anyone willing to help....
1.find the distance between the points A(7,2) and B(4,3)
2.Find the length of the line segment AB joining th points A(4,6) and B(5,8).
3.find the coordinates of bisection of the line segment PQ joining the points P(6,-2) and Q(3,9).
4.find the coordinates of trisection of the line segment BR joining the coordinates B(7,0) and R(5,-6)
Answers
Answer:
Find the distance along the y-axis. For the example points (3,2) and (7,8), in which (3,2) is Point 1 and (7,8) is Point 2: (y2 - y1) = 8 - 2 = 6. This means that there are six units of distance on the y-axis between these two points. Find the distance along
Step-by-step explanation:
1.d=√(x2-x1)²+(y2-y1)²
AB=√(4-7)²+(3-2)²=√(-3)²+1²=√9+1=√10 units.
2.d=√(x2-x1)²+(y2-y1)²
AB=√(5-4)²+(8-6)²=√1²+2²=√1+4=√5 units.
3.bisection=mid-point.
p(x,y)=[x1+x2/2,y1+y2/2]
p(x,y)=[6+3/2,9-2/2]=(9/2,7/2).
4. In trisection: the two ratios will be 1:2 and 2:1.
p(x,y)=[m1x2+m2x1/m1+m2, m1y2+m2y1/m1+m2]
1st point, p(x,y) with ratio 1:2 will be:
p(x,y)=[1(5)+2(7)/1+2, 1(-6)+2(0)/1+2]
=[5+14/3, -6/3]
=[19/3, -2]
2nd point, p(x,y) with ratio 2:1 will be:
p(x,y)=[2(5)+1(7)/2+1, 2(-6)+1(0)/2+1]
=[10+7/3,-12/3]
=[-17/3,-4]