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3 years ago

write the standard form of the line that is parallel to 2x 3y = 4 and passes through the point (1, -4).

1 answer:

3 0

Parallel lines has equal slopes.
For line, 2x + 3y = 4
3y = -2x + 4
y = -2/3 x + 4/3
i.e. slope is -2/3

Equation of a line is given by y - y1 = m(x - x1), where m is the slope and (x1, y1) is a point on the line.

y - (-4) = -2/3 (x - 1)
y + 4 = -2/3 x + 2/3
3y + 12 = -2x + 2
2x + 3y = 2 - 12

Required equation in standard form is 2x + 3y = -10

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Please help me solve this answer it all for me please

mestny [16]

Answer:

The slopes are

$m1=\dfrac{2}{5}, m2=-\dfrac{5}{2}$

Therefore, the equations are equations of Perpendicular Lines .

Step-by-step explanation:

Given:

$y=\dfrac{2}{5}\times x + 1$ ......................Equation ( 1 )

$5x+2y=-4\\\\\therefore y = \dfrac{-5}{2}\times x-2$ ..............Equation ( 2 )

To Find:

Slope of equation 1 = ?

Slope of equation 2 = ?

Solution:

On comparing with slope point form

$y=mx+c$

Where,

m = Slope

c = y-intercept

We get

Step 1.

Slope of equation 1 = m1 = $\dfrac{2}{5}$

Step 2.

Slope of equation 1 = m2 = $-\dfrac{5}{2}$

Step 3.

Product of Slopes = m1 × m2 = $\dfrac{2}{5}\times -\dfrac{5}{2}=-1$

Product of Slopes = m1 × m2 = -1

Which is the condition for Perpendicular Lines

The slopes are

$m1=\dfrac{2}{5},m2=-\dfrac{5}{2}$

Therefore, the equations are equations of Perpendicular Lines .

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3 years ago

2) Subtract and simplify. 11 7\10 − 7 3\10 =

Free_Kalibri [48]

Answer:

4 2/5

Step-by-step explanation:

11 7/10

- 7 3/10

----------------

4 4/10

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2 years ago

Help help I don’t really get this???

insens350 [35]

Answer:

Question 4: $y=\displaystyle -\frac{4}{5}x$

Question 5: $y=-5x-3$

Step-by-step explanation:

Hi there!

Linear equations are typically organized in slope-intercept form: $y=mx+b$ where m is the slope of the line and b is the y-intercept (the y-coordinate of the point where the line crosses the y-axis).

Question 4

1) Determine the slope (m)

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$ where two points that pass through the line are $(x_1,y_1)$ and $(x_2,y_2)$

In the graph, two easy-to-identify points on the line are (-5,4) and (5,-4). Plug these into the equation:

$\displaystyle m=\frac{-4-4}{5-(-5)}\\\\\displaystyle m=\frac{-4-4}{5+5}\\\\\displaystyle m=\frac{-8}{10}\\\\\displaystyle m=-\frac{4}{5}$

Therefore, the slope of the line is $\displaystyle -\frac{4}{5}$ . Plug this into $y=mx+b$ as the slope (m):

$y=\displaystyle -\frac{4}{5}x+b$

2) Determine the y-intercept (b)

On the graph, we can see that the line crosses the y-axis when y is 0. Therefore, the y-intercept (b) is 0. Plug this into $y=\displaystyle -\frac{4}{5}x+b$ :

$y=\displaystyle -\frac{4}{5}x+0\\\\y=\displaystyle -\frac{4}{5}x$

Question 5

1) Determine the slope (m)

$\displaystyle m=\frac{y_2-y_1}{x_2-x_1}$

Two easy-to-identify points are (-1,2) and (0,-3). Plug these into the equation:

$\displaystyle m=\frac{-3-2}{0-(-1)}\\\\\displaystyle m=\frac{-3-2}{0+1}\\\\\displaystyle m=\frac{-5}{1}\\\\m=-5$

Therefore, the slope is -5. Plug this into $y=mx+b$ :

$y=-5x+b$

2) Determine the y-intercept (b)

On the graph, we can see that the line crosses the y-axis at the point (0,-3). The y-coordinate of this point is -3. Therefore, the y-intercept (b) is -3. Plug this into $y=-5x+b$ :

$y=-5x+(-3)\\y=-5x-3$

I hope this helps!

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