Question

A (-1, 3),b (4,2),c( 3 - 2 )are the vertices of a triangle find the coordinates of the centroid of the triangle .Find the equation of the line through g and parallel to ac by brainly

Answer 1

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 :
$\text{Coordinates of centroid = } \bigg(\dfrac{x_{1} + x_{2} + x_{3} }{3},\dfrac{y_{1} + y_{2} + y_{3} }{3} \bigg)$
where $x_1 , x_2 ,x_3$ are the x coordinates of the vertices of triangle and $y_1 , y_2 ,y_3$ 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.
$\:$

Answer 2

$\sf {\huge {SOLUTION}}$ :

Vertices of triangle = a( - 1, 3 ), b( 4,2 ), c ( 3,-2 )

For centroid G of the triangle,

x = $\frac{x1 \:+\: x2\: +\:x3}{3}$

x = $\frac{-1 \:+\: 4 \: +\:3}{3}$

x = $\frac{6}{3}$

x = 2

y = $\frac{y1 \:+\: y2\: +\:y3}{3}$

y = $\frac{3 \:+\: 2\: +\:-2}{3}$

y = $\frac{3}{3}$

y = 1

Coordinate of Centroid G = ( x, y ) = ( 2, 1 )

For equation of line through G // to AC ,

Slope of AC = $\frac{y2 \:- \: y1} {x2 \:-\: x1}$

Here, y2 = - 2, y1 = 3, x2 = 3, x1 = - 1

Slope of AC = $\frac{-2 \:- \: 3} {3\:+\: 1 }$

Slope of AC = $\frac{-5} {4 }$

Equation of line parallel to AC,

( y - y1 ) = $\frac{y2 \:- \: y1} {x2 \:-\: x1}$ ( x - x1 )

( y - 1 ) = $\frac{-5}{4}$( x - 2 )

4 ( y - 1 ) = - 5 ( x - 2 )

4y - 4 = - 5x + 10

5x + 4y - 4 - 10 = 0

$\sf{\huge{5x \:+\:4y \:- \:14\: = \:0}}$