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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.
