Question

if A(-3,5),B(-1,1)and c(3,3) are the vertices of a triangle ABC,find the length of median AD.Also find the co- ordinates of the point which divides AD in the ratio 2:1

Answer 1

Given:

A(-3,5),B(-1,1)and c(3,3) are the vertices of a triangle ABC

To find:

• The length of the median AD
• The coordinates of the point which divides AD in the ratio 2:1

Solution:

AD is the median so D is the mid-point od AC

• Points of D = (x1+x2/2 , y1+y2/2)
• Points of D = (-1+3/2 , 1+3/2)
• D = (1, 2)

Length of AD:

$\sqrt{(x2-x1)^{2}+(y2-y1)^{2}}$

A = (-3, 5)

D = (1, 2)

• $\sqrt{(1+3)^{2}+(2-5)^{2}}$
• $\sqrt{4^{2}+(-3)^{2}}$
• $\sqrt{16+9}$
• $\sqrt{25}$
• 5 units

Coordinates of the point:

We will solve this by section formula

(x,y) = m.x2+n.x1/m+n , m.y2+n.y1/m+n

A = (-3, 5)

D = (1, 2)

m:n = 2:1

• (x,y) = m.x2+n.x1/m+n , m.y2+n.y1/m+n
• (x,y) = 2.1 + 1.-3/2+1 , 2.2 + 1.5/2+1
• (x,y) = 2-3/3 , 4+5/3
• (x,y) = -1/3 , 9/3
• (x,y) = -1/3 , 3

The coordinates of the point which divides AD in the ratio 2:1 is (-1/3, 3)

Answer 2

hope you will get it my answer