Question

PLEASE PLEASE SOLVE THIS WITH STEPS . IT'S VERY IMPORTANT . PLEASE SOLVE WITH STEPS . ( IRRELEVANT ANSWERS WILL BE DIRECTLY REPORTED)​

Answer 1

Question:

$\sf{D, \ E, \ F \ are \ midpoints \ of \ AB, \ BC, \ AC}$ $\sf{of \ \triangle ABC. \ If \ Coordinates \ of \ D,}$ $\sf{ E, \ F \ are \ (1, 2), \ (0, -1), (2, -1), \ then \ the}$ $\sf{coordinates \ of \ A \ are....}$ $\sf{a) \ (3,2) \ \ b) \ (-3,2) \ \ c) \ (-1,3) \ \ d) \ (2,1)}$

__________________________

Answer:

$\sf{a) \ (3,2) \ is \ the \ correct \ option.}$

Diagram:

$\setlength{\unitlength}{1 cm}\begin{picture}(0,0)\thicklines\qbezier(0,0)(0,0)(2,2)\qbezier(0,0)(0,0)(4,0)\qbezier(2,2)(4,0)(4,0)\put(2.7,1.1){ \boldsymbol / }\put(1,1){ \boldsymbol \backslash }\put(2,-0.1){ \boldsymbol | }\put(1.8,2.1){ \bf A\ (x_1,\ y_1) }\put(1.8,-0.5){ \bf E\ (0,\ -1) }\put(-0.3,-0.3){ \bf B\ (x_2,\ y_2) }\put(4,-0.3){ \bf C\ (x_3,\ y_3) }\put(2.9,1.2){ \bf F\ (2,\ -1) }\put(-0.4,1.1){ \bf D\ (1,\ 2) }\end{picture}$

Solution:

$\sf{Let \ coordinates \ of \ A, \ B \ and \ C}$

$\sf{be \ (x_{1},y_{1}), \ (x_{2},y_{2}) \ and \ (x_{3},y_{3}) \ respectively.}$

$\sf{By \ midpoint \ fotmula}$

$\boxed{\sf{x=\dfrac{x_{1}+x_{2}}{2}, \ y=\dfrac{y_{1}+y_{2}}{2}}}$

$\sf{Considering \ x \ coordinate}$

$\sf{For \ seg \ AB,}$

$\sf{\dfrac{x_{1}+x_{2}}{2}=1}$

$\sf{\leadsto{x_{1}+x_{2}=2...(1)}}$

$\sf{For \ seg \ BC,}$

$\sf{\dfrac{x_{2}+x_{3}}{2}=0}$

$\sf{\leadsto{x_{2}+x_{3}=0...(2)}}$

$\sf{For \ seg \ AC,}$

$\sf{\dfrac{x_{1}+x_{3}}{2}=2}$

$\sf{\leadsto{x_{1}+x_{3}=4...(3)}}$

$\sf{From \ equation (2), \ we \ get}$

$\sf{x_{2}+x_{3}=0}$

$\sf{\therefore{x_{2}=-x_{3}}}$

$\sf{Substitute \ x_{2}=-x_{3} \ in \ equation (1), \ we \ get}$

$\sf{x_{1}-x_{3}=2}$

$\sf{\therefore{x_{1}=2+x_{3}}}$

$\sf{Substitute \ x_{1}=2+x_{3} \ in \ equation (3), \ we \ get}$

$\sf{2+x_{3}+x_{3}=4}$

$\sf{\therefore{2x_{3}=2}}$

$\sf{\implies{\therefore{x_{3}=1}}}$

$\sf{Substitute \ x_{3}=1 \ in \ equation (2), \ we \ get}$

$\sf{x_{2}+1=0}$

$\sf{\implies{\therefore{x_{2}=-1}}}$

$\sf{Substitute \ x_{2}=-1 \ in \ equation (1), \ we \ get}$

$\sf{x_{1}-1=2}$

$\boxed{\sf{\therefore{x_{1}=3}}}$

$\sf{Considering \ y \ coordinate}$

$\sf{For \ seg \ AB,}$

$\sf{\dfrac{y_{1}+y_{2}}{2}=2}$

$\sf{\leadsto{y_{1}+y_{2}=4...(i)}}$

$\sf{For \ seg \ BC,}$

$\sf{\dfrac{y_{2}+y_{3}}{2}=-1}$

$\sf{\leadsto{y_{2}+y_{3}=-2...(ii)}}$

$\sf{For \ seg \ AC,}$

$\sf{\dfrac{y_{1}+y_{3}}{2}=-1}$

$\sf{\leadsto{y_{1}+y_{3}=-2...(iii)}}$

$\sf{From \ equation (2) \ we \ get,}$

$\sf{y_{2}+y_{3}=-2}$

$\sf{\therefore{y_{2}=-2-y_{3}}}$

$\sf{Substitute \ y_{2}=-2-y_{3} \ in \ equation (i), \ we \ get}$

$\sf{y_{1}-2-y_{3}=4}$

$\sf{\therefore{y_{1}=6+y_{3}}}$

$\sf{Substitute \ in \ y_{1}=6+y_{3} \ equation (iii), \ we \ get}$

$\sf{6+y_{3}+y_{3}=-2}$

$\sf{\therefore{2y_{3}=-8}}$

$\sf{\implies{\therefore{y_{3}=-4}}}$

$\sf{Substitute \ y_{3}=-4 \ in \ equation (ii), \ we \ get}$

$\sf{-4+y_{2}=-2}$

$\sf{\implies{\therefore{y_{2}=2}}}$

$\sf{Substitute \ y_{2}=2 \ in \ equation (1), \ we \ get}$

$\sf{y_{1}+2=4}$

$\sf{\boxed{\therefore{y_{1}=2}}}$

$\sf{But,}$

$\sf{A=(x_{1},y_{1})}$

$\sf{\therefore{A=(3,2)}}$

$\sf\purple{\tt{Hence, \ the \ coordinates \ of \ points}}$

$\sf\purple{\tt{A \ are \ x=3 \ and \ y=2.}}$

Answer 2

Answer:

hello............could u please inbox me ?