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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
Does a graph showing direct proportion always have to be a straight line?
Assuming both the scales on the graph are linear (that is to say that the numbers go up evenly) then YES, a graph which shows direct proportion must be a straight line. It must also pass through the origin (0,0). A straight line which does not pass through the origin is NOT showing direct proportion. Duncan
What is the perpendicular bisector equation of the line joined by the points 11 13 and 17 19 showing work and answer?
The equation of a line through point (x0, y0) with gradient m is given by:y - y0 = m(x - x0)The gradient (m) of a line between two points (x0, y0) and (x1, y1) is given by:m = change_in_y/change_in_x = (y1 - y0)/(x1 - x0)→ The equation of the line between (11, 13) and (17, 19) is given by:y - 13 = (19-13)/(17-11) (x - 11)→ y - 13 = 6/6 (x - 11)→ y - 13 = x - 11→ y = x + 2and its gradient is m = 1.The gradient (m') of a line perpendicular to a line with gradient m is such that mm' = -1, ie m' = -1/m→ The gradient of the perpendicular line to the line between (11, 13) and (17, 19) has gradient m' = -1/1 = -1.The perpendicular bisector goes through the point midway between (11, 13) and (17, 19) which is given by the average of the x and y coordinates: ((11+17)/2, (13+19)/2) = (14, 16)Thus the perpendicular bisector of the line joining (11, 13) to (17,19) is given by:y - 16 = -1(x - 14)→ y - 16 = -x + 14→ y + x = 30Which in its general form is: x+y-30 = 0
What is the equation of the tangent line that touches the circle x squared plus 10x plus y squared -2y -39 equals 0 at the point of 3 2 on the Cartesian plane showing work?
First find the slope of the circle's radius as follows:- Equation of circle: x^2 +10x +y^2 -2y -39 = 0 Completing the squares: (x+5)^2 + (y-1)^2 -25 -1 -39 = 0 So: (x+5)^2 +(y-1)^2 = 65 Centre of circle: (-5, 1) and point of contact (3, 2) Slope of radius: (1-2)/(-5-3) = 1/8 which is perpendicular to the tangent line Slope of tangent line: -8 Tangent equation: y-2 = -8(x-3) => y = -8x+26 Tangent equation in its general form: 8x+y-26 = 0
What does a hexagon and an octagon have alike?
Well, darling, a hexagon and an octagon are both polygons. They both have straight sides and angles. The only real difference is that a hexagon has six sides while an octagon has eight sides. So, in a nutshell, they're both just a bunch of straight lines showing off their angles.
What is the perpendicular distance from the coordinates of 7 and 5 to the straight line of 3x plus 4y -16 equals 0 showing key stages of work?
Points: (7, 5) Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Equations intersect at: (4, 1) Length of perpendicular line: 5
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
What is the perpendicular distance from the point of 2 4 on the Cartesian plane to the straight line equation of y equals 2x plus 10 showing key stages of work?
1 Equation: y = 2x+10 2 Perpendicular equation works out as: 2y = -x+10 3 Point of intersection: (-2, 6) 4 Distance is the square root of: (-2-2)2+(6-4)2 = 2 times sq rt of 5
What is the perpendicular distance from the point 7 5 to the staight line equation 3x plus 4y -16 equals 0 showing key stages of work?
Equation: 3x+4y-16 = 0 Perpendicular equation: 4x-3y-13 = 0 Both equations intersect at: (4, 1) Perpendicular distance: square root of (7-4)2+(5-1)2 = 5
What is the perpendicular distance from the point 19 29 that meets the straight line equation 3y equals 9x plus 18 on the Cartesian plane showing work and answer to an appropriate degree of accuracy?
If: 3y = 9x+18 then y = 3x+6 with a slope of 3 Perpendicular slope: -1/3 Perpendicular equation: y-29 = -1/3(x-19) => 3y = -x+106 Both equations intercept at: (8.8, 32.4) Perpendicular distance: square root of (8.8-19)^2+(32.4-29)^2 = 10.75 rounded
What is the perpendicular distance from the point 4 -2 to the line 2x -y -5 equals 0 showing key stages of work?
Points: (4, -2) Equation: 2x-y-5 = 0 Perpendicular equation: x+2y = 0 Equations intersect at: (2, -1) Perpendicular distance is the square root of: (2-4)2+(-1--2)2 = 5 Distance = square root of 5
What is the equation of a straight line that cuts through the middle of the points of -1 3 and -2 -5 at right angles on the Cartesian plane showing work?
The equation will be a perpendicular bisector equation of the given points:- Points: (-1, 3) and (-2, -5) Midpoint: (-3/2, -1) Slope: 8 Perpendicular slope: -1/8 Perpendicular equation: y--1 = -1/8(x--3/2) => y = -1/8x-3/16-1 Therefore the perpendicular bisector equation is: y = -1/8x -19/16
What is the perpendicular distance from the point of 2 4 to the equation y equals 2x plus 10 on the Cartesian plane showing work?
Equation: y = 2x+10 Point: (2, 4) Perpendicular slope: -1/2 Perpendicular equation: y-4 = -1/2(x-2) => 2y = -x+10 Both equations intersect at: (-2, 6) Using distance formula: (2, 4) to (-2, 6) = 2 times square root of 5
What is the perpendicular distance from the point 7 5 that meets the line of 4 1 and 0 4 at right angles on the Cartesian plane showing key aspects of work?
Points: (4, 1) and (0, 4) Slope: -3/4 Equation: 4y = -3x+16 Perpendicular slope: 4/3 Perpendicular equation: 3y = 4x-13 Both equations meet at: (4, 1) from (7, 5) at right angles Perpendicular distance: square root of [(4-7)squared+(1-5)squared)] = 5 units
What is the distance from the point 5 7 that is perpendicular to to the straight line equation of 3x-y plus 2 equals 0 on the Cartesian plane showing key aspects of work?
To find the perpendicular distance from a given point to a given line, find the equation of the line perpendicular to the given line which passes through the given point. Then the distance can be calculated as the distance from the given point to the point of intersection of the two lines, which can be calculated by using Pythagoras on the Cartesian coordinates of the two points. A line in the form y = mx + c has gradient m. If a line has gradient m, the line perpendicular to it has gradient m' such that mm' = -1, ie m' = -1/m (the negative reciprocal of the gradient). A line through a point (x0, y0) with gradient m has equation: y - yo = m(x - x0) Thus the equation of the line through (5, 7) that is perpendicular to 3x - y + 2 = 0 can be found. The intercept of this line with 3x - y + 2 = 0 can be calculated as there are now two simultaneous equations. → The perpendicular distance from (5, 7) to the line 3x - y + 2 = 0 is the distance form (5, 7) to this point of interception, calculated via Pythagoras: distance = √((change_in_x)^2 + (change_in_y)^2) This works out to be √10 ≈ 3.162
What is rhe perpendicular distance from the point 4 8 to the line of y equals 2x plus 10 on the Cartesian plane showing key aspects of work with an answer to an appropriate degree of accuracy?
Equation: y = 2x+10 and slope is 2 Perpendicular slope: -1/2 Perpendicular equation: 2y = -x+20 Both equations intersect at: (0, 10) from (4, 8) Distance: square root of (0-4)^2 plus (10-8)^2 = 4.472 to three decimal places
What is the equation and its perpendicular distance from the point -3 -5 whose line meets the line of 0 5 to 3 0 at its midpoint on the Cartesian plane showing work?
Points: (0, 5) and (3, 0) Midpoint: (1.5, 2.5) Slope: -5/3 Perpendicular slope: 3/5 Perpendicular equation: y--5 = 3/5(x--3) => 5y = 3x-16 Distance is the square root of (1.5--3)^2+(2.5--5)^2 = 8.746 to three decimal places
