Question

$\textbf{Solve it quickly }$​

Answer 1

$\textbf{Answer is in Attachment !}$

Answer 2

Let P (0, -1), Q (2, 1) and R (0, 3).

And X the mid point of PQ

Y the mid point of QR

Z the mid point of PR.

On joining these mid points (X, Y & Z) we get a ∆XYZ

We have to find the ratio of ∆PQR and ∆XYZ


Area of triangle =

$\dfrac{1}{2}$ |$x_{1} \: (y_{2} - y _{3}) \: + \: x_{2} \: (y _{3} - y_{1}) \: x _{3} \: (y _{1} - y_{2})$|


Ar of ∆PQR =

$\dfrac{1}{2}$ |$0 \: (1-3) \: + 2 \: (3+ 1) \: + 0 \: ( - 1- 1)$|

= $\dfrac{1}{2}$ |$0 \: ( - 2) \: + 2 \: ( + 4) \: + 0 \: ( - 2)$|

= $\dfrac{1}{2}$ | + 8|

= $\dfrac{1}{2}$ × 8

= $\boxed{4\: sq.\: units}$

Now...

$\underline{Coordinates\: of\: X}$

$\dfrac{x _{1} \: + \: x_{2} }{2}$, $\dfrac{x _{1} \: + \: x_{2} }{2}$

=> $\dfrac{0\:+\:2}{2}$, $\dfrac{-1\:+\:1}{2}$

=> (1, 0)

$\underline{Coordinates \:of\: Y}$

=> $\dfrac{0\:+\:0}{2}$, $\dfrac{-1\:+\:3}{2}$

=> (0, 1)

$\underline{Coordinates\: of\: Z}$

=> $\dfrac{2\:+\:0}{2}$, $\dfrac{1\:+\:3}{2}$

=> (1, 2)

Ar of ∆XYZ =

$\dfrac{1}{2}$ |$1 \: (1 - 2) \: + 0 \: (2 - 2) \: + 1\: ( - 0 - 1)$|

= $\dfrac{1}{2}$ |$-1\:0\:\:-1$|

= $\dfrac{1}{2}$ |-2|

= $\dfrac{1}{2}$ (+2)

= $\boxed{1 \:sq. \:units}$

Ratio..

$\dfrac{ \triangle \: XYZ}{ \triangle \: PQR}$ = $\dfrac{1}{4}$

$\textbf{1 : 4}$ (∆XYZ : ∆PQR)