x + 2y = 5
2y = -x + 5
y = -x/2 + 5/2
The gradient of the given line is '-1/2' ; the coefficient of 'x'.
For parallel lines the gradient is the same at '-1/2'
Hence substituting
y - 2 = ( -x -3) / 2 + 5/2
Notice you displace the given point against the corresponding (x,y)
Hence
y - 2 = -x/2 - 3/2 + 5/2
y - 2 = -x/2 + 2/2
y = -x/2 + 2 + 2/2 NB 2/2 = 1
y = -x/2 + 3
Multiplying through by '2'
2y = -x + 6
In the original form it is
x + 2y = 6 is the parallel line.
Write an equation in slope-intercept form for the line that passes through the given point and is parallel to the given line (-7,3); x=4
Parallel straight line equations have the same slope but with different y intercepts
Another name for the Playfair Axiom is the Euclid's Parallel Postulate. It states that given a line and a point not on that line, there is exactly one line parallel to the given line passing through the given point.
To find the equation of a line parallel to another line, we need the same direction vector. The direction vector of the given line is (2, -3). Therefore, the equation of the line parallel to it passing through (-1, 3) is x = -1 + 2t and y = 3 - 3t, where t is a parameter.
Any equation parallel to the x-axis is of the form:y = constant In this case, you can figure out the constant from the given point.
Any equation parallel to the x-axis is of the form:y = constant In this case, you can figure out the constant from the given point.
The equation is x = -7.
Write the equation of a line in slope-intercept form that has a slope of -2 and passes through the point (2, -8).
Given point: (6, 7) Equation: 3x+y = 8 Parallel equation: 3x+y = 25
Assuming that you are given another equation and need to find another equation in slope-intercept form passing through a given point. --- Ex. Find the equation of a line passing through (2, 5) and parallel to y = 3x -7. 3x means the slope is 3 and parallel means the new slope will be the same.use y = mx + b, substitute the point (2, 5) for x and y and substitute 3 for the slope, m. you get:y = mx + b5 = 3(2) + b5 = 6 + b-1 = bNow reuse y = mx + b and substitute m = 3 and b = -1 but leave y and x as variables to get the answer:y = 3x -1There are other ways to do this so you probably need to see if this is the method that your teacher wants.
Without an equality sign the given terms can't be considered to be an equation but in general when lines are parallel they have the same slope but different y intercepts.
The straight line equation would depend on the slope which has not been given.