In an isosceles triangle ABC , the vertex A is (6,1) and the equation of the base BC is 2x+y=4. Let the point B lie on the line x+3y=7. If (α,β) is the centroid △ABC, then 15(α+β) is equal to :
(A)39
(B)41
(C)51✓ Correct
(D)63
Answer: (C)
Solution
Point B lies on both 2x+y=4 and x+3y=7. Solving: From x+3y=7: x=7−3y. Substituting into 2x+y=4:
2(7−3y)+y=4⇒14−6y+y=4⇒y=2,x=1
So B=(1,2).
Let C=(h,4−2h) (on line 2x+y=4). Since the triangle is isosceles with AB=AC:
(6−1)2+(1−2)2=(6−h)2+(1−(4−2h))2
26=(6−h)2+(2h−3)2
26=36−12h+h2+4h2−12h+9
5h2−24h+19=0
(5h−19)(h−1)=0⇒h=519orh=1
Since h=1 gives C=B, we take h=519, giving C=(519,−518).