Question

The area of the triangle formed by the line y=X, X=5 and y=0

Answer 1

Answer:

Let equations of line

A : y - x = 0

B : x + y = 0

C : x = k , where k is constant .

first of all solve equations A and B

y - x = 0

x + y = 0

we get x = 0, and y = 0

hence, one point is ( 0, 0) of traingle .

again, solve equations B and C

x + y = 0

x = k

then, we get x = k And y = -k

so , ( k, - k) is 2nd point of traingle .

similarly solve equations C and A

y - x = 0 , x = k

we get ( k, k ) is third point of traingle.

now, we have three points

( 0, 0) , ( k, - k) and ( k, k)

we know, if three points of traingle are given then we have to use for finding area of traingle is 1/2 [x₁(y₂-y₃) + x₂(y₃-y₁) +x₃(y₁-y₂) ]

so, area of traingle = 1/2 [ 0( - k - k ) + k( k - 0) + k( 0 + k) ]

= 1/2 [ 0 + k² + k² ]

= 2k²/2

= k²

hence, area of traingle = k²

Answer 2

Answer:

12.5 sq.units.

Step-by-step explanation:

By solving the lines, we can get the vertices of the triangle as :

(00,);(5,0);(5,5).

Therefore, This is a right angled triangle, right angled at (5,0).

Here, Height = h = 5 units.

          Base = b = 5 units.

Therefore, area of the required triangle = $\frac{1}{2} bh$

                                                                   = $\frac{25}{2}$ = 12.5 sq.units.