Q. 3 : Find the co-ordinates of the points of trisection (I,e., points dividing into three equal parts) of the line segment joining the point A(2,-2) and B(-7,4).
NOTE
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Answers
Let P and Q be the points of trisection of AB , i.e., AP = PQ = QB
A.........P........ Q......... B
|--------|--------|----------|
(2,-2) ..................... (-7,4)
Therefore , P divides AB internally in the ratio 1:2
Let ( xˇ1 , y 1) = (2,-2)
(x2 , y2) = ( - 7 , 4 )
m1 : m2 = 1 : 2
Therefore , the coordinates of P , by applying the section formula ,
(m1x2 + m2x1 / m2 + m2 , m1y2 + m2y1 / m1 + m2 )
= [ 1(-7)+2(2) / 1 + 2 , 1(4) + 2(-2) / 1 + 2 ]
= (-3/3 , 0/3)
Similarly , Q also divides AB internally in ratio 2 : 1.
and the coordinates of Q by applying the section formula,
= [ 2(-7) + 1 (2) / 2+1 , 2(4) + 1 (-2) / 2+1 ]
= ( - 12/3 , 6/3 )
= ( -4 , 2 )
Hence , the coordinates of the points of trisection of the line segment joining A and B are (-1 , 0 ) and (-4 , 2).
Step-by-step explanation:
Given:- A line segment joining the points A(2,−2) and B(−7,4).
Let P and Q be the points on AB such that,
AP=PQ=QB
Therefore,
P and Q divides AB internally in the ratio 1:2 and 2:1 respectively.
As we know that if a point (h,k) divides a line joining the point (x
1
,y
1
) and (x
2
,y
2
) in the ration m:n, then coordinates of the point is given as-
(h,k)=(
m+n
mx
2
+nx
1
,
m+n
my
2
+ny
1
)
Therefore,
Coordinates of P=(
1+2
1×(−7)+2×2
,
1+2
1×4+2×(−2)
)=(−1,0)
Coordinates of Q=(
1+2
2×(−7)+1×2
,
1+2
2×4+1×(−2)
)=(−4,2)
Therefore, the coordinates of the points of trisection of the line segment joining A and B are (−1,0) and (−4,2).