1 Coordinates: (2, 4)
2 Equation: y = 2x+10
3 Perpendicular equation: y = -0.5+5
4 They intersect at: (-2, 6)
5 Distance is the square root of: (-2, -2)2+(6, -4) = 2*sq rt of 5 = 4.472 to 3 decimal places
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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
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
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
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
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.