One line passes through the points (−12,16) and (-1, 38). A second line has the equation 2y+x=23. Are these lines parallel, perp
endicular, or neither? Discuss your solution strategy and then implement it to demonstrate your answer.
1 answer:
The lines are ...
perpendicular.
_____
The solution strategy is to use linear regression to find the line through the given points, then plot the given line on the same graph. The result from the linear regression shows the slope of the line between the points is 2. The given line has slope -1/2. The product of these slopes is -1, so the lines are perpendicular.
The equation of the line through the points can also be written using the 2-point form of the equation for a line.
y = (y2 -y1)/(x2 -x1)*(x -x1) +y1
y = (38 -16)/(-1 -(-12))*(x -(-12)) +16
y = 22/11*(x +12) +16
y = 2x +40 . . . . . . . . slope = 2, y-intercept = 40.
When the equation for the given line is rearranged to solve for y, you see the slope of it is -1/2.
2y +x = 23
2y = -x +23
y = (-1/2)x +23/2 . . . . . . . . . slope = -1/2, y-intercept = 23/2.
perpendicular.
_____
The solution strategy is to use linear regression to find the line through the given points, then plot the given line on the same graph. The result from the linear regression shows the slope of the line between the points is 2. The given line has slope -1/2. The product of these slopes is -1, so the lines are perpendicular.
The equation of the line through the points can also be written using the 2-point form of the equation for a line.
y = (y2 -y1)/(x2 -x1)*(x -x1) +y1
y = (38 -16)/(-1 -(-12))*(x -(-12)) +16
y = 22/11*(x +12) +16
y = 2x +40 . . . . . . . . slope = 2, y-intercept = 40.
When the equation for the given line is rearranged to solve for y, you see the slope of it is -1/2.
2y +x = 23
2y = -x +23
y = (-1/2)x +23/2 . . . . . . . . . slope = -1/2, y-intercept = 23/2.
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