All categories

3 years ago

Find slope: (6,-9) and (-5,-9)

Mathematics

2 answers:

Orlov [11]3 years ago

4 0

Y= (-9+9)/
(-5-6)
so
0 over negative 11.
y is equal to -9
there is no slope.

horsena [70]3 years ago

4 0

Slope Formula: y2-y1 / x2-x1

I am assuming that (6,-9) is our (x1,y1) and (-5,-9) is our (x2,y2)

So (-9)-(-9) / (-5)-(6) = 0/-11 = 0

Hope this helped.

You might be interested in

Let $A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$. There exists a point $X$ and a constant $k$ such that for any point $P$,

goldfiish [28.3K]

Answer:

k=32

Step-by-step explanation:

Given the points:

$A = (4,-1)$, $B = (6,2)$, and $C = (-1,2)$.$

The first step is to find the Centroid of the triangle.

Centroid, X

$=\left(\dfrac{x_1+x_2+x_3}{3} ,\dfrac{y_1+y_2+y_3}{3} \right)\\=\left(\dfrac{4+6+(-1)}{3} ,\dfrac{-1+2+2}{3} \right)\\=\left(\dfrac{9}{3} ,\dfrac{3}{3} \right)=(3,1)$

Next, let P be a point (x,y)

Using the distance formula, $\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

$PA^2=(x-4)^2+(y-(-1))^2\\PB^2=(x-6)^2+(y-2)^2\\PC^2=(x-(-1))^2+(y-2)^2\\PX^2=(x-3)^2+(y-1)^2\\$

On Substitution into: $PA^2 + PB^2 + PC^2 = 3PX^2 + k$

$(x-4)^2+(y-(-1))^2+(x-6)^2+(y-2)^2+(x-(-1))^2+(y-2)^2=3[(x-3)^2+(y-1)^2]+k$

Let us simplify the LHS first

$\\LHS: x^2-8x+16+y^2+2y+1+x^2-12x+36+y^2\\-4y+4+x^2+2x+1+y^2-4y+4\\=3x^2-18x+3y^2-6y+62$

Also, the Right Hand Side

$RHS:3[(x-3)^2+(y-1)^2]+k\\=3[x^2-6x+9+y^2-2y+1]+k\\=3x^2-18x+27+3y^2-6y+3+k\\=3x^2+3y^2-18x-6y+30+k$

Therefore:

$3x^2-18x+3y^2-6y+62=3x^2+3y^2-18x-6y+30+k\\k=3x^2-18x+3y^2-6y+62-3x^2-3y^2+18x+6y-30\\k=3x^2-3x^2+3y^2-3y^2-18x+18x+62-30\\k=32$

7 0

3 years ago

Which angles are vertical to

ArbitrLikvidat [17]

GAH is vertical to FAE

4 0

3 years ago

Read 2 more answers

2a. A bicycle is $85. The tax is 6%. How much is the tax? 7

iVinArrow [24]

Answer = $5.10
85 x 0.06 = $5.10

8 0

3 years ago

Find the value of x. Please help me!!!!!!

Katen [24]

Answer:

I think this will help you.

Step-by-step explanation:

hi This will help you to do this

answer is 5

3 0

3 years ago

A triangle has two sides of length 1 and 2. What compound inequality describes the possible lengths for the third side, x?Write

klemol [59]

Step-by-step explanation:

since in a triangle each side must be shorter than the sum of the other 2 sides (otherwise the end points cannot connect, and there is no triangle), the necessary inequality condition must be

side < 1 + 2 = 3

so,

side < 3

for a lower limit let's go through the cases

1 < 2 + side (is always true)

2 < 1 + side

1 < side (side must be larger than 1)

and again

side < 1 + 2 = 3

side < 3

so the full restriction for the third side is

1 < side < 3

7 0

3 years ago