Answer: (1,-1)
Step-by-step explanation:
Midpoint of BC=(6+4)/2, (3–1)/2. =(5,1)
Slope of BC is (3+1)/4–6)= 4/-2 = -2
Slope of perpendicular bisector of BC =+1/2
Eqn of perpendicular bisector is : Y-1 =1/2 (x-5)
Y=1/2 •(x-5) +1
Midpoint of AB. (6–2)/2, (3–1)/2 ={2,1)
Slope of AB is(3+1)/(-2–6) = 4/-8 =-1/2
Slope of perpendicular bisector = +2
Eqn of perpendicular bisector is Y-1. =2( X-2)
Y=2X-4+1 = 2X -3
Solving Y=(X-5)/2 +1
& Y=2X-3
2X-3 =(x-5/2)+1
2X-4 =(x-5/2)
4X-8 = x-5
3X =3
X=1
Y= 2×1–3= -1
Circumcentre is(1,-1)
Given:
The figure of a triangle LMN.
P is the centroid of triangle LMN.
To find:
14. Find the value of PN if QN=30.
15. Find the value of PN if QN=9.
Solution:
We know that the centroid in the intersection of medians of a triangle and centroid divides each median in 2:1.
Since P is the centroid it means NQ is the median from vertex N. It means P divides the median NQ in 2:1. So, PN:PQ=2:1.
14. We have QN=30.




Therefore, the value of PN is 20 when QN=30.
15. We have QN=9.




Therefore, the value of PN is 6 when QN=9.
X = 0
X=- 3 1/3 or x= -3.3