Question

THE LINE SEGMENT JOINING A(6,3) and B(-1,-4) is doubled in length by having half its length added to each end.Find the coordinates of the new ends.

Answer 1

The formula to find the coordinates when a segment is divided in a ratio m :n is

$(\frac{mx_{2}+nx_{1}  }{m+n},\frac{my_{2}+ny_{1}  }{m+n})$

Here A(6,3) & B(-1,-4), Let $C(x_{1},y_{1} )$

CA : AB = 1:2 = m:n

We get

$(6,3)=(\frac{2x_{1}-1 }{2+1},\frac{2y_{1}-4 }{2+1})=(\frac{2x_{1}-1 }{3},\frac{2y_{1}-4 }{3})$

Simplifying we get $x_{1}=\frac{19}{2},y_{1}=\frac{13}{2}$

Also

AB : BD = 2:1

Applying the formula we get

$(-1,-4)=(\frac{2x_{2}+6 }{2+1},\frac{2y_{2}+3 }{2+1})=(\frac{2x_{2}+6 }{3},\frac{2y_{2}+3 }{3})$

Simplifying we get

$x_{2}=\frac{-9}{2},y_{1}=\frac{-15}{2}$

The coordinates of the new end are

$C(\frac{19}{2},\frac{13}{2}),D(\frac{-9}{2},\frac{-15}{2})$

Answer 2

Answer:

section formula.

Step-by-step explanation:

AB:AC=1:2=m:n.

A(6,3)=[2x1-1/2+1,2y1-4/2+1]

(6,3)=[2x1-1/3,2y1-4/3]

equating on both side,

2x1-1/3=6;2y1-4/3=3

2x1-1=18;2y1-4=9.

2x1=18+1;2y1-4=9+4

2x1=19;2y1=13

x1=19/2;y1=13/2.

AB:BD=2:1

by formula

(-1,-4)=[2x2+6/2+1,2y2+3/2+1]

(-1,-4)=(2x2+6/3,2y2+3/3)

equating on both sides.

2x3+6/3=-1;2y2+3/3=-4

2x2+6=-3;2y2+3=-12

2x2=-3-6;2y2=-12-3

2x2=-9;2y2=-15

x2=-9/2;y2=-15/2.

therefore C=(19/2,13/2)

D=(-9/2,-15/2)..

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