Find equation of a line passing through the centroid of the triangle ABC with A(2,3),B(-4,5),C(1,2).
Answers
Answer:
It is given that the coordinates of a triangle are A( - 1 , 3 ) , b( 4 , 2 ) and c( 3 , - 2 ).
From the properties of triangle, we know :
where are the x coordinates of the vertices of triangle and are the y co ordinates of the vertices of the triangle.
Thus,
= > Centroid of this triangle = [ ( - 1 + 4 + 3 ) / 3 , ( 3 + 2 - 2 ) / 3 ]
= > Centroid of this triangle = [ 6 / 3 , 3 / 3 ]
= > Centroid of this triangle = ( 2 , 1 )
We have to find the equation of the line which is passing through G( centroid ) and parallel to AC.
As that line is parallel to AC, slope of AC should be equal to the slope of that line.
Therefore,
= > Slope of that line = Slope of AC
= > Slope of the line( which is parallel to AC ) = [ ( - 2 - 3 ) / ( 3 + 1 ) ]
= > Slope of the line( parallel to AC ) = - 5 / 4
Thus the equation of the line which is parallel to AC should be :
= > y - 1 = ( - 5 / 4 ) ( x - 2 )
= > 4y - 4 = - 5x + 10
= > 5x + 4y - 4 - 10 = 0
= > 5x + 4y - 14 = 0
Hence,
Equation of the line which is parallel to AC and passing through centroid is 5x + 4y - 14 = 0.