Draw the graph of each of the following linear equations in two variables:
(i) x + y = 4
(ii) x - y = 2
(iii) -x + y = 6
(iv) y = 2x
(v) 3x + 5y = 15
(vi) $\frac{x}{2}-\frac{y}{3}=2$
(vii) $\frac{x-2}{3}=y-3$
(viii) 2y = -x +1
(i) We are given, x + y = 4
We get, y = 4 - x,
Now, substituting x = 0 in y = 4 - x,
We get y = 4
Substituting x = 4 in y = 4 - x, we get y = 0
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given table
(ii) We are given, x - y = 2
We get, y = x – 2
Now, substituting x = 0 in y= x – 2, we get y = - 2
Substituting x = 2 in y = x – 2, we get y = 0
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
(iii) We are given, - x + y = 6
We get, y = 6 + x
Now, substituting x = 0 in y = 6 + x,
We get y = 6
Substituting x = - 6 in y = 6 + x, we get y = 0
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation.
(iv) We are given, y = 2x
Now, substituting x = 1 in y = 2x, we get y = 2
Substituting x = 3 in y = 2x, we get y = 6
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
(v) We are given, 3x + 5y = 15
We get, 15 - 3x = 5y
Now, substituting x = 0 in 5y = 15 - 3x,
We get; 5y = 15 y = 3
Substituting x = 5 in 5y = 15 – 3x we get 5y = 0
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
(vi) we are given.
$\frac{x}{2}-\frac{y}{3}=2$
$\frac{3 x-2 y}{6}=2$
3x - 2y = 12
We get,
$\frac{3 x-12}{2}=y$
We get y = - 6
Substituting x = 4 in
$\frac{3 x-12}{2}=y$
We get y = 0 Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
(vii) We are given,
$\frac{x-2}{3}=y-3$
We get, x - 2 = 3(y - 3)
x - 2 = 3y - 9
x = 3y - 7
Now, substituting x = 5 in x = 3y - 7,
We get; y = 4
Substituting x = 8 in x = 3y - 7,
We get; y = 5
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation
(viii) We are given, 2y = - x +1
We get, 1 – x = 2y
Now, substituting x = 1 in 1– x = 2y, we get y = 0
Substituting x = 5 in 1 – x = 2y, we get y = - 2
Thus, we have the following table exhibiting the abscissa and ordinates of points on the line represented by the given equation