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Find the equation of a straight line perpendicular to the line 4y-3x12?
Wiki User
∙ 15y ago
There are infinitely many lines perpendicular to this line. All of them have the slope of -4/3, if that fact is of any help to you.
Wiki User
∙ 15y ago
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Does an equation allow you to find the x and y coordinates of any point on the xy plane?
Yes if it is a straight line equation
What is the formula used to find slope?
The slope of a straight line equation is: y2-y1/x2-x1
How do you form an equation for the perpendicular bisector of the line segment joining the points of p q and 7p 3q showing all details of your work?
First find the midpoint the slope and the perpendicular slope of the points of (p, q) and (7p, 3q) Midpoint = (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope = (3q-q)/(7p-p) = 2q/6p = q/3p Slope of the perpendicular is the negative reciprocal of q/3p which is -3p/q From the above information form an equation for the perpendicular bisector using the straight line formula of y-y1 = m(x-x1) y-2q = -3p/q(x-4p) y-2q = -3px/q+12p2/q y = -3px/q+12p2/q+2q Multiply all terms by q and the perpendicular bisector equation can then be expressed in the form of:- 3px+qy-12p2-2q2 = 0
How do you find the equation of the perpendicular line bisecting the line segment of -2 5 and -8 -3?
if u don't now then i don't nowImproved answer as follows:-First find the mid-point of (-2, 5) and (-8, -3) which is (-5, 1)Then find the slope or gradient of (-2, 5) and (-8, -5) which is 4/3The perpendicular slope is the negative reciprocal of 4/3 which is -3/4So the perpendicular bisector passes through (-5, 1) and has a slope of -3/4Use y -y1 = m(x -x1)y -1 = -3/4(x- -5)y = -3/4x-11/4 which can expressed in the form of 3x+4y+11 = 0So the equation of the perpendicular bisector is: 3x+4y+11 = 0
Area of a trapezoid from vertices?
Use the coordinates of the vertices to establish which two sides are parallel.Find the lengths of the two parallel sides (X and Y).Find the equation of a perpendicular to one of these lines at a point P.Find the point where this perpendicular line meets the other parallel line (Q).Find the distance PQ = H.Area = 1/2*(X + Y)*H
What is the equation of the straight line that passes through (82) and (8 -3)?
Find an equati find an equation for the line perpendicular to the line 8x - 8 y equals negative 2 having the same Y intercept as -6x + 2 y equals negative 8
What is 2a equals b?
2a = b Is an example of an equation with linear dependence between the variable a and b (b is twice a)If you know any a you can find the bIf you graph this equation with a on one axis and b on the other (perpendicular) you will get a straight line
Find the slope and y - intercept of the linear equation 5x-10 equals -20?
5x - 10 = -20This equation can be restated as 5x = -10 : x = -2This is the equation of a straight line perpendicular to the x axis and passing through the point x = -2. There is no y intercept and the slope is indeterminate.
What is the perpendicular distance from the point of 7 and 5 that meets the straight line equation of 3x plus 4y equals 0 on the Cartesian plane showing work?
Equation: 3x+4y = 0 => y = -3/4x Perpendicular slope: 4/3 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (2.08, -1.56) Distance from (7, 5) to (2.08, -1.56) = 8.2 units using the distance formula
How do you find the straight line equation that passes through the point 3 -4 and is perpendicular to the line 5x -2y equals 3?
The equation 5x -2y = 3 is the same as y = 2.5x -1.5 The perpendicular slope or gradient is the negative reciprocal of 2.5 which is minus 1/2.5 To find the perpendicular equation use y -y1 = m(x -x1) and the point (3, -4) y - (-4) = -1/2.5(x -3) y = -1/2.5x +6/5 -4 y = -1/2.5x -14/5 which can be rearranged in the form of 2x +5y +14 = 0
Find the equation of the line perpendicular to -9x 2y-3 that contains the poins 3-1?
"http://wiki.answers.com/Q/Find_the_equation_of_the_line_perpendicular_to_-9x_2y-3_that_contains_the_poins_3-1"
How do you find the equation oa straight line?
It depends on what information you have to start with.
Find the equation of the line perpendicular when the slope is zero?
when the slope is 0, the graph is a horizontal line on the x axis so the y axis is perpendicular to it, which can be written x=0
How do you work out an equation for the perpendicular bisector of the line segment AB when A is at -4 8 and B is at 0 -2?
First find the midpoint of the line segment AB which is: (-2, 3) Then find the slope of AB which is: -5/2 The slope of the perpendicular bisector is the positive reciprocal of -5/2 which is 2/5 Then by using the straight line formula of y-y1 = m(x-x1) form an equation for the perpendicular bisector which works out as:- y-3 = 2/5(x-(-2)) y = 2/5x+4/5+3 y = 2/5x+19/5 => 5y = 2x+19 So the equation for the perpendicular bisector can be expressed in the form of:- 2x-5y+19 = 0
An equation allows you to find the x- and coordinates of any point on the xy - plane?
Yes if it is a straight line equation
How do you work out and find the perpendicular bisector equation meeting the straight line segment of p q and 7p 3q?
First find the mid-point of the line segment which will be the point of intersection of the perpendicular bisector. Then find the slope or gradient of the line segment whose negative reciprocal will be the perpendicular bisector's slope or gradient. Then use y -y1 = m(x -x1) to find the equation of the perpendicular bisector. Mid-point: (7p+p)/2 and (3q+q)/2 = (4p, 2q) Slope or gradient: 3q-q/7p-p = 2q/6p = q/3p Slope of perpendicular bisector: -3p/q Equation: y -2q = -3p/q(x -4p) y = -3px/q+12p2/q+2q Multiply all terms by q to eliminate the fractions: qy = -3px+12p2+2q2 Which can be expressed in the form of: 3px+qy-12p2-2q2 = 0
How do you form an equation for the perpendicular bisector of the line segment joining -2 5 to -8 -3?
First find the midpoint of (-2, 5) and (-8, -3) which is (-5, 1) Then find the slope of (-2, 5) and (-8, -3) which is 4/3 Slope of the perpendicular bisector is the negative reciprocal of 4/3 which is -3/4 Now form an equation of the straight line with a slope of -3/4 passing through the point (-5, 1) using the formula y-y1 = m(x-x1) The equation works out as: 3x+4y+11 = 0
