Can someone please help me with 15 and 16

Write the equation of the line that passes through the points

sesenic [268]

Answer:

$y+5=2(x-3)$ or $y+9=2(x-1)$

Step-by-step explanation:

step 1

Find the slope of the line

The formula to calculate the slope between two points is equal to

$m=\frac{y2-y1}{x2-x1}$

we have the ordered pairs

(3,-5) and (1,-9)

substitute the values

$m=\frac{-9+5}{1-3}$

$m=\frac{-4}{-2}$

$m=2$

step 2

Find the equation in point slope form

$y-y1=m(x-x1)$

Analyze two cases

First case

we have

$m=2$

$point\ (3,-5)$

substitute

$y+5=2(x-3)$

Second case

we have

$m=2$

$point\ (1,-9)$

substitute

$y+9=2(x-1)$

Note: The equation of the line in point slope form varies according to the point you choose, in contrast to the slope-intercept form

6 0

3 years ago

What is the perimeter of △ABC?

timofeeve [1]

<h3>Given</h3>

ΔABC
A(-3, -1), B(0, 3), C(1, 2)

the length of the perimeter of ΔABC to the nearest tenth

<h3>Solution</h3>

The perimeter of a triangle is the sum of the lengths of its sides. The length of each side can be found using the Pythagorean theorem. Effectively, each pair of points is treated as the end-points of the hypotenuse of a right triangle with legs parallel to the x- and y-axes. The leg lengths are the differences betweeen the x- and y- coordinates of the points.

The difference of the x-coordinates of segment AB are 0-(-3) = 3. The y-coordinate difference is 3-(-1) = 4. So, the leg lengths of the right triangle whose hypotenuse is segment AB are 3 and 4. The Pythagorean theorem tells us

... AB² = 3² +4² = 9 +16 = 25

... AB = √25 = 5

You may recognize this as the 3-4-5 triangle often introduced as one of the first ones you play with when you learn the Pythagorean theorem.

LIkewise, segment AC has coordinate differences of ...

... C - A = (1, 2) -(-3, -1) = (4, 3)

These are the same leg lengths (in the other order) as for segment AB, so its length is also 5.

Segment BC has coordinate differences ...

... C - B = (1, 2) -(0, 3) = (1, -1)

The length of the line segment is figured as the root of the sum of squares, even though one of the coordinate differences is negative. The leg lengths of the right triangle used for finding the length of BC are the absolute value of these differences, or 1 and 1. Then the length BC is

... BC = √(1² +1²) = √2 ≈ 1.4

So the perimeter of the triangle ABC is

... AB + BC + AC = 5 + 1.4 + 5 = 11.4 . . . . perimeter of ∆ABC in units

_____

Please be aware that the advice to "round each step" is bad advice, in general. For real-world math problems, you only round the final result. You always carry at least enough precision in the numbers to ensure that there will not be any error in the final rounding.

In this problem, the only number that is not an integer is √2, so it doesn't really matter.