Question
If A (20, 10), B(0, 20) are given, find the coordinates of the points which divide segment AB into five congruent parts.
✓
Answer
Let the points dividing AB be C,D,E,F
AC:CD:DE:EF:FB∷1:1:1:1:1
A point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
For C m:n :: 1:4
$x=\frac{(20) 4+(0) 1}{1+4}=16$
$y=\frac{10(4)+(20) 1}{1+4}=12$
For D m:n :: 2:3
$x=\frac{(20) 3+(0) 2}{2+3}=12$
$y=\frac{(10) 3+(20) 2}{2+3}=14$
For E m:n :: 3:2
$x=\frac{(20) 2+(0) 3}{2+3}=8$
$y=\frac{(10) 2+(20) 3}{2+3}=16$
For F m:n :: $4: 1$
$x =\frac{(20) 1+(0) 3}{1+4}=4$
$y =\frac{(10) 1+(20) 4}{1+4}=18$
AC:CD:DE:EF:FB∷1:1:1:1:1
A point P(x,y) divides the line formed by points (a,b) and (c,d) in the ratio of m:n, then the coordinates of the point P is given by
$x =\frac{ an + cm }{ m + n } \text { and } y =\frac{ bn + dm }{ m + n }$
For C m:n :: 1:4
$x=\frac{(20) 4+(0) 1}{1+4}=16$
$y=\frac{10(4)+(20) 1}{1+4}=12$
For D m:n :: 2:3
$x=\frac{(20) 3+(0) 2}{2+3}=12$
$y=\frac{(10) 3+(20) 2}{2+3}=14$
For E m:n :: 3:2
$x=\frac{(20) 2+(0) 3}{2+3}=8$
$y=\frac{(10) 2+(20) 3}{2+3}=16$
For F m:n :: $4: 1$
$x =\frac{(20) 1+(0) 3}{1+4}=4$
$y =\frac{(10) 1+(20) 4}{1+4}=18$
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