A Ajbunsua Leecher Oct 12, 2020 #1 inscribe in the triangles whose sides lie on the lines 3x-y-6=0, x+3y-3=0 and x-3y+6=0 sample
F faye422 Forum Veteran Oct 12, 2020 #3 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?
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?