Find the slope-intercept equation of the line that satisfies the given conditions. Then graph the line.
Contains points P=( -1,4);parallel to the line with equation 3x+y =5
y = mx + b
3x + y = 5
3x - 3x + y = -3x + 5
y = -3x + 5
m = -3
y - y1 = m(x -x1)
y - 4 = -3(x - (-1))
y -4 =-3(x+1)
y - 4 = -3x -3
y -4 + 4 = -3x -3 + 4
y = -3x + 1
To find the slope-intercept equation of a line parallel to the given line, we need to determine the slope of the given line.
The equation of the given line is 3x + y = 5. To write it in slope-intercept form (y = mx + b), where m is the slope and b is the y-intercept, we need to isolate y.
Start by subtracting 3x from both sides of the equation:
y = -3x + 5
Now we can see that the slope of the given line is -3.
Since the line we are looking for is parallel to the given line, it will have the same slope of -3.
To find the slope-intercept equation of the line passing through the point P(-1,4) with a slope of -3, we can substitute the point into the slope-intercept form (y = mx + b) and solve for b.
Using the point (-1, 4), we have 4 = -3(-1) + b.
Simplifying, we get 4 = 3 + b.
Subtracting 3 from both sides gives us b = 4 - 3 = 1.
Now we know that the slope (m) is -3 and the y-intercept (b) is 1.
Therefore, the slope-intercept equation of the line that satisfies the given conditions is y = -3x + 1.
To graph the line, use the slope-intercept equation to plot the y-intercept at (0, 1). Then use the slope (-3) to find one additional point on the line, and connect the two points to draw the line.