Find the Area of the Traingle Formed by joining the middle Points of the sides of the Traingle Whose Vertices are (0, -1), (2, 1) and (0, 3) | Bzziii.com

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Find the area of the traingle formed by joining the middle points of the sides of the traingle whose vertices are (0, -1), (2, 1) and (0, 3)







Let the vertices of triangle be A(0,-1) B(2,1) C(0,3)

Let the mid-point of

AB be P
BC be Q
AC be R

Now, joining the Points P,Q,R

We get ΔPQR

We need to find area of ΔABC and area of ΔPQR

Now, finding area of ΔABC

Area of ΔABC = `1/2` [`"x"_1("y"_2-"y"_3) + "x"_2("y"_3-"y"_1) + "x"_3("y"_1-"y"_2)`]

Here,

`"x"_1` = 0 , `"y"_1` = -1
`"x"_2` = 2 , `"y"_2` = 1
`"x"_3` = 0 , `"y"_3` = 3


Putting values

Area of Traingle = `1/2` [`"x"_1("y"_2-"y"_3) + "x"_2("y"_3-"y"_1) + "x"_3("y"_1-"y"_2)`]

Area of ΔABC = `1/2` [0 (1 - 3) + 2 (3 - (-1) + 0 ( -1-1)]

= `1/2` [0 + 2(3 + 1) + 0]

= `1/2` [0 + 2(4) + 0]

= `1/2` `\times` 8

= 4

Finding area of triangle PQR

First we have to find coordinates of P, Q, R


Since P is the mid-point of AB

Coordinates of P

= `(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})`

= `(\frac{0 + 2}{2},\frac{-1+1}{2})`

= `(\frac{ 2}{2},\frac{0}{2})`

= (1, 0)


Similarly, Q is the mid -point of BC

Coordinates of Q

= `(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})`

= `(\frac{2 + 0}{2},\frac{1+3}{2})`

= `(\frac{ 2}{2},\frac{4}{2})`

= (1, 2)

Similarly, R is the mid -point of AC

Coordinates of R

= `(\frac{x_{1}+x_{2}}{2},\frac{y_{1}+y_{2}}{2})`

= `(\frac{0 + 0}{2},\frac{-1+3}{2})`

= `(\frac{ 0}{2},\frac{2}{2})`

= (0, 1)

So, the coordinates are P (1, 0), Q (1, 2), R (0, 1)



Now, we need to find out the area of

∆PQR Area of triangle

PQR =`1/2` [`"x"_1("y"_2-"y"_3) + "x"_2("y"_3-"y"_1) + "x"_3("y"_1-"y"_2)`]

Here

`"x"_1` = 1 , `"y"_1` = 0
`"x"_2` = 1 , `"y"_2` = 2
`"x"_3` = 0 , `"y"_3` = 1  

Putting values Area of triangle PQR

= `1/2` [1(2 − 1) + 1(1 − 0) + 0(0 − 2)]

= `1/2` [1(1) + 1(1) + 0]

= `1/2` [1 + 1]

= `1/2` `\times` 2 = 1 square unit


Hence, the required ratio

= `"𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝑃𝑄𝑅"/"𝐴𝑟𝑒𝑎 𝑜𝑓 ∆𝐴𝐵𝐶"`

= `1/4`

 ∴ Required Ratio is 1 : 4

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