Question

Find the co-ordinates of in-centre of the triangle whose vertices are (0, 6), (8, 12) and (8, 0)

• A. (5, 6)
• B. (5, 7)
• C. (6, 5)
• D. (7, 5)

Answer 1

The correct option is A (5, 6)

Let A(0, 6), B(8, 12) and C(8, 0) be the vertices of triangle ABC
Then c = AB = $\sqrt{{(0-8)}^2+{(6-12)}^2}$ = 10, b = CA = $\sqrt{{(0-8)}^2+{(6-0)}^2}$ = 10 and a = BC = $\sqrt{{(8-8)}^2+{(12-0)}^2}$ = 12
The co-ordinates of the in-centre are $(\frac{a{x}_1+b{x}_2+c{x}_3}{a+b+c};\frac{a{y}_1+b{y}_2+c{y}_3}{a+b+c})$
or $(\frac{12\times 0+10\times 8+10\times 8}{12+10+10},\frac{12\times 6+10\times 12+10\times 0}{12+10+10})$
or $(\frac{160}{32},\frac{192}{32})or(5,6)$