Question 2:
We have that point C(6,10) partitions A(3,5) and B(a,b) in the ratio m:n=1:3
We use the section formula:
[tex](x,y)=(\frac{mx_2+nx_1}{m+n}, \frac{my_2+ny_1}{m+n})[/tex]
We substitute the values to get:
[tex](6,10)=(\frac{1*a+3*3}{1+3}, \frac{1*b+3*5}{1+3})[/tex]
We simplify to get:
[tex](6,10)=(\frac{a+9}{4}, \frac{b+15}{4})[/tex]
This implies that:
[tex](6=\frac{a+9}{4},10= \frac{b+15}{4})[/tex]
[tex](24=a+9,40= b+15)[/tex]
[tex](a=15,b=25)[/tex]
The coordinates of B are (15,25)
Question 3:
This time we have G(3,2) and H(15,8), and we want to find C(x,y) that partitions GH in a m;n=4:1 ratio:
We apply the section formula to get:
[tex](x,y)=(\frac{4*15+1*3}{4+1}, \frac{4*8+1*2}{4+1})[/tex]
[tex](x,y)=(\frac{60+3}{5}, \frac{32+2}{5})[/tex]
[tex](x,y)=(\frac{63}{5}, \frac{34}{5})[/tex]
Question 4
We want to find the coordinates of the midpoint of the line segment with endpoints B (9, -8) and C (2, -5),
The midpoint is given by:
[tex](x,y)=(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})[/tex]
We substitute the endpoints to obtain:
[tex](x,y)=(\frac{9+2}{2}, \frac{-1+-5}{2})[/tex]
This simplifies to:
[tex](x,y)=(\frac{11}{2}, \frac{-6}{2})[/tex]
The midpoint is:
[tex](5\frac{1}{2}, -3)[/tex]
Question 5
A directed line segment has length and direction. That means there is a beginning point and an endpoint.
Partitions can occur on a directed line segments. This does not mean that all directed line segment must be partitioned.
Ans: False
