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Question: Incentre of the triangle whose vertices are (6, 0)(0, 6)and(7, 7) is-...

Incentre of the triangle whose vertices are (6, 0)(0, 6)and(7, 7) is-

A

(92,92)\left( \frac{9}{2},\frac{9}{2} \right)

B

(72,72)\left( \frac{7}{2},\frac{7}{2} \right)

C

(112,112)\left( \frac{11}{2},\frac{11}{2} \right)

D

None of these

Answer

(92,92)\left( \frac{9}{2},\frac{9}{2} \right)

Explanation

Solution

A (6, 0) ; B(0, 6) ; C (7, 7)

AB=72=62\mathrm { AB } = \sqrt { 72 } = 6 \sqrt { 2 } BC=50=52\mathrm { BC } = \sqrt { 50 } = 5 \sqrt { 2 } CA=50=52\mathrm { CA } = \sqrt { 50 } = 5 \sqrt { 2 }

\ in centre

ŗ (6×52+0×52+7×6262+52+52,0×52+6×52+7×6262+52+52)\left( \frac{6 \times 5\sqrt{2} + 0 \times 5\sqrt{2} + 7 \times 6\sqrt{2}}{6\sqrt{2} + 5\sqrt{2} + 5\sqrt{2}},\frac{0 \times 5\sqrt{2} + 6 \times 5\sqrt{2} + 7 \times 6\sqrt{2}}{6\sqrt{2} + 5\sqrt{2} + 5\sqrt{2}} \right)

ŗ (722162,722162)\left( \frac{72\sqrt{2}}{16\sqrt{2}},\frac{72\sqrt{2}}{16\sqrt{2}} \right) ŗ (92,92)\left( \frac{9}{2},\frac{9}{2} \right)