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3 years ago

If the midpoint of segment AB is M(1,-3) and B(5,-10), find the coordinates of point A.

Mathematics

1 answer:

azamat3 years ago

6 0

Answer:

Therefore the required point A(x₁ , y₁) = A( -3 , 4 )

Step-by-step explanation:

Given:

The midpoint of segment AB is M(1,-3)

and B(5,-10),

Let

point A( x₁ , y₁)

point B( x₂ , y₂) ≡ (5 , -10)

M(x , y) = (1 , -3 )

To Find:

point A( x₁ , y₁) = ?

Solution:

M is the midpoint of segment AB. {Given}

BY Mid point Formula we have

$M(x,y)=(\frac{x_{1}+x_{2} }{2}, \frac{y_{1}+y_{2} }{2})$

Substituting the given values in above equation we get

$M( 1, -3)=(\frac{x_{1}+5 }{2}, \frac{y_{1}-10 }{2})\\\\\textrm{on equating the X coordinate and Y coordinate of M we get}\\1 =\frac{x_{1}+5 }{2}\\\\and\\-3=\frac{y_{1}-10 }{2}\\\\\therefore x_{1}+5=2\ and\ y_{1}-10 =-6\\\\\therefore x_{1} = 2-5=-3\ and\ y_{1}=10-6=4\\\\\therefore x_{1} = -3\ and \ y_{1}=4\\\\\therefore A(x_{1},y_{1})=A(-3,4)$

Therefore the required point A(x₁ , y₁) = A( -3 , 4 )

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We split up the domain of integration into 10 subintervals,

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where the left and right endpoints for the $i$ -th subinterval are, respectively,

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$m_i=\dfrac{\ell_i+r_i}2=\dfrac{2i-1}4$

with $1\le i\le10$ .

Over each subinterval, we interpolate $f(x)=\sqrt{1+x^8}$ with the quadratic polynomial,

$p_i(x)=f(\ell_i)\dfrac{(x-m_i)(x-r_i)}{(\ell_i-m_i)(\ell_i-r_i)}+f(m_i)\dfrac{(x-\ell_i)(x-r_i)}{(m_i-\ell_i)(m_i-r_i)}+f(r_i)\dfrac{(x-\ell_i)(x-m_i)}{(r_i-\ell_i)(r_i-m_i)}$

Then

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\sum_{i=1}^{10}\int_{\ell_i}^{r_i}p_i(x)\,\mathrm dx$

It turns out that the latter integral reduces significantly to

$\displaystyle\int_0^5f(x)\,\mathrm dx\approx\frac56\left(f(0)+4f\left(\frac{0+5}2\right)+f(5)\right)=\frac56\left(1+\sqrt{390,626}+\dfrac{\sqrt{390,881}}4\right)$