i need help on math question please?...?

Question

Show that the points A( 6,-1 ) , B( 0,2 ) , and C( -2,3 ) lie on a straight line.

Ok, now i know, to get the answer, you have to use the formula of the area of the triangle ( Under the topic coordinate of geometry ) and if the answer is 0, then it means that the three points lies on a straight line.

But i just can't get the answer to 0.

Answer 1

Area = (1/2)(y1(x2-x3) + y2(x3-x1) + y3(x1-x2))
=(1/2)((-1)(0-(-2)) + 2(-2-6) + 3(6-0))
=(1/2)(-2 - 16 + 18)
=(1/2)(-16 + 16)
=(1/2)(0)
=0

Answer 2

You can determine that the 3 points lie on the same line by showing that the area of the "triangle" ABC = 0, but that's not the easiest way.

Two methods a), and b) co-ordinate geometry

a) Take the slope between A and C

(3-(-1)) / (-2-6) = 4/-8 = -1/2

The equation to the line passing through A nd C is

y = (-1/2)x +b

plugging in (6,-1) gives

-1 = -6/2 +b so b = 2

so equation is

y = -x/2 +2

now plug x = 0, which gives y =2

So the point B = (0,2) is on the line too.

b) Co-ordinate geometry

Use distance formula to compute distance AB, BC and AC.

AB+BC = AC, That is:

sqrt((6-0)^2+(-1-2)^2) + sqrt((0-2)^2+(2-3)^2) =

sqrt((6-(-2))^2 +(-1-3)^2)

sqrt(36+9) +sqrt(4+1) = sqrt(64+16) True

6.708+2.236 = 8.944

Answer 3

u can approach this problem in many ways. one of the ways to do it is to prove that the area of the triangle formed with the three points as vertices is 0. u will get the area to be zero. u can get the area easily using the determinant method.
the method is as follows

let (x1,y1),(x2,y2),(x3,y3) be the vertices of a triangle
then the area is given by
A= 1/2* |x1 x2 x3 x1|
|y1 y2 y3 y1|
the expansion is 1/2*(x1*y2-y1*x2+x2*y3-y2*x3+x3*y1-y3*x1)

substituting the given vertices, it is
1/2* |6 0 -2 6|
|-1 2 3 -1|
on expanding it, we get 1/2*(12-0+0+4+2-18)=1/2*(18-18)=0
so, we just proved that the area of the triangle is zero. so, they lie on a straight line.

the above method to find the area can be extended to any number of vertices, ie., it can be applied to all the polygons.

another method is to find the equation of the line passing through any two points and to prove that the third point also lies on the same line.
for eg., take (6,-1) and (0,2)
the equation of line is x+2y-4=0
substitute the third point (-2,3)
-2+2*3-4=-6+6=0
so, it satisfies the equation. so, it lies on the same line, which make the three points collinear.

Answer 4

you look for the equation of a line passing on 2 of the points suppose B and C
the equationof a straight line is y = a x + b
with the point b you have 2 = 0*x +b hence b=2
with the point C we have 3 = a * (-2) + 2 hence a = -0.5

the straight line is y = -0.5*x +2
if you put 6 in that equation you find y = -1
and you see that this point A is on the straight line

Answer 5

u can use the formula for area of triangle for points (x1,y1) (x2,y2) (x3,y3)
area = ( -x2*y1 + x3*y1 + x1*y2 - x3*y2 - x1*y3 + x2*y3) / 2
thus area = ( -(-1*0) + (-2)*(-1) + 6*2 - (-2)*2 - 6*3 + 0*3)
= 0 + 2 + 12 - (-4) -18 + 0
= 14 + 4 -18
=0
Since area of triangle = 0
the points lie on straight line.

Answer 6

You could find the gradients of the line segments AB and AC, then provided the gradients are the same, ABC must be a straight line since AB and AC would be parallel and also go through same point (A).

I will leave you to do the calculations yourself but I have checked it and it works.

Answer 7

you should draw it. After that you can show it using the similarity of the triangles.

Answer 8

the formula is abs of (x1-x2)(x1-x3)-(y1-y2)(y1-y3)=6*8-(-3)(-4)

so theres a mistake in question

Answer 9

use your distance formula from point A-B, from point B-C and from point C-A... and then u can use the area of the triangle..