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Find the slope of a line perpendicular to x - y equals 16?
Slope of a line perpendicular to x-y=16
How do you write an equation that is perpendicular?
Line B is perpendicular to Line A if its slope is the negative reciprocal of the slope of Line A.When a linear equation is in the formy = mx + b,m is the slope, and b is the y-intercept. So, for example,y = (2/3)x + 5is perpendicular toy = (-3/2)x + 7.(The y-intercepts in these two equations are random numbers.)
Graph the line perpendicular to 7x plus 10y equals 4.5?
7x + 10y = 4.5 : 10y = -7x + 4.5 : y = -x.7/10 + 0.45, the gradient of this line is -7/10 Two straight lines are perpendicular if the product of their gradients is -1. Let the equation for the perpendicular line be y = mx + c Then m x -7/10 = -1 : m = 10/7 The equation for the perpendicular line is y = x.10/7 + c If the values of x and y for the point of intersection are provided then these can be substituted in the perpendicular line equation and the value of c obtained. If appropriate, the equation can then be restructured to a format similar to the original equation.
What would be the perpendicular line for y equals -x-5?
If the gradient of a line is m, the gradient of a line perpendicular to it (m') is such that mm' = -1, that is: m' = -1/m So for: y = -x - 5 m = -1, so the gradient of a perpendicular line is: m' = -1/-1 = 1 So a general line perpendicular to y = -x - 5 would be: y = x + c A specific line perpendicular to y = -x - 5 at point (xo, -xo-5) on the line is: y - (-xo-5) = 1(x - xo) ⇒ y = x - 2xo - 5
When a point lies on the y-axis what do you know about it's x-coordinate .?
The x-coordinate of any point on the y-axis is 0. The y-axis is a line perpendicular to the x-axis. Any point on a line perpendicular to the x-axis has the same x-coordinate. The y-axis is the line perpendicular to the x-axis through 0, and has the equation x = 0; similarly, the x-axis is the line perpendicular to the y-axis through 0 and has the equation y = 0.
Find the slope of a line perpendicular to x - y equals 16?
Slope of a line perpendicular to x-y=16
How do you write an equation that is perpendicular?
Line B is perpendicular to Line A if its slope is the negative reciprocal of the slope of Line A.When a linear equation is in the formy = mx + b,m is the slope, and b is the y-intercept. So, for example,y = (2/3)x + 5is perpendicular toy = (-3/2)x + 7.(The y-intercepts in these two equations are random numbers.)
Graph the line perpendicular to 7x plus 10y equals 4.5?
7x + 10y = 4.5 : 10y = -7x + 4.5 : y = -x.7/10 + 0.45, the gradient of this line is -7/10 Two straight lines are perpendicular if the product of their gradients is -1. Let the equation for the perpendicular line be y = mx + c Then m x -7/10 = -1 : m = 10/7 The equation for the perpendicular line is y = x.10/7 + c If the values of x and y for the point of intersection are provided then these can be substituted in the perpendicular line equation and the value of c obtained. If appropriate, the equation can then be restructured to a format similar to the original equation.
What would be the perpendicular line for y equals -x-5?
If the gradient of a line is m, the gradient of a line perpendicular to it (m') is such that mm' = -1, that is: m' = -1/m So for: y = -x - 5 m = -1, so the gradient of a perpendicular line is: m' = -1/-1 = 1 So a general line perpendicular to y = -x - 5 would be: y = x + c A specific line perpendicular to y = -x - 5 at point (xo, -xo-5) on the line is: y - (-xo-5) = 1(x - xo) ⇒ y = x - 2xo - 5
What is the slope intercept form of the equation of the line that is perpendicular to 7x - 8y equals 12 and contains the point P at -3 1?
The slope m' of a line perpendicular to a line with slope m is such that m'm = -1 → m' = -1/m A line with slope m' through point (X, Y) has equation: y - Y = m'(x - X) → The line perpendicular to 7x - 8y = 12 through p (-3, 1) is found: 7x - 8y = 12 → 8y = 7x - 12 → y = (7/8)x - 3/2 → The slope of 7x - 8y = 12 is m = 7/8 → The slope of the perpendicular line is m' = -1/m = -1/(7/8) = -8/7 → The line through P with this slope m' is: y - 1 = -8/7(x - -3) → y = (-8/7)x - 24/7 + 1 → y = (-8/7)x - 17/7 This can be rearranged into: y = (-8/7)x - 17/7 → 7y = -8x - 17 → 7y + 8x + 17 = 0
Is y equals -2 a parallel line or a perpendicular line?
y=-2 is parallel to the x-axis and perpendicular to the y-axis.
When a point lies on the y-axis what do you know about it's x-coordinate .?
The x-coordinate of any point on the y-axis is 0. The y-axis is a line perpendicular to the x-axis. Any point on a line perpendicular to the x-axis has the same x-coordinate. The y-axis is the line perpendicular to the x-axis through 0, and has the equation x = 0; similarly, the x-axis is the line perpendicular to the y-axis through 0 and has the equation y = 0.
What is the perpendicular bisector equation of the line y equals x plus 5 spanning the circle x2 plus 4x plus y2 -18y plus 59 equals 0?
Equation of line: y = x+5 Equation of circle: x^2 +4x +y^2 -18y +59 = 0 The line intersects the circle at: (-1, 4) and (3, 8) Midpoint of line (1, 6) Slope of line: 1 Perpendicular slope: -1 Perpendicular bisector equation: y-6 = -1(x-1) => y = -x+7 Perpendicular bisector equation in its general form: x+y-7 = 0
The line given by the equation y equals -8x is perpendicular to what line?
It would be perpendicular to a line with the equation Y = 1/8 X.
How do you find the midpoint the slope the perpendicular slope and the equation for the perpendicular bisector of the line segment joining the points of 3 5 and 7 7?
Midpoint = (3+7)/2, (5+7)/2 = (5, 6) Slope of line segment = 7-5 divided by 7-3 = 2/4 = 1/2 Slope of the perpendicular = -2 Equation of the perpendicular bisector: y-y1 = m(x-x1) y-6 =-2(x-5) y = -2x+10+6 Equation of the perpendicular bisector is: y = -2x+16
Which lines are perpendicular to the line y equals -2?
Any line of the form x = some_number. eg all the lines: x = -2 x = 0 x = 1.2 are perpendicular to y = -2.
How do you graph x equals 7?
You draw a vertical line at x=7. Does not matter what y value you pick, x is always 7. So find 7 on the x axis and draw a vertical line that is perpendicular to the x axis at the point (7,0)
