Math, asked by ramadevipenneru0, 2 months ago

The vertices of a triangle are A(a, 0), B(0, b) and C(a, b)
a b b
The centroid of AABC is
a)
2'2
8.
(2a
9. The mid point of AB is
b
)
129
3
10.
The mid point of BC is
c) a,
a
11. The mid point of AC is
d)
2.
74
Var​

Answers

Answered by krishnaveni5418
0

Answer:

The given vertices represent a right angled triangle with right angle at C(a,b) and the hypotenuse AB, as AB

2

=AC

2

+BC

2

So that orthocentre of the triangle is at C(a,b).

Circumcentre of the triangle is the middle point of the hypotenuse AB i.e. (a/2,b/2)

And centroid of the triangle is

(

3

a+a+0

,

3

0+b+b

)=(

3

2a

,

3

2b

)

Let P(x,y) be the foot of the altitude from C.

Then P lies on AB whose equation is

a

x

+

b

y

=1 ( 1 )

Also CP is perpendicular to AB and its equation

⇒y−b=(a/b)(x−a) ( 2 )

Solving (1) & (2) we get

x=

a

2

+b

2

a

3

,y=

a

2

+b

2

b

3

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