🎓 Academic UNANG MAKASAGOT NITO LOAD 50, ASAP SANA

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Ajbunsua

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inscribe in the triangles whose sides lie on the lines 3x-y-6=0, x+3y-3=0 and x-3y+6=0

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Yung equation ba ng circle need mo lods? Gawin mo lang din yung ginawa sa problem 1.

3x-y-6=0 , x+3y-3=0 , x-3y+6=0

d¹= (3h-k-6)/√10
d²= (h+3k-3)/√10
d³= (h-3k+6)/√10

d²=d³
(h+3k-3)/√10 = (h-3k+6)/√10
h+3k-3 = h-3k+6
6k=9
k=1.5

-d¹=d² and k=1.5
-(3h-k-6)/√10 = (h+3k-3)/√10
-3h+1.5+6 = h+3(1.5)-3
-3h + 7.5 = h + 1.5
4h = 6
h=1.5

C(1.5,1.5) center of inscribed circle
Solve for r,
r=d¹=d²=d³
r=d²= 1.5+3(1.5)-3 /√10
r= 3/√10

Equation of a circle (x-h)²+(y-k)²=r²
(X-1.5)²+(Y-1.5)²=9/10

Tama ba lods? 😂
 
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