Question

1. Show that the triangle with vertices A (8,5). B (1, -2) and C(-3, 2) is a right angled triangle. Find its area.​

Answer 1

Answer:

Step-by-step explanation: (1)

Given vertices are A(3,4), B(2,-1) and C(4,-6).

We need to calculate the length of the sides AB,BC,AC.

We have to distance formula which can tell the distance between 2 points.

d = √(x2 - x1)^2 + (y2 - y1)^2.

(1)

AB = √(2 - 3)^2 + (-1 - 4)^2

    = √(-1)^2 + (5)^2

    = √1 + 25

    = √26.

(2)

BC = √(4 - 2)^2 + (-6 + 1)^2

    = √(2)^2 + (5)^2

    = √4 + 25

    = √29

(3)

AC = √(4 - 3)^2 + (-6 - 4)^2

    = √(1)^2 + (-10)^2

    = √1 + 100

    = √101

Therefore, the length of the sides of the triangle are √26,√29 and √101.

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(2)

Let the given points be A(2,-2), B(-2,1) and C(5,2).

Using the distance formula,w e find that

⇒ AB = √(-2 - 2)^2 + (1 + 2)^2

        = √16 + 9

        = √25.

⇒ BC = √(5 + 2)^2 + (2 - 1)^2

         = √49 + 1

         = √50.

⇒ AC = √(5 - 2)^2 + (2 + 2)^2

         = √9 + 16

         = √25.

     

Now,

⇒ AB^2 + AC^2

⇒ (5)^2 + (5)^2

⇒ 25 + 25

⇒ 50.

⇒ (BC)^2.

Therefore, AB^2 + AC^2 = BC^2.

.

Answer 2

Answer:

• 14 units²

Step-by-step explanation:

Plot the points.

According to the attached we can see the triangle is right.

Use distance formula to find the length of AB and BC:

• $AB=\sqrt{(1-8)^2+(-2-5)^2}=\sqrt{7^2+7^2}=7\sqrt{2}$
• $BC=\sqrt{(-3-1)^2+(2+2)^2}=\sqrt{4^2+4^2}=2\sqrt{2}$

Find the area:

• $A = 1/2(AB*BC)$
• $A = 1/2(7\sqrt{2}*2\sqrt{2}) = 14$