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Answer 1

Answer:

Vedic Mathematics is a book written by the Indian monk Bharati Krishna Tirtha, and first published in 1965. It contains a list of mathematical techniques, which the author stated were retrieved from the Vedas and supposedly contained all mathematical knowledge.

Answer 2

$\large\underline{\sf{Solution-}}$

Let us consider a object as Square formed with four lines,

which are as follow :

$\rm :\longmapsto\:x + y = 1 - - - (1)$

$\rm :\longmapsto\: - x + y = 1 - - - (2)$

$\rm :\longmapsto\: - x - y = 1 - - - (3)$

$\rm :\longmapsto\: x - y = 1 - - - (4)$

Let plot the lines on the graph paper,

Consider Line (1)

$\rm :\longmapsto\:x + y = 1$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 + y = 1$

$\rm :\longmapsto\:y = 1$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x + 0 = 1$

$\rm :\longmapsto\:x = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points.

➢ See the attachment graph.

Consider Line (2)

$\rm :\longmapsto\: - x + y = 1$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\: - 0 + y = 1$

$\rm :\longmapsto\: y = 1$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\: - x + 0 = 1$

$\rm :\longmapsto\: - x= 1$

$\rm :\longmapsto\: x= - \: 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 1 \\ \\ \sf - 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.

Consider Line (3)

$\rm :\longmapsto\: - x - y = 1$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\: - 0 - y = 1$

$\rm :\longmapsto\: - y = 1$

$\rm :\longmapsto\: y = - \: 1$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\: - x - 0 = 1$

$\rm :\longmapsto\: - x = 1$

$\rm :\longmapsto\: x = - \: 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 1 \\ \\ \sf - 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points (0 , 4), (1 , 2) & (2 , 0)

➢ See the attachment graph.

Consider Line (4)

$\rm :\longmapsto\:x - y = 1$

Substituting 'x = 0' in the given equation, we get

$\rm :\longmapsto\:0 - y = 1$

$\rm :\longmapsto\: - y = 1$

$\rm :\longmapsto\: y = - \: 1$

Substituting 'y = 0' in the given equation, we get

$\rm :\longmapsto\:x - 0 = 1$

$\rm :\longmapsto\:x = 1$

Hᴇɴᴄᴇ,

➢ Pair of points of the given equation are shown in the below table.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf - 1 \\ \\ \sf 1 & \sf 0 \end{array}} \\ \end{gathered}$

➢ Now draw a graph using the points

➢ See the attachment graph.

Hence, the point of intersection of lines are as follow

Line 1 and Line 2 is A (0, 1)

Line 2 and Line 3 is B (- 1, 0)

Line 3 and Line 4 is C (0, - 1)

Line 4 and Line 1 is D (1, 0).