Question

In a triangle PQR, vertices of P(3,5) and Q(8,4) and centroid at (5,6). Then find the co-ordinate of R.​

Answer 1

$\blue{\bold{\underline{\underline{Answer:}}}}$

$\green{\tt{\therefore{Coordinate\:of\:R=(4,9)}}}$

$\orange{\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

$\green{\underline \bold{Given:}} \\ \tt: \implies Coordinate \: of \: P = (3,5) \\ \\ \tt: \implies Coordinate \: of \: Q= (8,4) \\ \\ \tt: \implies Centroid \: of \:triangle = (5,6) \\ \\ \red{\underline \bold{To \: Find:}} \\ \tt: \implies Coordinate \: of \: R = ?$

• According to given question :

$\bold{As \: we \: know \: that : } \\ \text{Centroid \: formula--} \\ \\ \tt: \implies x = \frac{ x_{1} + x_{2} + x_{3} }{3} \\ \\ \tt: \implies 5 = \frac{3 + 8 + x_{3}}{3} \\ \\ \tt: \implies 5 \times 3 = 11 + x_{3} \\ \\ \tt: \implies x_{3} = 15 - 11 \\ \\ \green{\tt: \implies x_{3} =4} \\ \\ \bold{For \: y \: ordinate} \\ \tt: \implies y = \frac{ y_{1} + y_{2} + y_{3} }{3} \\ \\ \tt: \implies 6= \frac{5+ 4 + y_{3}}{3} \\ \\ \tt: \implies 6\times 3 = 9+ y_{3} \\ \\ \tt: \implies y_{3} = 18 -9 \\ \\ \green{\tt: \implies y_{3} =9} \\ \\ \green{\tt{\therefore Coordinate \: of \: R\: is \: (4,9)}}$

Answer 2

Step-by-step explanation:

Aloha !

Given,

P(3,5) & Q (8,4) & R (x³ , y³) And given centroid point (5,6)

G= (x¹+x²+x³/3, y¹+y²+y³/3)

(5,6)=(3+8+x³/3 , 5+4+y³/3)

(5,6)= ( 11+x³/3 , 9+y³/3)

Now,

5=11+x³/3

15=11+x³

x³=4

And,

6=9+y³/3

18=9+y³

y³=9

Therefore R(4,9)

Thank you

@ Twilight Astro ✌️☺️♥️