Find the point of the line passing through A (-1, -1) and B (4, 4) which is:

a. Twice as far from A as from B (2 cases)

b. 3 times as far from B as from A (2 cases)

I hope someone can help me with this problem, thank you.

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- Nov 24th 2009, 02:22 AMpedrobasureroDivision of a Line Segment Problem
Find the point of the line passing through A (-1, -1) and B (4, 4) which is:

a. Twice as far from A as from B (2 cases)

b. 3 times as far from B as from A (2 cases)

I hope someone can help me with this problem, thank you. - Nov 24th 2009, 03:20 AMstapel
Find the distance between the two points, using

**the Distance Formula**.

Find the slope of the line through the two points, using**the slope formula**.

Pick one of the points (it doesn't matter which) and, using also the slope you just found, find**the line equation**.

a) Since one distance is twice the other distance, then the length of the segment is divided into three parts, two of which go with point A and the other with point B. The unknown points are of the form (x, y) = (x, mx + b). Plug this form (using the values of m and b that you obtained above) into the Distance Formula, with the distance of (x, mx + b) from A being twice (times 2) the distance from B.

Square both sides, and solve the resulting equation for x. Plug these x-values into your line equation to obtain the corresponding y-values.

b) This one works just line (a), other than dividing the length of the segment by 4 rather than by 3. (Wink)

If you get stuck, please reply showing your steps and reasoning so far. Thank you! - Nov 24th 2009, 03:31 AMHallsofIvy
Actually, you don't need the distance formula! Think "similar triangles". Draw a right triangle with the line from A to B as hypotenuse and lines parallel to the x and y axes as legs. Extend the hypotenuse and mark your point "P", say, as the desired point that is "Twice as far from A as from B" on opposite side of B as A is. And draw legs parallel to the x and y axes. You have two triangles with the same angles and so "similar triangles". You should be able to see that the hypotenuse of the new triangle is twice as long as the hypotenuse of the original triangle so the its legs are twice as long as the corresponding legs on the original triangle. You can get the coordinates of the point from that.

The other case is a little more complicated. This point will lie**between**A and B. Draw the right triangle there. Since the point is "twice as far from A as from B", it lies**2/3**of the way from A to B. That means that the legs of the new triangle are 2/3 as long as those of the original triangle. Again, you can get the coordinates of the point from that. - Nov 24th 2009, 04:39 AMpedrobasurero
thank you stapel and HallsofIvy for the response, i made it and came to an answer of

1st case A

(9,9)

2nd case A

(7/3, 7/3)

@HallofIvy, now my problem the letter B, how will I answer this one? - Nov 24th 2009, 05:13 AMHallsofIvy
Good! Those are correct.

@HallofIvy, now my problem the letter B, how will I answer this one?[/QUOTE]

Same reasoning. Extending the line past A, in order that the distance from B be 3 times the distance from A, the distance from A to this new point, P, must be half the distance from B to A. That way the distance from B to P is 1+ 1/2= 3/2 the distance from B to A: and 3/2 is 3 times 1/2. That is, the hypotenuse on your new right triangle is 3/2 as long as the hypotenuse on the original triangle and so the legs must be 3/2 as long.

For the second case, the point between A and B, the point that is 1/4 of the way fro A to B divides the interval into**four**equal segments and so is "three times as far from B as it is from A".