Solution:

s

A(2,1)\(\rightarrow\)A’(-1,2)

B(4,3)\(\rightarrow\)B’(-3,4)

C(1,4)\(\rightarrow\)C’(-4,1)

So, the coordinates of the vertices of image \(\triangle\)A’B’C’ are A(-1,2), B(-3,4) and C(-4,1).

 

Solution:

Here, when input is 1, output = 4\(\rightarrow\)1 + 3 \(\rightarrow\) input + 3

When input is 2, output = 5\(\rightarrow\)2 + 3 \(\rightarrow\)input +3

When input is 3, output = 6\(\rightarrow\)3 + 3\(\rightarrow\)input+3

Similarly, when input is x, output = x+3

\(\therefore\) The required relation is y = x + 3.

Now, when x = 4, y = 4+3 = 7

When x= 5, y = 5+3 = 8

When x = 6, y = 6+3 = 9

When x = 7, y = 7+3 = 10.

Solution:

Here, the relation is y = 2x – 1

When x = 1, y = 2×1-1 =`1; when x = 2, y = 2×2-1 = 3

When x = 3, y = 2×3-1 = 5; when x = 4, y = 2×4-1 = 7

Input (x) 1 2 3 4
Output (y) 1 3 5 7