A(2,-4) B(3,3) C(-1,5) are the vertices of triangle ABC Find the equation of 1) The median of triangle through A 2) The altitude of triangle through B
Answers
Answer:
Equation of the median through A is 8x + y - 12 = 0
Equation of altitude of triangle through B is x - 3y + 6 = 0
Step-by-step explanation:
It is given that the points A( 2 , - 4 ) , B( 3 , 3 ) and C( - 1 , 5 ) are the vertices of a triangle.
( 1 ) : Equation of the median through A.
First, we have to find the co ordinates of the point where the median through A bisects BC( side opposing A, that's why we have taken side BC ).
Let the point be A', since the point A' is an end of the median of BC, its coordinates are mid the point of line BC.
By Mid Point Formula : -
= > Coordinates of BC = [ { 3 + ( - 1 ) } / 2 , { 3 + 5 } / 2 ]
= > Coordinates of BC = [ ( 3 - 1 ) / 2 , 8 / 2 ]
= > Coordinates of BC = ( 1 , 4 )
= > Slope of line AA' = [ 4 - ( - 4 ) ] / [ 1 - 2 ]
= > Slope of line AA' = ( 4 + 4 ) / ( - 1 )
= > Slope of line AA' = - 8
= > Equation of the line AA' ( median through A )
= > y - y₁ = m( x - x₁ )
= > y - ( - 4 ) = - 8( x - 2 )
= > y + 4 = - 8x + 16
= > 8x + y - 12 = 0
Hence, equation of the median through A is 8x + y - 12 = 0
( 2 ) : Equation of the altitude of the triangle through B
In this question, first we have to find the slope of the altitude through B.
As we know, altitude is the height( or perpendicular ), and in this question altitude is being produced from B, so it will be perpendicular to the side opposing point B i.e. AC.
Let the point where altitude through B intersects AC be B'.
Since, altitude through B is perpendicular to AC,
Slope of AC = - 1 / slope of BB'
= > Slope of AC = [ 5 - ( -4 ) ] / [ - 1 - 2 ]
= > Slope of AC = ( 5 + 4 ) / ( - 3 )
= > Slope of AC = - 3
Thus,
= > Slope of BB' = - 1 / ( - 3 )
= > Slope of BB' = 1 / 3
Therefore,
= > Equation of the line BB'
= > y - 3 = ( x - 3 )
= > 3( y - 3 ) = ( x - 3 )
= > 3y - 9 = x - 3
= > x - 3y + 6 = 0
Hence,
Equation of altitude of triangle through B is x - 3y + 6 = 0 .