Question

1 given the end - points a=(5, - 2) and b=(0,3) of line segment ab, what are the coordinates of the point p that divides segment ab in the ratio 3:2 from point b?
my calculations
a=(5, - 2),b=(0,3). = 3:2
x_1 = 5,y_1=-2,x_2 = 0,y_2 = 3,m = 3,n = 2
x=\frac{3\times5 + 2\times0}{3 + 2}=\frac{15}{5}=3
y=\frac{3\times(-2)+2\times3}{3 + 2}=\frac{-6 + 6}{5}=0
p=(3,0)
2 given the end - points a=(2,1) and b=(6,9) of line segment ab, what are the coordinates of the point p that divides segment ab in the ratio 3:1 from point a?
my calculations
a=(2,1),b=(6,9)=3:1
x_1 = 2,y_1 = 1,x_2 = 6,y_2 = 9,m = 3,n = 1
x=\frac{3\times6+1\times2}{3 + 1}=\frac{18 + 2}{4}=5
y=\frac{3\times9+1\times1}{3 + 1}=\frac{27 + 1}{4}=7
p=(5,7)
3 if the point p is located \frac{2}{3} of the way along segment ab from point a, what are the coordinates of point p if a=(-3,4) and b=(6,1) are the end - points of segment ab?
my calculations
a=(-3,4)b=(6,1)=2:1
x_1=-3,y_1 = 4,x_2 = 6,y_2 = 1,m = 2,n = 1
x=\frac{2\times6+1\times(-3)}{2 + 1}=\frac{12-3}{3}=3
y=\frac{2\times1+1\times4}{2 + 1}=\frac{2 + 4}{3}=2
4 if the point p is located \frac{1}{4} of the way along segment ab from point b, what are the coordinates of point p if a=(5, - 1) and b=(-3,7) are the end - points of segment ab?
5 a point p=(7,1) divides segment ab in a ratio of 2:1 from end - point a=(5,3). what are the coordinates of point b?
6 a point p=(4,2) divides segment ab in a ratio of 1:3 from end - point b=(6,3). what are the coordinates of point a?

Explanation:

Step1: 분할 비율 공식 정의

선분을 $m:n$ 비율로 분할하는 점의 좌표 공식은 $x=\frac{mx_2+nx_1}{m + n}$, $y=\frac{my_2+ny_1}{m + n}$입니다. 여기서 $(x_1,y_1)$과 $(x_2,y_2)$는 선분의 두 끝점 좌표입니다.

Step2: 4번 문제 풀이

$A=(5,-1)$, $B=(-3,7)$이며, $P$는 $B$로부터 $AB$의 $\frac{1}{4}$ 지점에 있습니다. 즉, $BP:PA = 1:3$, $m = 1$, $n = 3$, $x_1=-3$, $y_1 = 7$, $x_2=5$, $y_2=-1$입니다.
$x=\frac{1\times5+3\times(-3)}{1 + 3}=\frac{5 - 9}{4}=\frac{-4}{4}=-1$
$y=\frac{1\times(-1)+3\times7}{1 + 3}=\frac{-1 + 21}{4}=\frac{20}{4}=5$

Step3: 5번 문제 풀이

$A=(5,3)$, $P=(7,1)$이며, $AP:PB = 2:1$, $m = 2$, $n = 1$, $x_1=5$, $y_1 = 3$, $x = 7$, $y = 1$입니다.
$x=\frac{mx_2+nx_1}{m + n}$에서 $7=\frac{2x_2+1\times5}{2 + 1}$, $21=2x_2+5$, $2x_2=16$, $x_2 = 8$
$y=\frac{my_2+ny_1}{m + n}$에서 $1=\frac{2y_2+1\times3}{2 + 1}$, $3=2y_2+3$, $2y_2=0$, $y_2 = 0$

Step4: 6번 문제 풀이

$B=(6,3)$, $P=(4,2)$이며, $BP:PA = 1:3$, $m = 1$, $n = 3$, $x_1=6$, $y_1 = 3$, $x = 4$, $y = 2$입니다.
$x=\frac{mx_2+nx_1}{m + n}$에서 $4=\frac{1\times x_2+3\times6}{1 + 3}$, $16=x_2+18$, $x_2=-2$
$y=\frac{my_2+ny_1}{m + n}$에서 $2=\frac{1\times y_2+3\times3}{1 + 3}$, $8=y_2+9$, $y_2=-1$

Answer:

4번: $P=(-1,5)$
5번: $B=(8,0)$
6번: $A=(-2,-1)$