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xeze [42]
1 year ago
8

The midpoint of AB is M(1, 1). If the coordinates of A are (-2,5), what are the

Mathematics
1 answer:
Ronch [10]1 year ago
3 0

The coordinates of B are (4 , -3) .

A cartesian coordinate system, which uses signed distances between two fixed perpendicular oriented lines and the point measured in the same unit of length, uniquely identifies any point in a plane by a pair of numerical coordinates.

  • The origin of each reference coordinate line, known as an axis of the system or simply an axis (plural axes), is the intersection of the ordered pairs (0, 0).
  • The coordinates can also be determined by looking at the locations of the perpendicular projections of the point onto the two axes, which are shown as signed distances from the origin.

Let the coordinates of B be (x, y)

As M is the mid point of AB , we know from the distance formula that the coordinates of M will be:

abscissae of M = (-2 + x) /2

or, 1 =   (-2 + x) /2

or, x = 4

Ordinate of M = (5 + y)/2

or , 1 = (5 + y)/2

or, y = -3

Therefore the coordinates of B are (4 , -3)

To learn more about coordinates visit:

brainly.com/question/27749090

#SPJ1

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Find dy/dx if y =x^3+5x+2/x²-1
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<u>Differentiate using the Quotient Rule</u> –

\qquad\pink{\twoheadrightarrow \sf \dfrac{d}{dx} \bigg[\dfrac{f(x)}{g(x)} \bigg]= \dfrac{ g(x)\:\dfrac{d}{dx}\bigg[f(x)\bigg] -f(x)\dfrac{d}{dx}\:\bigg[g(x)\bigg]}{g(x)^2}}\\

According to the given question, we have –

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Let's solve it!

\qquad\green{\twoheadrightarrow \bf \dfrac{d}{dx}\bigg[ \dfrac{x^3+5x+2 }{x^2-1}\bigg]} \\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1) \dfrac{d}{dx}(x^3+5x+2) - ( x^3+5x+2)  \dfrac{d}{dx}(x^2-1)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{(x^2-1)(3x^2+5)  -  ( x^3+5x+2) 2x}{(x^2-1)^2 }\\

\qquad\pink{\sf \because \dfrac{d}{dx} x^n = nx^{n-1} }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-(2x^4+10x^2+4x)}{(x^2-1)^2 }\\

\qquad\twoheadrightarrow \sf \dfrac{3x^4+5x^2-3x^2-5-2x^4-10x^2-4x}{(x^2-1)^2 }\\

\qquad\green{\twoheadrightarrow \bf \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}\\

\qquad\pink{\therefore  \bf{\green{\underline{\underline{\dfrac{d}{dx} \dfrac{x^3+5x+2 }{x^2-1}}  =  \dfrac{x^4-8x^2-4x-5}{(x^2-1)^2 }}}}}\\\\

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