if a (4,1),b(-2,3)and c(0,5)and vertices of triangle abc and ad is its median then if P(2/3 ,3) is a point on ad ,then the ratio ap:dp is
Answers
Answer:
i)
Median is the line joining the midpoint of one side of a triangle to the opposite vertex. So, the coordinates of D would be(
2
6+1
,
2
5+4
)::(
2
7
,
2
9
)
ii)
P divides AD in the ratio 2:1.
A(x
1
,y
1
)=(4,2), D(x
2
,y
2
)=(
2
7
,
2
9
)
m:n=2:1
Using section formula, we get the coordinates of P.
P(x,y)=(
m+n
nx
1
+mx
2
,
m+n
ny
1
+my
2
)
=
⎝
⎜
⎜
⎛
2+1
1⋅4+2⋅
2
7
,
2+1
1⋅2+2⋅
2
9
⎠
⎟
⎟
⎞
=(
3
11
,
3
11
)
iii)
Coordinates of E will be (
2
5
,3) and the coordinates of F=(5,
2
7
).
Coordinates of Q=(
m+n
nx
1
+mx
2
,
m+n
ny
1
+my
2
)
=
⎝
⎜
⎜
⎛
2+1
1⋅6+2⋅
2
5
,
2+1
1⋅5+2⋅3
⎠
⎟
⎟
⎞
=(
3
11
,
3
11
)
Coordinates of R=(
m+n
nx
1
+mx
2
,
m+n
ny
1
+my
2
)
=
⎝
⎜
⎜
⎛
2+1
1⋅1+2⋅5
,
2+1
1⋅4+2⋅
2
7
⎠
⎟
⎟
⎞
=(
3
11
,
3
11
)
iv)
The coordinates of P,Q and R are the same which is (
3
11
,
3
11
).
This point is called the centroid, denoted by G.
v)
Centroid of triangle ABC=(
3
x
1
+x
2
+x
3
,
3
y
1
+y
2
+y
3
)
Step-by-step explanation:
Given :-
A (4,1),B(-2,3) and C(0,5) are the vertices of triangle ABC and AD is its median then if P(2/3 ,3) is a point on AD.
To find :-
Find AP : DP ?
Solution :-
Given points are :A (4,1),B(-2,3) and C(0,5)
AD is the median of the triangle ABC .
Let the coordinates of the point D be (x,y)
Since AD is the median , It divides BC into two equal parts.
=> D is the mid point of BC
We know that
The coordinates of the mid point of a linesegment joining the points (x1, y1) and (x2, y2) is
((x1+x2)/2, (y1+y2)/2)
Let (x1, y1) = B(-2,3) => x1 = -2 and y1 = 3
Let (x2,y2) = C(0,5) => x2 = 0 and y2 = 5
The coordinates of the point D = (x,y)
=> ( (-2+0)/2 , (3+5)/2 )
=> (-2/2, 8/2)
=> (-1,4)
Therefore, D = (-1,4)
Now ,
The point on AD = P(2/3,3)
Let P divides AD in the ratio = m1:m2
=> AP:PD = m1:m2
Let (x1, y1) = A(4,1) => x1 = 4 and y1 = 1
Let (x2, y2) = D(-1,4) => x2 = -1 and y2 = 4
We know that
By Section formula
({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))
On substituting these values in the above formula then
=> ({(m1)(-1)+(m2)(4)}/(m1+m2) , {(m1)(4)+(m2)(1)}/(m1+m2))
=> ({-m1+4m2}/(m1+m2) , {4m1+m2}/(m1+m2))
Now ,
(2/3,3)=({-m1+4m2}/(m1+m2) , {4m1+m2}/(m1+m2))
On comparing both sides then
=> {-m1+4m2}/(m1+m2) = 2/3
=> 3(-m1+4m2) = 2(m1+m2)
=> -3m1 +12m2 = 2m1+2m2
=> -3m1-2m1 = 2m2 -12m2
=> -5m1 = -10m2
=> 5m1 = 10m2
=> m1/m2 = 10/5
=> m1/m2 = 2/1
=> m1:m2 = 2:1
or
{4m1+m2}/(m1+m2) = 3
=> 4m1+m2 = 3(m1+m2)
=> 4m1+m2 = 3m1+3m2
=> 4m1-3m1 = 3m2-m2
=> 1 m1 = 2m2
=> m1/m2 = 2/1
=> m1 : m2 = 2:1
So, AP : DP = 2:1
Answer:-
The value of AP : DP for the given problem is 2:1
Used formulae:-
Mid point Formula :-
→ The coordinates of the mid point of the linesegment joining the points (x1, y1) and (x2, y2) is ( (x1+x2)/2, (y1+y2)/2 )
Section Formula :-
→ The coordinates of the point P(x,y) which divides the linesegment joining the points (x1, y1) and (x2, y2) in the ratio m1:m2 is P(x,y)
({m1x2+m2x1}/(m1+m2) , {m1y2+m2y1}/(m1+m2))