Q) Two rails are represented by the linear equations x + 2y – 4 = 0 and 2x + 4y – 12 = 0. Represent this situation geometrically.

Ans: 

Let’s try to find the intersection points on X – axis and Y – axis for each of the lines:

I. For linear equation x + 2y – 4 = 0:

For X – axis: y = 0 and x = 4 – 2 y = 4 – 2 x 0 = 4

∴ point on X – axis: (4,0)

For Y – axis: x = 0 and y = \frac{4 - x}{2} = \frac {4 - 0}{2} = 2

∴ point on Y – axis: (0,2)

II. For linear equation 2x + 4y – 12 = 0:

For X – axis: y = 0 and x = \frac{12 - 4 y}{2} = 6 -2y = 6 – 2 x 0 = 6

∴ point on X – axis: (6,0)

For Y – axis: x = 0 and y = \frac{12 - 2 x}{4} = \frac {6 - x}{2} = 3

∴ point on Y – axis: (0,3)


Q) Two rails are represented by the linear equations x + 2y - 4 = 0 and 2x + 4y - 12 = 0. Represent this situation geometrically.
Ans: 
Let's try to find the intersection points on X - axis and Y - axis for each of the lines:
I. For linear equation x + 2y - 4 = 0:
For X - axis: y = 0 and x = 4 - 2 y = 4 - 2 x 0 = 4
∴ point on X - axis: (4,0)
For Y - axis: x = 0 and y = <img decoding= = 2 ∴ point on Y – axis: (0,2) II. For linear equation 2x + 4y – 12 = 0: For X – axis: y = 0 and x = \frac{12 - 4 y}{2} = 6 -2y = 6 – 2 x 0 = 6 ∴ point on X – axis: (6,0) For Y – axis: x = 0 and y = \frac{12 - 2 x}{4} = \frac {6 - x}{2} = 3 ∴ point on Y – axis: (0,3) To represent the equations graphically, we plot the points P(0,3) and Q (6,0) to get the line PQ. Similarly, we plot the points R(0,2) and S (4,0) to get the line RS. Here, the lines do not intersect each other i.e. they are parallel” width=”300″ height=”255″ />

To represent the equations graphically, we plot the points P(0,3) and Q (6,0) to get the line PQ.

Similarly, we plot the points R(0,2) and S (4,0) to get the line RS.

Here, the lines do not intersect each other i.e. they are parallel

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