Question

Let A(0,0), B(3,4), C(6,0) be the co-ordinate of triangle ABC. A point R inside the triangle is such that triangle RAB, RBC AND RAC are of equal area. Find the product of the coordinates of R

Answer 1

Answer:

• first method

$\huge \red{ \bf{solution}} \\ \\ \\ according \: to \: the \: question \: and \: attatchment \\ \\ area \: of \: \triangle \: ARC= \frac{1}{2} (area \: of \triangle \: ABC) \\ \\ \implies \: \frac{1}{3} \times \frac{1}{2} \times 4 \times 6 \\ \\ \implies \: 4 \: units \: \: \: ....(1)\: \\ now \\ area \: of \: \triangle \: ARC = \frac{1}{2} \times h \times 6 \\ \\ \implies \: 3h \: \: \: ...(2) \\ \\ from \: (1) \: and \: (2) \\ \\ \implies \: 4 = 3h \\ \implies \: h = \frac{4}{3} \\ \\ therefore \: co -\: ordinates \: of \: R \: (3 \: , \frac{4}{3} )$

• second method

now ,answer is ,

product of co ordinate of R is

(X1+X2+X3)/3, (y1+y2+y3)/3

= (3,4/3)

product of this is ,

= 4