is found using slope intercept form

4 0

3 years ago

The set of types of matter<br>well difined or not

Tresset [83]

Answer:

It is not well defined.

Step-by-step explanation:

Because set is a collection of well defined objet. And matter contains of solid, liquid and gas

3 0

3 years ago

1.

Eddi Din [679]

Answer:

a) $y=\dfrac{5}{2}x$

b) yes the two lines are perpendicular

c) $y=\dfrac{5}{4}x+6$

Step-by-step explanation:

a) All this is asking if to find a line that is perpendicular to 2x + 5y = 7 AND passes through the origin.

so first we'll find the gradient(or slope) of 2x + 5y = 7, this can be done by simply rearranging this equation to the form $y = mx + c$

$5y = 7 - 2x$

$y = \dfrac{7 - 2x}{5}$

$y = \dfrac{7}{5} - \dfrac{2}{5}x$

$y = -\dfrac{2}{5}x+\dfrac{7}{5}$

this is changed into the $y = mx + c$ , and we easily see that -2/5 is in the place of m, hence $m = \frac{-2}{5}$ is the slope of the line 2x + 5y = 7.

Now, we need to find the slope of its perpendicular. We'll use:

$m_1m_2=-1$ .

here both slopes $m_1$ and $m_2$ are slopes that are perpendicular to each other, so by plugging the value -2/5 we'll find its perpendicular!

$\dfrac{-2}{5}m_2=-1$ .

$m_2=\dfrac{5}{2}$ .

Finally, we can find the equation of the line of the perpendicular using:

$(y-y_1)=m(x-x_1)$

we know that the line passes through origin(0,0) and its slope is 5/2

$(y-0)=\dfrac{5}{2}(x-0)$

$y=\dfrac{5}{2}x$ is the equation of the the line!

b) For this we need to find the slopes of both lines and see whether their product equals -1?

mathematically, we need to see whether $m_1m_2=-1$ ?

the slopes can be easily found through rearranging both equations to $y=mx+c$

Line:1

$2x + 3y =6$

$y =\dfrac{-2x+6}{3}$

$y =\dfrac{-2}{3}x+2$

Line:2

$y = \dfrac{3}{2}x + 4$

this equation is already in the form we need.

the slopes of both equations are

$m_1 = \dfrac{-2}{3}$ and $m_2 = \dfrac{3}{2}$

using

$m_1m_2=-1$

$\dfrac{-2}{3} \times \dfrac{3}{2}=-1$

$-1=-1$

since the product does equal -1, the two lines are indeed perpendicular!

c)if two perpendicular lines have the same intercept, that also means that the two lines meet at that intercept.

we can easily find the slope of the given line, y = − 4 / 5 x + 6 to be $m=\dfrac{-4}{5}$ and the y-intercept is $c=6$ the coordinate at the y-intercept will be (0,6) since this point only lies in the y-axis.