Line A is perpendicular to line B. If the slope of line A is multiplied by 4, what must the slope of line B be multiplied by in order for the lines to still be perpendicular?
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Solution
The slopes of perpendicular lines are negative reciprocals of each other. Therefore, if the slope of line A is m, the slope of line B is –1⁄m. If the slope of line A is multiplied by 4, it becomes 4m. The negative reciprocal of 4m is –\frac{1}{4m}. To change the original slope of line B, –1⁄m, to this new slope,–\frac{1}{4m}, you must multiply by 1⁄4.
Which of the following is the product of the slopes of perpendicular lines?
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Solution
The slopes of perpendicular lines are negative reciprocals of each other. Therefore, if the slope of one line is m, the slope of the other line is –1⁄m. The product of these slopes is (m)(–1⁄m) = –1.
What is the midpoint of a line with endpoints at (2x + 3,y – 4) and (10x – 1,3y + 6)?
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Solution
The midpoint of a line is equal to the average of the x values of the endpoints and the average of the y values of the endpoints: =\left ( \frac{2x+3+10x-1}{2},\frac{y-4+3y+6}{2} \right )=\left (\frac{12x+2}{2},\frac{4y+2}{2} \right ) (6x + 1,2y + 1).
Two perpendicular lines intersect at the point (1,5). If the slope of one line is 3, what is the equation of the other line?
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Solution
The slopes of perpendicular lines are negative reciprocals of each other. The slope of the other line is –1⁄3. The equation of the other line is y =–1⁄3(x) + b, where b is the y-intercept of the line.Use the point (1,5) to find the y-intercept; 5 =–1⁄3(1) + b, b = \frac{16}{3}. Therefore, the equation of the line is y = –1⁄3 + \frac{16}{3}.
What is the distance from the point (–x,y) to the point (x,–y)?
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Solution
To find the distance between two points, use the distance formula:
D = \sqrt{((x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2})}
D = \sqrt{((x-(-x))^{2}+((-y)-y)^{2})}
D = \sqrt{((2x)^{2}+(-2y)^{2})}
D = \sqrt{(4x^{2}+4y^{2})}
D = \sqrt{4(x^{2}+y^{2})}
D = 2\sqrt{(x^{2}+y^{2})}