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Yuki888 [10]
3 years ago
7

Point M is the midpoint of PQ. The coordinates of P and M are given below.

Mathematics
1 answer:
Zepler [3.9K]3 years ago
7 0

The coordinates of Q are (-1,6)

Further Explanation:

Given points are:

(x_m, y_m) = (5, -2)\\(x_p,y_p) = (11, -10)

The formula for mid-pint is:

(x_m, y_m) = (\frac{x_p+x_q}{2} , \frac{y_p+y_q}{2})\\From\ this\ we\ get\\x_m = \frac{x_p+x_q}{2}\\Putting\ values\\5 =\frac{11+x_q}{2}\\5*2 = 11+x_q\\10 = 11+x_q\\10-11 = x_q\\x_q = -1\\AND\\y_m = \frac{y_p+y_q}{2}\\-2 = \frac{-10+y_q}{2}\\-2*2 = -10+y_q\\-4 = -10+y_q\\-4+10 = y_q\\y_q = 6

The coordinates of Q are (-1,6)

Keywords: Coordinate geometry, mid-point

Learn more about mid-point at:

  • brainly.com/question/11018983
  • brainly.com/question/11175936

#LearnwithBrainly

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\dfrac{\partial\left(\mu\left(x^2y+e^x\right)\right)}{\partial y} = \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu \\\\ \dfrac{\partial\left(-\mu x^2\right)}{\partial x} = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu \\\\ \implies \left(x^2y+e^x\right)\dfrac{\partial\mu}{\partial y} + x^2\mu = -x^2\dfrac{\partial\mu}{\partial x} - 2x\mu

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x^2\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} - 2x\mu \\\\ (x^2+2x)\mu = -x^2\dfrac{\mathrm d\mu}{\mathrm dx} \\\\ \dfrac{\mathrm d\mu}{\mu} = -\dfrac{x^2+2x}{x^2}\,\mathrm dx \\\\ \dfrac{\mathrm d\mu}{\mu} = \left(-1-\dfrac2x\right)\,\mathrm dx \\\\ \implies \ln|\mu| = -x - 2\ln|x| \\\\ \implies \mu = e^{-x-2\ln|x|} = \dfrac{e^{-x}}{x^2}

The modified DE,

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\dfrac{\partial\left(e^{-x}y+\frac1{x^2}\right)}{\partial y} = e^{-x} \\\\ \dfrac{\partial\left(-e^{-x}\right)}{\partial x} = e^{-x}

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