Given the midpoint (1.5,1.5) and the endpoint (5,7) where is the other endpoint located

Mathematics

1 answer:

earnstyle [38]3 years ago

8 0

The formula of a midpoint:

$M_{AB}\left(\dfrac{x_A+x_B}{2},\ \dfrac{y_A+y_B}{2}\right)$

We have:

$M(1.5,\ 1.5)\to x_M=1.5,\ y_M=1.5\\A(5,\ 7)\to x_A=5,\ y_A=7$

Substitute

$\dfrac{5+x_B}{2}=1.5\qquad|\cdot2\\\\5+x_B=3\qquad|-5\\\\x_B=-2\\\\\dfrac{7+y_B}{2}=1.5\qquad|\cdot2\\\\7+y_B=3\qquad|-7\\\\y_B=-4$

<h3>Answer: (-2, -4)</h3>

You might be interested in

Help I’ll mark you if your tight

vovikov84 [41]

Answer: 2
Explanation:

7 0

3 years ago

Solve the differential equations dy/dx=((xy^2)+x)/y

arsen [322]

$\displaystyle \dfrac{dy}{dx}=\dfrac{xy^2+x}{y}\\ \dfrac{dy}{dx}=\dfrac{x(y^2+1)}{y}\\ \dfrac{y}{y^2+1}\, dy=x\, dx\\ \int \dfrac{y}{y^2+1}\, dy=\int x\, dx\\ \dfrac{\ln (y^2+1)}{2}=\dfrac{x^2}{2}+C\\ \ln(y^2+1)=x^2+C\\ y^2+1=e^{x^2+C}\\ y^2=e^{x^2+C}-1\\ y=\sqrt{e^{x^2+C}-1} \vee y=-\sqrt{e^{x^2+C}-1}\\ \boxed{y=\sqrt{Ce^{x^2}-1} \vee y=-\sqrt{Ce^{x^2}-1}}\\$

7 0

3 years ago

PLEASE HELP ITS URGENT!

Rudik [331]

Answer:

Step-by-step explanation:

The two equations are similar, because if you multiply the first one by 3 it will be equal to the second one, so when that happens we say that there are infinite solutions.

8 0

3 years ago

Please help they’re 2 questions

IgorLugansk [536]

i think it’s is a for the first one

4 0

4 years ago

What is the image of (12,-9) after a dilation by a scale factor of 1/3 centered at the origin

vladimir2022 [97]

Answer:

(4, -3)

Step-by-step explanation:

You just divide each coordinate by 3.

7 0

3 years ago