Question

Find the length of the medians of the triangle whose vertices are (1,-1), (0,-4) and (-5,3).

Answer 1

${\blue{\text{Length of medians are :}}}$

$AD = \frac{\sqrt{130}}{2}$

$BE = \sqrt{13}$

$CF = \frac{\sqrt{130}}{2}$

${\large{\underline{\underline{\text{\red{Given :}}}}}}$

${\text{Vertices of triangle}}\:\: (1,-1)(0,4)(-5,3)$

${\large{\underline{\underline{\text{\red{To Find :}}}}}}$

${\text{Length of medians of the triangle}}$

${\large{\underline{\underline{\text{\red{Solution :}}}}}}$

Let the vertices of the triangle ABC be A(1,-1),B(0,4) and C(-5,3).

Let D,E,F are the midpoints of the sides BC,CA,AB respectively then AD,BE,CF are the medians.

$\setlength{\unitlength}{1cm}\begin{picture}(20,15)(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(0,0)(0,0)(2,3)\qbezier(4,0)(4,0)(2,3)\qbezier(0,0)(0,0)(3,1.5)\qbezier(2,0)(2,0)(2,3) \qbezier(4,0)(4,0)(1,1.5)\put(-0.2,-0.3){\bf B(0,4)}\put(2,-0.3){\bf D}\put(4,-0.3){\bf C(-5,3)}\put(3.2,1.5){\bf E}\put(2,3.2){\bf A(1,-1)}\put(0.7,1.5){\bf F}\end{picture}$

${\purple{\text{Mid - Point formula}} = ( \frac{x_2-x_1}{2},\frac{y_2-y_1}{2})}$

$\bigstar\: {\pink{\text{D is the midpoint of BC}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0, 0)(0,0)(4,0)\put(-0.5,-0.3){\bf B(0,4)}\put(2,-0.3){\bf D}\put(4,-0.3){\bf C(-5,3)}\end{picture}$

$\therefore D = ( \frac{0-5}{2} , \frac{4+3}{2})$

${\green{D = (\frac{-5}{2},\frac{7}{2})}}$

$\bigstar\:{\pink{\text{E is the midpoint of CA}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0, 0)(0,0)(4,0)\put(-0.5,-0.3){\bf C(-5,3)}\put(2,-0.3){\bf E}\put(4,-0.3){\bf A(-1,1)}\end{picture}$

$\therefore E = (\frac{-5+1}{2},\frac{3-1}{2}) \implies ( \frac{-4}{2},\frac{2}{2})$

${\green{E = (-2,1)}}$

$\bigstar\:{\pink{\text{F is the midpoint of AB}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0, 0)(0,0)(4,0)\put(-0.5,-0.3){\bf A(1,-1)}\put(2,-0.3){\bf F}\put(4,-0.3){\bf B(0,4)}\end{picture}$

$\therefore F = (\frac{1+0}{2},\frac{3-1}{2})$

$\green{F= (\frac{1}{2},\frac{3}{2})}$

${\purple{\text{Distance formula}}=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}$

${\text{\pink{Length of the median AD}}}$

$A (1,-1) \:\:\: D(\frac{-5}{2},\frac{7}{2})$

$\implies \sqrt{(\frac{-5}{2}-1)^2 + (\frac{7}{2}+1)^2}$

$\implies \sqrt {(\frac{-5-2}{2})^2+(\frac{7+2}{2})^2}$

$\implies \sqrt{(\frac{-7}{2})^2+(\frac{9}{2})^2}$

$\implies \sqrt{(\frac{49}{2}+\frac{81}{2})}$

$\boxed{AD= \sqrt{\frac{130}{4}}}$

${\text{\pink{Length of median BE}}}$

$B(0,4),E(2,1)$

$\implies \sqrt{(2-0)^2+(1-4)^2}$

$\implies \sqrt{4+(-3)^2}$

$\implies \sqrt{4+9}$

${\boxed{BE = \sqrt{13}}}$

${\text{\pink{Length of median CF}}}$

$c(-5,3),F(\frac{1}{2},\frac{3}{2})$

$\implies \sqrt{(\frac{1}{2}+5)^2+(\frac{3}{2}-3)^2}$

$\implies \sqrt{(\frac{1+10}{2})^2+(\frac{3-6}{2})^2}$

$\implies \sqrt{(\frac{11}{2})^2+(\frac{-3}{2})^2}$

$\implies \sqrt{\frac{121}{4}+(\frac{3}{2})^2}$

$\implies \sqrt {\frac{121}{4}+\frac{9}{4}}$

${\boxed{CF=\frac{\sqrt{130}}{2}}}$

( Kindly visit the answer in web (Brainly.in) To view the display of triangle and sides )

Answer 2

${\blue{\text{Length of medians are :}}}$

$AD = \frac{\sqrt{130}}{2}$

$BE = \sqrt{13}$

$CF = \frac{\sqrt{130}}{2}$

${\large{\underline{\underline{\text{\red{Given :}}}}}}$

${\text{Vertices of triangle}}\:\: (1,-1)(0,4)(-5,3)$

${\large{\underline{\underline{\text{\red{To Find :}}}}}}$

${\text{Length of medians of the triangle}}$

${\large{\underline{\underline{\text{\red{Solution :}}}}}}$

Let the vertices of the triangle ABC be A(1,-1),B(0,4) and C(-5,3).

Let D,E,F are the midpoints of the sides BC,CA,AB respectively then AD,BE,CF are the medians.

$\setlength{\unitlength}{1cm}\begin{picture}(20,15)(0,0)\thicklines\qbezier(0,0)(0,0)(4,0)\qbezier(0,0)(0,0)(2,3)\qbezier(4,0)(4,0)(2,3)\qbezier(0,0)(0,0)(3,1.5)\qbezier(2,0)(2,0)(2,3) \qbezier(4,0)(4,0)(1,1.5)\put(-0.2,-0.3){\bf B(0,4)}\put(2,-0.3){\bf D}\put(4,-0.3){\bf C(-5,3)}\put(3.2,1.5){\bf E}\put(2,3.2){\bf A(1,-1)}\put(0.7,1.5){\bf F}\end{picture}$

${\purple{\text{Mid - Point formula}} = ( \frac{x_2-x_1}{2},\frac{y_2-y_1}{2})}$

$\bigstar\: {\pink{\text{D is the midpoint of BC}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0, 0)(0,0)(4,0)\put(-0.5,-0.3){\bf B(0,4)}\put(2,-0.3){\bf D}\put(4,-0.3){\bf C(-5,3)}\end{picture}$

$\therefore D = ( \frac{0-5}{2} , \frac{4+3}{2})$

${\green{D = (\frac{-5}{2},\frac{7}{2})}}$

$\bigstar\:{\pink{\text{E is the midpoint of CA}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0, 0)(0,0)(4,0)\put(-0.5,-0.3){\bf C(-5,3)}\put(2,-0.3){\bf E}\put(4,-0.3){\bf A(-1,1)}\end{picture}$

$\therefore E = (\frac{-5+1}{2},\frac{3-1}{2}) \implies ( \frac{-4}{2},\frac{2}{2})$

${\green{E = (-2,1)}}$

$\bigstar\:{\pink{\text{F is the midpoint of AB}}}$

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0, 0)(0,0)(4,0)\put(-0.5,-0.3){\bf A(1,-1)}\put(2,-0.3){\bf F}\put(4,-0.3){\bf B(0,4)}\end{picture}$

$\therefore F = (\frac{1+0}{2},\frac{3-1}{2})$

$\green{F= (\frac{1}{2},\frac{3}{2})}$

${\purple{\text{Distance formula}}=\sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}}$

${\text{\pink{Length of the median AD}}}$

$A (1,-1) \:\:\: D(\frac{-5}{2},\frac{7}{2})$

$\implies \sqrt{(\frac{-5}{2}-1)^2 + (\frac{7}{2}+1)^2}$

$\implies \sqrt {(\frac{-5-2}{2})^2+(\frac{7+2}{2})^2}$

$\implies \sqrt{(\frac{-7}{2})^2+(\frac{9}{2})^2}$

$\implies \sqrt{(\frac{49}{2}+\frac{81}{2})}$

$\boxed{AD= \sqrt{\frac{130}{4}}}$

${\text{\pink{Length of median BE}}}$

$B(0,4),E(2,1)$

$\implies \sqrt{(2-0)^2+(1-4)^2}$

$\implies \sqrt{4+(-3)^2}$

$\implies \sqrt{4+9}$

${\boxed{BE = \sqrt{13}}}$

${\text{\pink{Length of median CF}}}$

$c(-5,3),F(\frac{1}{2},\frac{3}{2})$

$\implies \sqrt{(\frac{1}{2}+5)^2+(\frac{3}{2}-3)^2}$

$\implies \sqrt{(\frac{1+10}{2})^2+(\frac{3-6}{2})^2}$

$\implies \sqrt{(\frac{11}{2})^2+(\frac{-3}{2})^2}$

$\implies \sqrt{\frac{121}{4}+(\frac{3}{2})^2}$

$\implies \sqrt {\frac{121}{4}+\frac{9}{4}}$

${\boxed{CF=\frac{\sqrt{130}}{2}}}$

( Kindly visit the answer in web (Brainly.in) To view the display of triangle and sides )