Question

solve ittttt...........​

Answer 1

Answer:

Step-by-step explanation:

Given in picture

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Answer 2

$\huge\underline\bold\blue{Given}$

$\bold{Vertices \: of \: triangle}$

$\bold{A(1,-1), \: C(0,4), \: B(-5,3)}$

$\huge\underline\bold\blue{To \: find}$

$\textbf{1) Length of Median of triangle}$

$\textbf{2) Check whether it is scalene,}$$\textbf{equilateral or isosceles triangle}$

$\large\bold\green{NOTE:-}$

$\textbf{Scalene triangle - All sides are different}$

$\textbf{Equilateral triangle - All sides are equal}$

$\textbf{Isosceles triangle - Two sides are equal}$

$\huge\underline\bold\blue{Formulae}$

$\bold{Distance \: Formula}$

$\sqrt{(x2-x1)^2+(y2-y1)^2}$

$\bold{Midpoint \: Formula}$

$\large\frac{x1+x2}{2}$$\frac{y1+y2}{2}$

$\huge\underline\bold\blue{Solution}$

$\large\bold\red{Distance \: of \: AC}$

$\bold{AC \: =}$$\sqrt{(x2-x1)^2+(y2-y1)^2}$

$\bold{AC \: =}$$\sqrt{(0-1)^2+(4+1)^2}$

$\bold{AC \: =}$$\sqrt{(-1)^2+(5)^2}$

$\bold{AC \: =}$$\sqrt{1+25}$

$\bold{AC \: =}$$\sqrt{26}$$\bold{unit}$

$\large\bold\red{Distance \: of \: BC}$

$\bold{BC \: =}$$\sqrt{(0+5)^2+(4-3)^2}$

$\bold{BC \: =}$$\sqrt{(5)^2+(1)^2}$

$\bold{BC \: =}$$\sqrt{25+1}$

$\bold{BC \: =}$$\sqrt{26}$ unit

$\large\bold\red{Distance \: of \: AB}$

$\bold{AB \: =}$$\sqrt{(-5-1)^2+(3+1)^2}$

$\bold{AB \: =}$$\sqrt{(-6)^2+(4)^2}$

$\bold{AB \: =}$$\sqrt{36+16}$

$\bold{AB \: =}$$\sqrt{52}$

$\bold{AB \: =}$${2}$$\sqrt{13}$unit

$\bold{AC=BC≠AB}$

$\textbf{So, it is an isosceles triangle}$

$\textbf{because it has two equal sides}$

$\large\bold\blue{According \: to \: midpoint \: formula}$

$\large\bold\red{Coordinate \: of \: D}$

$\bold{(x,y)=}$ $\large\frac{x1+x2}{2}{,}\large\frac{y1+y2}{2}$

$\bold{(x,y)=}$$\large\frac{1-5}{2}{,}\large\frac{-1+3}{2}$

$\bold{x \: =}$$\large\frac{1-5}{2}$

$\bold{x \: =}$$\large\frac{-4}{2}$

$\bold{x \: = \: -2}$

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$\bold{y \: =}$$\large\frac{-1+3}{2}$

$\bold{y \: =}$$\large\frac{2}{2}$

$\bold{y \: = \: 1}$

$\large\bold\red{So, \: coordinate \: of \: D \: (-2,1)}$

$\textbf{Now, length of median}$

$\textbf{Again using distance formula}$

$\large\bold\red{Distance \: of \: CD}$

$\bold{CD \: =}$$\sqrt{(-2-0)^2+(1-4)^2}$

$\bold{CD \: =}$$\sqrt{(-2)^2+(-3)^2}$

$\bold{CD \: =}$$\sqrt{4+9}$

$\bold{CD \: =}$$\sqrt{13}$ unit