1. ## Lines

Hey guys, just need help with lines today:

1. Calculate the distance between each of the following pair of points. (You may give your answer with the square root sign if necessary).

a) A (1,3) and B (5,6)
b) C (2,12) and D (14,7)
c) E (8,0) and F (0,6)
d) O (0,0) and H (9,-12)
e) I (8,2) and J (-2,-3)
f) K (-1,-3) and L (-8,1)
g) M (0,-5) and N (0,-1)

2. Find the perimeter of triangle ABC correct to 1 decimal place if the
coordinates of the vertices are

a) A (7,-9), B (0,0) and C (-2,-3)
b) A (-1,-1), B (-6,2) and C (3,-2)

3. If the vertices of triange PQR are P (-3,0), Q (1,-1) and R (0,3), show that triangle PQR is isosceles.

4. The points A (1,6), B (-4,3) and C (4,1) are the vertices of a triangle.

a) show that AB $^2$+AC $^2$=BC $^2$
b) Is triangle ABC a right-angled triangle?

5.
a) Given three points A (0,2), B (2,5) and C (4,8), calculate the lengths of AB, BC and AC.
b) Do the points A, B and C lie on the same straight line? Give reason.

2. Originally Posted by Rocher
1. Calculate the distance between each of the following pair of points. (You may give your answer with the square root sign if necessary).

a) A (1,3) and B (5,6)
b) C (2,12) and D (14,7)
c) E (8,0) and F (0,6)
d) O (0,0) and H (9,-12)
e) I (8,2) and J (-2,-3)
f) K (-1,-3) and L (-8,1)
g) M (0,-5) and N (0,-1)
This isn't much more than "busy work."

Given any two points $(x_1, y_1) \text{ and } (x_2, y_2)$ the distance between them is
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

-Dan

3. Originally Posted by Rocher
2. Find the perimeter of triangle ABC correct to 1 decimal place if the
coordinates of the vertices are

a) A (7,-9), B (0,0) and C (-2,-3)
b) A (-1,-1), B (-6,2) and C (3,-2)
This is similar to 1. Find the distance between points A and B, the distance between points B and C, and the distance between points C and A. Then add these three distances to get your perimeter.

-Dan

4. Originally Posted by Rocher
3. If the vertices of triange PQR are P (-3,0), Q (1,-1) and R (0,3), show that triangle PQR is isosceles.
By definition, an isosceles triangle is formed when at least two sides of the triangle are equal. So find the lengths PQ, PR, and QR. If any two of them are equal, then the triangle is isosceles.

-Dan

5. Originally Posted by Rocher
4. The points A (1,6), B (-4,3) and C (4,1) are the vertices of a triangle.

a) show that AB $^2$+AC $^2$=BC $^2$
b) Is triangle ABC a right-angled triangle?
Part a) should be old hat to you by now. Calculate the distances and plug them in.

For b) if the relation in a) holds, then the triangle must be a right triangle, since this relationship only holds for right triangles. (This is the Pythagorean theorem.)

-Dan

6. Originally Posted by Rocher
5.
a) Given three points A (0,2), B (2,5) and C (4,8), calculate the lengths of AB, BC and AC.
b) Do the points A, B and C lie on the same straight line? Give reason.
Again, part a) ought to come easily by now.

If A, B, and C were on the same line, then we must have that AB + BC = AC. (The sum of the two smaller distances must equal the largest distance.) Make sure you understand why this has to be so.

-Dan

7. Originally Posted by Rocher
, that's confusing...
I'm posting too many at the same time. Which one is too confusing? (And please watch your language!)

-Dan

8. The word I used was not a bad word XD. But nonetheless, what is confusing is the first post you made.

9. Do you think you can show me question A in #1? I don't get it...

10. Originally Posted by Rocher
a) A (1,3) and B (5,6)
Originally Posted by topsquark
Given any two points $(x_1, y_1) \text{ and } (x_2, y_2)$ the distance between them is
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Originally Posted by Rocher
Do you think you can show me question A in #1? I don't get it...
A (1,3) and B (5,6)
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

$d = \sqrt{(5 - 1)^2 + (6 - 3)^2}$

$d = \sqrt{4^2+3^2}$

$d = \sqrt{16+9}$

$d = \sqrt{25}$

$d = 5$

Don't over think it, it's just the Pythagorean theorem. Each set of two points can be represented with a right triangle, where the length between the two points is the hypotenuse, the change in y is the height, and the change in x is the length. So i's just $a^{2}+b^{2}=c^{2}$

If you have a hard time seeing this, try graphing the two points, then drawing a line straight down from the higher point, and then straight across (left or right) from the lower point to intersect the line created by the higher point, you will see it has created a right triangle. Then you can calculate the distance between the two by squaring the x distance and the y distance to get the hypotenuse squared. Then you just square root that to get the hypotenuse.