There are an infinity of lines passing through the point whose coordinates are (2,2), each with a different slope [gradient]. The equation of the line will be of the form (y - 2) = m*(x - 2) where m is the gradient.
A linear equation.
Rearranging the original equation, we get y=-(2/3)x+12. Since 12 is the constant, this is the point that the line of this equation will cut the y-axis if x=0. Therefore, -(2/3) is the gradient and for an equation to produce a parallel line, the gradient must be equal. Summing up, y=-(2/3)+c (where c equals any real number) would be parallel
15
Change the number in front of the X, as that is the gradient.
There are an infinity of lines passing through the point whose coordinates are (2,2), each with a different slope [gradient]. The equation of the line will be of the form (y - 2) = m*(x - 2) where m is the gradient.
A linear equation.
Rearranging the original equation, we get y=-(2/3)x+12. Since 12 is the constant, this is the point that the line of this equation will cut the y-axis if x=0. Therefore, -(2/3) is the gradient and for an equation to produce a parallel line, the gradient must be equal. Summing up, y=-(2/3)+c (where c equals any real number) would be parallel
An equation such as y = mx + c is said to be in standard form. From such an equation, Gradient = coefficient of x = 3
yes
That would be a linear equation.
3
If you have the equation, yes. If the equation is given in terms of x and y, make y the subject of the equation. That is, expres the equation in the form y = mx + c where m and c are constants. Then the gradient is m.
If necessary, rearrange the linear equation so that it is in the slope-intercept form: y = mx + c Then the gradient of the line is m.
2y= 3x+6
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1