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Write the equations for two medians in slope-intercept form and solve the system of equations.
(1, 53), see solution.
We are given the coordinates of the triangle vertices A, B, and C. Let's plot these points on a coordinate plane and draw the triangle â–³ ABC.
Now, we can recall that a centroid of a triangle is a point of intersection of the triangle medians. Let's draw two medians and find their point of intersection.
In order to draw a median of a triangle, we need to choose a vertex and find a midpoint of the opposite to it side. Let's choose A. We can find the midpoint of BC by substituting the coordinates of B and C into the Midpoint Formula.
Substitute ( 2,5) & ( 4,- 3)
a+(- b)=a-b
Add and subtract terms
Calculate quotient
Now let's plot the midpoint of BC, which we can name N, and draw the median AN.
Substitute ( - 3,3) & ( 3,1)
a-(- b)=a+b
Add and subtract terms
Put minus sign in front of fraction
a/b=.a /2./.b /2.
Now that we know both the slope and y-intercept of AN, we can write its equation in slope-intercept form. y= mx+ b ⇒ y= - 13x+ 2
Similarly, we can draw the second median from the vertex B. We can find the midpoint of the opposite side AC by substituting the coordinates of A and C into the Midpoint Formula.
Substitute ( - 3,3) & ( 4,- 3)
a+(- b)=a-b
Add and subtract terms
Calculate quotient
Let's now plot the midpoint of AC, name it M, and draw the median BM.
We need to find the equation of BM in slope-intercept form. First, we can calculate the slope by substituting the coordinates of B and M into the Slope Formula.
Substitute ( 2,5) & ( 12,0)
a = 2* a/2
Subtract terms
- a/- b=a/b
a/b=a * 2/b * 2
So far the equation is the following. y= mx+b ⇒ y= 10/3x+b To find the y-intercept b, we will substitute point B(2,5) into the above equation and solve it for b.
x= 2, y= 5
a/c* b = a* b/c
LHS-20/3=RHS-20/3
a = 3* a/3
Subtract fractions
Rearrange equation
The equation of BM in slope-intercept form is as follows. y=10/3x+ b ⇒ y=10/3x+( - 5/3)
We have found two equations of the medians of â–³ ABC. Using them, we can form a system of equations. y=- 13x+2 y= 103x+(- 53) If we solve it we will find the common solution for both equations. It is a point of intersection of these medians and the centroid that we were asked to find. Let's use the Substitution Method.
(II): a+(- b)=a-b
(I): y= 103x- 53
(I): LHS+ 13x=RHS+ 13x
(I): LHS+ 53=RHS+ 53
(I): a/b=a * 3/b * 3
(I): Add fractions
(I): .LHS / 113.=.RHS / 113.
(II): x= 1
(II): a * 1=a
(II): Subtract fractions
The coordinates of the centroid of â–³ ABC are (1, 53).