coordinate geometry?

Question

can anybody help...?
A(3,6) B(7,4) and C(1,2) are the vertices of a triangle. Show that ABC is a right-angled isosceles triangle
thanx

Answer 1

In order to do this, two legs must be perpendicular to each other and have the same length.

Take the length of each side:
AB = √[(4-6)² + (7-3)²]
AB = √20

BC = √[(2-4)² + (1-7)²]
BC = √40

AC = √[(2-6)² + (1-3)²]
AC = √20

AB and AC are the two legs, and have equal lengths. Now we must prove that the two are perpendicular, thus forming a right angle, and making ABC a right triangle.

m(AB) = -1/m(AC)
(4-6)/(7-3) = -(1-3)/(2-6)
-2/4 = 2/-4
-1/2 = -1/2
True.
Therefore, these segments are on perpendicular lines, making angle A a right angle, and triangle ABC a right triangle.

Since triangle ABC is a right triangle with two legs of equal length, ABC is an isosceles right triangle.

Answer 2

AB² = (7 - 3)² + (4 - 6)² = 16 + 4 = 20
AC² = (3 - 1)² + (6 - 2)² = 4 + 16 = 20
BC² = (1 - 7)² + (2 - 4)² = 36 + 4 = 40

BC² = AB² + AC²

Triangle is a right angled and isosceles triangle in which AB = AC and angle BAC = 90°

Answer 3

1) An isosceles triangle has (at least) two equal sides.

2) On this triangle, AB = AC = sqrt(20); therefore, this is an isosceles triangle

3) Is it a right-angled isosceles triangle? The answer is yes; because, the angle between AB and AC is 90 degrees

4) One way to prove 3) is to show that the scalar (aka dot) product of the vectors AB and AC equal zero. If the dot product of any two non-zero vectors is zero, then, the angle between them must be 90 degrees.
a) Vab = vector from A to B = 4i - 2j
b) Vac = vector from A to C = -2i - 4j
c) dot(Vab, Vac) = 0;

5) There are other ways to prove 3) w/o using vectors. If the height of this isosceles triange equals half the base then the angle between AB and AC is 90 degrees.

6) The given triangle is a right-angled isosceles triangle.

Answer 4

AB is sqrt(16+4) = sqrt20 units long.
AC is sqrt(4+16) = sqrt20 units long.
BC is sqrt(36+4) = sqrt40 units long.

AB is congruent to AC, and the sum of the squares of these two is 40, which is the length of BC squared. QED.

You can also prove it like this: Let D=(1,6) and E=(7,6). Then triangles ADC and BEA are congruent (right triangles, legs 2 and 4 units long). This proves AC=AB.

Note that angles DAC and DCA are complementary. Since

Answer 5

the length of AB is the square
root of the sum of the difference
in coordinates squared of the
points A and B

hence,
AB=sqrt{(3-7)^2+(6-4)^2}
=sqrt{(-4)^2+(2)^2}
=sqrt20
likewise,
AC=sqrt{(3-1)^2+(6-2)^2}
=sqrt(2^2+4^2}
=sqrt20
finally,
BC=sqrt{(7-1)^2+(4-2)^2}
=sqrt{6^2+2^2}
=sqrt40

now AB=AC, therefore,
triangle ABC is isosceles

but, if triangle ABC is a right-angled
triangle,(BC)^2 should equal the sum
of(AB)^2+(AC)^2{pythagoras}

(AB)^2+(AC)^2
=(sqrt20)^2+(sqrt20)^2
=20+20=40

now,(BC)^2=(sqrt40)^2=40

therefore,
(BC)^2=(AB)^2+(AC)^2
hence,triangle ABC is
right-angled

we have therefore shown that
triangle ABC is a right-angled
isosceles triangle

i hope that this helps

Answer 6

AB = sqrt((7-3)^2 +(4-6)^2) = 2sqrt(5)
AC = sqrt((1-3)^2 +(2-6)^2 = 2sqrt(5)
BC = sqrt(1-7)^2 + (2-4)^2 = 2sqrt (10)
AB = AC because they both = 2sqrt(5)

Does AB^2 +AC^2 = BC^2?
(2sqrt(5))^2 + (2sqrt(5))^2 =? (2sqrt(10))^2
20 + 20 =? 40 Yes it does.

So the triangle is isosceles becaues AB = AC and it is a right triangle because AB^2 +AC^2 = BC^2

Answer 7

OKay, first to show its right-angled.
The slope of AC is (2-6)/(1-3) = -4/-2 = 2.
And the slope of AB is (4-6)/(7-3) = -2/4 = -1/2.
Now, since slope of AC is the negative inverse of AB, that means its right-angled.

Now Distance of AC is sqr((2-6)^2+(1-3)^2)) = sqr(16+4)=sqr(20).
And distance of AB is sqr((4-6)^2+(7-3)^2) = sqr(4+16)=sqr(20).
So, since AC equals AB, that means its isoceles.