Question

Can anyone do this for me plzz and fast. Unnecessary answers will be reported. Correct answer will be marked as brianliest ​

Answer 1

Solution:

Please see the attachment.

Drop a perpendicular from S(-5,2) on line PQ and name it M(-2,2).

Similarly, drop a perpendicular from S'(5,2) on line P'Q' and name it M'(2,2).

w.k.t., distance between two points (x₁,y₁) and (x₂,y₂) is $\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}$.

⇒Distance between P(-2,5) and Q(-2,-1) = $\sqrt{(-2-(-2))^{2}+(-1-5)^{2}}$ = 6

Similarly, distance between Q(-2,-1) and R(-5,-4) = $\sqrt{(-5-(-2))^{2}+(-4-(-1))^{2}}$ = $3\sqrt{2}$

Distance between R(-5,-4) and S(-5,2) = $\sqrt{(-5-(-5))^{2}+(2-(-4))^{2}}$ = 6

Distance between S(-5,2) and P(-2,5) = $\sqrt{(-2-(-5))^{2}+(5-2)^{2}}$ = $3\sqrt{2}$

Distance between S(-5,2) and M(-2,2) = $\sqrt{(-2-(-5))^{2}+(2-2)^{2}}$ = 3

Distance between S'(5,2) and M'(2,2) = $\sqrt{(2-5)^{2}+(2-2)^{2}}$ = 3

w.k.t., mid-point of line joining two points (x₁,y₁) and (x₂,y₂) is ($\frac{x_{1}+x_{2}}{2} , \frac{y_{1}+y_{2}}{2}$).

⇒Mid-point of P(-2,5) and R(-5,-4) = ($\frac{-2-5}{2} , \frac{5-4}{2}$) = ($\frac{-7}{2} , \frac{1}{2}$)

Similarly, mid-point of Q(-2,-1) and S(-5,2) = ($\frac{-2-5}{2} , \frac{-1+2}{2}$) = ($\frac{-7}{2} , \frac{1}{2}$)

Mid-point of P'(2,5) and R'(5,-4) = ($\frac{2+5}{2} , \frac{5-4}{2}$) = ($\frac{7}{2} , \frac{1}{2}$)

Mid-point of Q'(2,-1) and S'(5,2) = ($\frac{2+5}{2} , \frac{-1+2}{2}$) = ($\frac{7}{2} , \frac{1}{2}$)

⇒Both PQRS and P'Q'R'S' are parallelograms.

w.k.t., area of parallelogram with height h and base b = hb

Area of parallelogram PQRS = (PQ)(SM) = 6×3 = 18 sq units

Area of parallelogram P'Q'R'S' = (P'Q')(S'M') = 6×3 = 18 sq units

From the above, we infer that on taking reflection of a fiqure in y-axis, the are remains unchanged. Also, the reflected points differ by sign of x-coordinates.

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