F3 coordinate question

(a)in-centre(∠ bisector)cosA=4/5sin(A/2)=√[(1-cosA)/2]=√[(1-4/5)/2]=1/√10=√10/10cos(A/2)=√[(1+cosA)/2]=√[(1+4/5)/2]=3/√10=3√10/10Let O=centerΔOAB sine law: O=180-45-A/2=116.565(deg)4/sinO=OA/sin45OA=4*sin45/sin116.565=3.162Ox=OA*cos(A/2)=3.162*3√10/10=3Oy=OA*sin(A/2)=3.162√10/10=1So O=(3,1).......ans
(b)circumcentre( ⊥bisector)Ox=(Ax+Bx)/2=(2,0)Oy=(By+Cy)/2=(0,3/2)=> O=(2,3/2).........ans
(c)orthocentre(altitude/height)O=B=(4,0)........ans
(d)centroid(median)O=(A+B+C)/3=(8/3,1).........ans

Since ABC is a right - angled triangle.
(b) Circumcentre is the mid - point of AC = (2, 1.5), call it M.
(c) Orthocentre is B = (4,0)
(d) Centroid divides B and M in the ratio 1 : 2.
x = [4(1) + 2(2)]/(1 + 2) = 8/3
y = [0(1) + 1.5(2)]/(1 + 2) = 3/3 = 1.
So centroid is (8/3, 1).
(a)
Let radius of the inscribed circle be r, so the in - centre is (4 - r, r).
By considering the area of triangle ABC
Area = AC(r)/2 + BC(r)/2 + AB(r)/2 = 5r/2 + 3r/2 + 4r/2 = 12r/2 = 6r = (3)(4)/2
so r = 1.
so in - centre is (4 - 1, 1) = (3,1).