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3 years ago

Find the equation of the line.

Mathematics

1 answer:

anygoal [31]3 years ago

7 0

Answer:

y=4x + 7

Step-by-step explanation:

Line=y=ax+b

b = where the line crosses the y-axis (where x=0). Here, it's 7.

a= vertical distance ÷ horizontal distance.

You can take any pair of distances to a point on the line, for example (4,1) is easy to read in the graph.

x=0 to 1 (horizontal distance=1-0=1)

and

y= 0 to 4 (vertical distance= 4-0=4)

a=4÷1=4.

So, the answer is y=4x + 7 as a= 4 and b=7.

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Gemiola [76]

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5 0

3 years ago

Write an equation in standard form of the line passing through the points (12, 3) and (-3, 8).

aliya0001 [1]

Answer:

x + 3y = 21

Step-by-step explanation:

$(x_{1},y_{1})=(12,3) \\\\(x_{2},y_{2})=(-3,8)\\\\Slope= \dfrac{y_{2}-y_{1}}{x_{2}-x_{1}}\\\\= \dfrac{8-3}{-3-12}= \dfrac{5}{-15}\\\\= \dfrac{-1}{3}$

Equation of the line: y =mx+ b

$y = \dfrac{-1}{3}x+b\\\\\\Plugin \ x = 12 \ and \ y=3 \ in \ the \ above \ equation\\\\3= \dfrac{-1}{3}*12+b\\\\3= -4+b\\\\3+4 = b\\\\b = 7\\\\\\\y= \dfrac{-1}{3}x+7\\\\$

Multiply the equation by 3

3y = -x + 21

x + 3y = 21

6 0

3 years ago

Lesson 4.10 - Analytic Geometry

Nadusha1986 [10]

Geometric shapes are mathematical shapes that include squares and triangles

<h3>How to prove the shape is a square</h3>

The side lengths of a square are congruent, and the adjacent sides are perpendicular.

So, we start by calculating the side lengths using the following distance formula

$d = \sqrt{(x_1 -x_2)^2 + (y_1 -y_2)^2}$

Using the above formula, we have:

$AB = \sqrt{(3-2)^2 + (4+2)^2} = \sqrt{37$

$BC = \sqrt{(2+4)^2 + (-2+1)^2} = \sqrt{37$

$CD = \sqrt{(-4+3)^2 + (-1-5)^2} = \sqrt{37}$

$DA = \sqrt{(-3-3)^2 + (5-4)^2} = \sqrt{37}$

The above shows that the side lengths of the square are congruent.

Next, calculate the slope of the sides using:

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

So, we have:

$m_{AB} = \frac{4 +2}{3 -2} = 6$

$m_{BC} = \frac{-2 + 1}{2+4} = -\frac 16$

$m_{CD} = \frac{-1 -5}{-4+3} = 6$

$m_{DA} = \frac{5 -4}{-3-3} = -\frac 16$

Notice that the opposite slopes are congruent, and the adjacent slopes are opposite reciprocal.

The above highlight, and the equal side lengths show that the figure (1) is a square

<h3>How to prove the shape is a right isosceles triangle</h3>

The legs of a right isosceles triangle are congruent, and the legs are perpendicular.

So, we start by calculating the lengths of the legs using the following distance formula

$d = \sqrt{(x_1 -x_2)^2 + (y_1 -y_2)^2}$

Using the above formula, we have:

$XY = \sqrt{(5-4)^2 + (-1-4)^2} = \sqrt{26$

$XZ = \sqrt{(5-0)^2 + (-1+2)^2} = \sqrt{26$

The above shows that the legs of the right isosceles triangle are congruent.

Next, calculate the slope of the legs using:

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

So, we have:

$m_{XY} = \frac{5-4}{-1 -4} = -\frac{1}{5}$

$m_{XZ} = \frac{5-0}{-1+2} = 5$

Notice that the slopes are opposite reciprocal.

The above highlight, and the equal legs show that the figure (2) is a right isosceles triangle

Read more about geometric shapes at:

brainly.com/question/14285697

7 0

3 years ago

Find the area of the rectangle.

umka2103 [35]

The area is length x width which means it would be 14 x 8

14 x 8 = 112 so the area of the rectangle is 112 square ft.

8 0

3 years ago

Help please!!<br> .<br> .<br> .<br><br> .

rjkz [21]

Answer:

2nd Triangle:

1st line: Given, but if you were given that JK and MP were parallel, you'd use the phrase "If 2 parallel lines are cut by a transversal, alternate interior angles are congruent".

2nd line: Given. If you were given something like "JP bisects KM, or "JP and KM bisect each other", you'd write "bisectors cut lines equally in half".

3rd line: Vertical angles are congruent. Angles opposite each other (like JLK and MLP are vertical and always congruent.

4th line: SAS (or side angle side) You have an angle, a side, and an angle all in order. Therefore it's SAS.

Show me the full screenshots of the other 2 and I'll happily solve them.

3 0

3 years ago

Find the equation of the line.​

Find the equation of the line.