Question

15. If the vertices of a 3ABC are A(-4,4 ), B(8 ,4) and C(8,10). Find the equation of the
straight line along the median from the vertex A.

Answer 1

Solution:-

given by:- A(-4,4 ), B(8 ,4) and C(8,10).


we have :-

》Slope of the line BC:

          
》  m = (y₂ - y₁)/(x₂ - x₁)

》     = (10-4)/(8-8)
       
》   = 6/0

》       = 0

Slope of AD = -1/0

Equation of AD:

by fromula

》 (y - y₁) = m(x - x₁)

》(y - 4) = (-1/0) (x - (-4))

》0× (y - 4) = -1 (x + 4)

》0 = - x - 4

(x + 4 = 0)

equation of the straight line ( x+4=0) ans

☆i hope its help☆

Answer 2

Solution :

Given , A(-4,4),B(8,4) and C(8,10)

are vertices of ∆ABC ,

AM is the median .

M is the midpoint of B(8,4)= (x1,y1)

and C(8,10) = (x2 , y2 )

Coordinates of M = [ (x1+x2)/2,(y1+y2)/2]

= [ (8+8)/2 , (4+10)/2]

= ( 8 , 7 )

ii ) Equation the median , passing

through A(-4,4) = ( x1, y1) and

M(8,7) = ( x2 , y2 )

y - y1 = [ (y2-y1)/(x2 - x1) ] ( x - x1 )

=> y-4 = [(7-4)/(8+4)](x+4)

=> y-4 = (3/12)(x+4)

=> y - 4 = (1/4)( x + 4 )

=> 4( y - 4 ) = x + 4

=> 4y - 16 = x + 4

=> x + 4 - 4y + 16 = 0

=> x - 4y + 20 = 0

Therefore ,

Equation of the median is

x - 4y + 20 = 0

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