Please help. I've been trying to solve these for more than an hour now but I can't seem to get it.

1) The segment joining (-4,7), (5,-2) is divided into two segments, one of which is 5 times as long as the other. Find the point of division.

2) The segment joining (4,0), (3, -2) is extended each way a distance equal to three times its own length. Find the terminal points.

Please show the solutions so I can study them. Thanks in advance :).

1)

Naturally, there are two places to make the division: one near (-4,7) and the other near (5,-2)

Taking the one near (-4,7), we want a point which is 1/6 the way to (5,-2).

(-4 + 1/6 (5+4),7 + 1/6 (-2-7)) = (-5/2,11/2)

2)
in the same way as above, figure the difference between the points and go out 3 times that distance in each direction:

(4,0) + 3((4-3),(0+2)) = (4,0)+(3,6) = (7,6)

(3,-2) + 3((3-4),(-2-0)) = (3,-2)+(-3,-6) = (0,-8)

x= -5/2

y= 11/2

#1.

x=-4 + 5/6(5+4)=7/2
y=7 + 5/6(-2-7)=-1/2
P(7/2,-1/2)

#2.

x=4 + 3(4-3)=7
y=0 + 3(0+2)=6
p(7,6)
x=3 + 3(3-4)=0
y=-2 + 3(-2-0)=-8
p(0,-8)

Sure, I'd be happy to help you solve these problems. Let's start with the first one:

1) The segment joining (-4,7), (5,-2) is divided into two segments, one of which is 5 times as long as the other. We need to find the point of division.

To solve this problem, we can use the concept of the midpoint formula. The midpoint of a line segment is the point that is equidistant from both endpoints.

Let's denote the point of division as (x,y), where x represents the x-coordinate and y represents the y-coordinate. By applying the midpoint formula, we can set up the following equation:

(x-coordinate of midpoint) = (average of x-coordinates of endpoints)
y = (average of y-coordinates of endpoints)

Using the given coordinates of the endpoints, we have:

(x + (-4))/2 = (-4 + 5)/2 (for the x-coordinate)
y = (7 + (-2))/2 (for the y-coordinate)

Simplifying these equations, we get:

(x - 4)/2 = 1/2 (for the x-coordinate, after simplification)
y = 5/2 (for the y-coordinate)

To solve for x, we can cross-multiply and simplify the equation:

x - 4 = 1
x = 1 + 4
x = 5

Therefore, the x-coordinate of the point of division is 5. Substituting this value into one of the original equations, we can solve for y:

(y - 7)/2 = 1/2
y - 7 = 1
y = 1 + 7
y = 8

So, the y-coordinate of the point of division is 8.

Thus, the point of division is (5, 8).

Moving on to the second problem:

2) The segment joining (4,0), (3,-2) is extended each way a distance equal to three times its own length. We need to find the terminal points.

To solve this problem, we can use the concept of slope and distance formula.

First, let's find the length of the line segment by using the distance formula:

Length = sqrt((x2 - x1)^2 + (y2 - y1)^2)

Substituting the given coordinates:
Length = sqrt((3 - 4)^2 + (-2 - 0)^2)
Length = sqrt((-1)^2 + (-2)^2)
Length = sqrt(1 + 4)
Length = sqrt(5)

The length of the line segment is sqrt(5).

Given that the line segment is extended in both directions by three times its own length, we can calculate the new endpoints by using the slope formula:

Slope = (y2 - y1)/(x2 - x1)

The slope of the line segment is:

m = (-2 - 0)/(3 - 4)
m = (-2)/(-1)
m = 2

Now that we have the slope and one point on the line segment, we can calculate the new endpoints by extending the segment three times its length in both directions.

Starting from the first point (4,0):

For the extension in the positive direction, we move three times the length of the segment in the positive direction. Since the slope is positive, we move upwards. The new point is:

(x, y) = (4, 0) + 3 * (slope)
= (4, 0) + 3 * (2, 1)
= (4, 0) + (6, 3)
= (10, 3)

For the extension in the negative direction, we move three times the length of the segment in the negative direction. Since the slope is positive, we move downwards. The new point is:

(x, y) = (4, 0) - 3 * (slope)
= (4, 0) - 3 * (2, 1)
= (4, 0) - (6, 3)
= (-2, -3)

Therefore, the terminal points are (10, 3) and (-2, -3).

I hope this explanation helps you understand how to solve these types of problems.