Question

Please solve this question fast (◍•ᴗ•◍)❤​

Answer 1

Gɪᴠᴇɴ :-

• Two vertices of ∆ABC are A(6,4) & B(-2,2) .
• centroid G = (3,4).

Tᴏ Fɪɴᴅ :-

• vertices of Third vertex.
• Area od ∆ABC ?

Fᴏʀᴍᴜʟᴀ ᴜsᴇᴅ :-

• The centroid of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3) has the coordinates (x1+x2+x3)/3, (y1+y2+y3)/3.
• Area of ∆ vertices A(x1, y1), B(x2, y2) and C(x3, y3) is :- |(1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)]|

Sᴏʟᴜᴛɪᴏɴ :-

→ A = (x1,y1) = (6,4)

→ B = (x2,y2) = (-2,2)

→ C = Let (x3,y3)

Than, Putting values in centroid Vertices first, we get,

→ 3 = (x1 + x2 + x3)/3

→ 3*3 = (6 - 2 + x3)

→ 9 = 4 + x3

→ x3 = 5 .

Similarly,

→ 4 = (y1 + y2 + y3)/3

→ 4*3 = (4 + 2 + x3)

→ 12 = 6 + x3

→ y3 = 6 .

Hence, vertices of Third vertex C are (5,6) .

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Now, Putting all values in Area we get :-

→ Area of ∆ABC = |(1/2) [x1 (y2- y3 )+x2 (y3-y1 )+x3(y1-y2)]|

→ Area = (1/2)[6(2-6) + (-2)(6-4) + 5(4-2)]

→ Area = (1/2)[ 6 * (-4) + (-2)*2 + 5*2 ]

→ Area = (1/2)[ (-24) + (-4) + 10 ]

→ Area = (1/2) [ (-28) + 10 ]

→ Area = (1/2) [ (-18) ]

→ Area = | (-9) l

→ Area = 9 unit². (Ans.)

Hence, Area of ∆ABC is 9 unit².

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Note :- The two vertical bars means we have to use "Absolute Value". Or, it is always positive even if the formula produced a negative result. Polygons can never have a Negative Area.

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Answer 2

$\sf\blue{Question}$

$\sf{Two \ vertices \ of \ \triangle \ ABC \ are \ A(6,4)}$

$\sf{and \ B(-2,2). \ If \ the \ Centroid \ is \ G(3,4),}$

$\sf{find \ the \ third \ vertiex \ of \ the \ triangle}$

$\sf{and \ also \ find \ area \ of \ \triangle \ ABC}$

________________________________

$\sf\red{\underline{\underline{Answer:}}}$

$\sf{1. \ The \ third \ vertex \ of \ triangle \ is \ C(5,6).}$

$\sf{2. \ Area \ of \ \triangle \ ABC \ is \ 9 \ sq \ units}$

$\sf\orange{Given}$

$\sf{\implies{Two \ vertices \ of \ \triangle \ ABC \ are}}$

$\sf{A(6,4) \ and \ B(-2,2).}$

$\sf{\implies{The \ Centroid \ is \ G(3,4)}}$

$\sf\pink{To \ find:}$

$\sf{1. \ Third \ vertex \ of \ triangle.}$

$\sf{2. \ Area \ of \ \triangle \ ABC}$

$\sf\green{\underline{\underline{Solution:}}}$

$\sf{1.}$

$\sf{Let \ the \ Co-ordinates \ of \ vertex \ C \ be \ (x3,y3)}$

$\sf{Here, \ x=3, \ y=4, \ x1=6, \ y1=4, \ x2=-2 \ and \ y2=2}$

$\boxed{\sf{Co-ordinates \ of \ Centroid (x,y)=\frac{x1+x2+x3}{3},\frac{y1+y2+y3}{3}}}$

$\sf{For \ x \ co-ordinate}$

$\sf{\therefore{3=\frac{6+(-2)+x3}{3}}}$

$\sf{9=6-2+x3}$

$\sf{x3=9-4}$

$\sf{x3=5}$

$\sf{For \ y \ co-ordinate}$

$\sf{4=\frac{4+2+y3}{3}}$

$\sf{12=6+y3}$

$\sf{y3=12-6}$

$\sf{y3=6}$

$\sf{\therefore{C(x3,y3)=(5,6)}}$

$\sf\purple{\tt{\therefore{The \ third \ vertex \ of \ triangle \ is \ C(5,6).}}}$

$\sf{2.}$

$\sf{Area \ of \ \triangle \ ABC}$

$\sf{=\frac{1}{2}[x1(y2-y3)+x2(y3-y1)+x3(y1-y2)]}$

$\sf{=\frac{1}{2}[6(2-6)+(-2)(6-4)+5(4-2)]}$

$\sf{=\frac{1}{2}[-24-4+10]}$

$\sf{=\frac{1}{2}[-28+10]}$

$\sf{=| \ -9 \ |}$

$\sf{=9 \ sq \ units}$

$\sf\purple{\tt{\therefore{Area \ of \ \triangle \ ABC \ is \ 9 \ sq \ units}}}$