Question

find the mid point of line joining the points (-5,7) and (-1,3).​

Answer 1

$\setlength{\unitlength}{14mm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,0){5}}\put(5.1, - 0.3){\sf B}\put( - 0.2, - 0.3){\sf A}\put(5.2, 0){\sf (-1,3)}\put( - 1, 0){\sf (-5,7)}\put(2.3, 0.2){\sf P}\put(2.1, - 0.3){\sf (x,y)}\put(5, 0){\circle*{0.1}}\put(2.4, 0){\circle*{0.1}}\put(0, 0){\circle*{0.1}}\end{picture}$

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☯ Let's consider Mid point of line joining the points A (-5,7) and B (-1,3) be P (x,y).

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$\underline{\bigstar\:\boldsymbol{Using\:Mid-point\:formula\::}}$

$\star\;{\boxed{\sf{\pink{(x,y) = \bigg(\dfrac{x_1 + x_2}{2}\:,\: \dfrac{y_1 + y_2}{2}\bigg)}}}}$

$\sf Here \begin{cases} & \sf{x_1 , x_2 = -5,-1} \\ & \sf{y_1 , y_2 = 7,3} \end{cases}$

$\dag\;{\underline{\frak{Now,\:Putting\:values\:in\:formula,}}}$

$:\implies\sf (x,y) = \bigg( \dfrac{-5 + (-1)}{2}\:,\: \dfrac{7 + 3}{2} \bigg)\\ \\ \\ :\implies\sf (x,y) = \bigg( \dfrac{-5 - 1}{2}\:,\: \dfrac{10}{2} \bigg)\\ \\ \\ :\implies\sf (x,y) = \bigg( \dfrac{-6}{2}\:,\: \dfrac{10}{2} \bigg)\\ \\ \\ :\implies{\underline{\boxed{\sf{\purple{(x,y) = (-3,5)}}}}}\;\bigstar$

$\therefore\:{\underline{\sf{Hence,\:The\:coordinate\:of\:mid-point\:is\:{\textsf{\textbf{(-3,5)}}}.}}}$

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$\qquad\qquad\boxed{\underline{\underline{\pink{\bigstar \: \bf\:More\:to\:know\:\bigstar}}}}$

• $\sf Distance\:Formula,\:d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

• $\sf Section\: Formula = (x,y) = \bigg( \dfrac{x_1 m_2 + x_2 m_1}{m_1 + m_2}\:,\: \dfrac{y_1 m_2 + y_2 m_1}{m_1 + m_2} \bigg)$

Answer 2

Step-by-step explanation:

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$\text{\Large\underline{\red{Question:-}}}$

● Find the mid point of line joining the points (-5,7) and (-1,3).

$\text{\Large\underline{\orange{To\ find:-}}}$

● In this question we have to find the mid-point of the line.

$\text{\Large\underline{\green{Given:-}}}$

$\sf We\ have = \begin{cases} \sf\pink{Two\ mid-points\ that\ is:-} \\ \\ \sf\pink{1.\ (-5,\ 7)\ and\ 2.\ (-1,\ 3).} \end{cases}$

$\text{\Large\underline{\purple{Solution:-}}}$

The mid-points are given as:-

● (-5, 7)

● (-1, 3)

$\sf\underline{\ddag\qquad According\ to\ the\ question \qquad\ddag}$

$\sf \: \: \: \: \: \: \: : \implies {\bigg (\dfrac{x_{1}\ +\ x_{2}}{2}\ ,\ \dfrac{y_{1}\ +\ y_{2}}{2} \bigg)}$

$\sf \: \: \: \: \: \: \: : \implies {\bigg (\dfrac{-5\ +\ (-1)}{2}\ ,\ \dfrac{7\ +\ 3}{2} \bigg)}$

$\sf \: \: \: \: \: \: \: : \implies {\bigg (\dfrac{-5-1}{2}\ ,\ \dfrac{10}{2} \bigg)}$

$\sf \: \: \: \: \: \: \: : \implies {\bigg (\dfrac{-6}{2}\ ,\ \dfrac{10}{2} \bigg)}$

$\sf \: \: \: \: \: \: \: : \implies {\purple{\underline{\fbox{(-3,\ 5)}}}}\bigstar$

Hence:-

● The coordinate of mid-point is (-3, 5).

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