A line with slope m has a perpendicular with slope m' such that:mm' = -1→ m' = -1/mThe line segment with endpoints (p, q) and (7p, 3q) has slope:slope = change in y / change in x→ m = (3q - q)/(7p - p) = 2q/6p = q/3p→ m' = -1/m = -1/(q/3p) = -3p/qThe perpendicular bisector goes through the midpoint of the line segment which is at the mean average of the endpoints:midpoint = ((p + 7p)/2, (q + 3q)/2) = (8p/2, 4q/2) = (4p, 2q)A line through a point (X, Y) with slope M has equation:y - Y = M(x - x)→ perpendicular bisector of line segment (p, q) to (7p, 3q) has equation:y - 2q = -3p/q(x - 4p)→ y = -3px/q + 12p² + 2q→ qy = 12p²q + 2q² - 3pxAnother Answer: qy =-3px +12p^2 +2q^2
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7p + 2q = 46 . . . . (A) 5p + 3q = 36 . . . . (B) 3*(A): 21p + 6q = 138 2*(B): 10p + 6q = 72 Subtracting gives 11p = 66 so that p = 6 Substitute for p in (A): 7*6 + 2q = 46 or 42 + 2q = 46 which gives 2q = 4 so that q = 2 Solution: (p, q) = (6,2)
15 + 3 = 7p - 4p ie 18 = 3p so p = 6
3p - 8 = 13 - 4p : Add 4p to both sides 7p - 8 = 13 : Add 8 to both sides 7p = 21 : Divide both sides by 7 p = 3.
Points: (p, q) and (7p, 3q) Midpoint: (4p, 2q) Slope: q/3p Perpendicular slope: -3p/q Perpendicular bisector equation:- => y-2q = -3p/q(x-4p) => qy-2q^2 = -3p(x-4p) => qy-2q^2 = -3px+12p^2 => qy = -3px+12p^2+2q^2 In its general form: 3px+qy-12p^2-2q^2 = 0
A line with slope m has a perpendicular with slope m' such that:mm' = -1→ m' = -1/mThe line segment with endpoints (p, q) and (7p, 3q) has slope:slope = change in y / change in x→ m = (3q - q)/(7p - p) = 2q/6p = q/3p→ m' = -1/m = -1/(q/3p) = -3p/qThe perpendicular bisector goes through the midpoint of the line segment which is at the mean average of the endpoints:midpoint = ((p + 7p)/2, (q + 3q)/2) = (8p/2, 4q/2) = (4p, 2q)A line through a point (X, Y) with slope M has equation:y - Y = M(x - x)→ perpendicular bisector of line segment (p, q) to (7p, 3q) has equation:y - 2q = -3p/q(x - 4p)→ y = -3px/q + 12p² + 2q→ qy = 12p²q + 2q² - 3pxAnother Answer: qy =-3px +12p^2 +2q^2
3p +12= 7p 7p-3p=12 12?3=4 p=4
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
-7p + 5r - 6p + 3r = -7p - 6p + 5r + 3r = -(7 + 6)p + (5 + 3)r = -13p + 8r
2p+4q=16 (now divide the equation by two) p+2q=8 (now subtract 2q) p=8-2q 7p+12q=52 (substitue the answer you got for p in the previous equation) 7(8-2q)+12q=52 (multiply the first equation by 7) 56-14q+12q=52 (subtract 14q from 12q) 56-2q=52 (subtract the 56 from 52) -2q=-4 (multiply by -1) 2q=4 (divide by 2) q=2 p=8-2q (substitute the value of q) p=8-2(2) (multiply) p=8-4 (subtract) p=4 2p+4q=16 (check your answers with the new values of p and q) 2(4)+4(2)=16 8+8=16 true 7p+12q=52 7(4)+12(2)=52 28+24=52 true
In its general form it works out as: 3px+qy-12p2-2q2 = 0Improved Answer:-Points:(p, q) and (7p, 3q)Slope: q/3pPerpendicular slope: -3p/qMidpoint: (4p, 2q)Equation: y-2q = -3p/q(x-4p) => yq = -3px+12p2+2q2Perpendicular bisector equation in its general form: 3px+qy-12x2-2q2 = 0
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
-3p
5p + 15p + 7p = 27p 5p - 15p + 7p = -3p 5p x 15p + 7p = 75p2 + 7p 5p/15p + 7p = 7p + 1/3
The perpendicular bisector of a line has a gradient (m') which is the negative reciprocal of the gradient (m) of the line, and passes through the mid-point of the line.The equation of a line with gradient m through a point (x0, y0) has an equation of the form: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)Thus the line through (p, q) and (7p, 3q) has gradient:m = (3q - q) / (7p - p) = 2q / 6p = q/3pand the perpendicular bisector has gradient:m' = -1 / m = -1 / (q/3p) = -3p/qThe midpoint of the line through (p, q) and (3p, 3q) is:midpoint = ((p + 7p)/2), (q + 3q)/2) = (4p, 2q)Thus the perpendicular bisector of the line between (p, q) and (7p, 3q) has equation:y - 2q = -3p/q (x - 4p)→ qy - 2q² = -3px + 12p²→ qy + 3px = 12p² + 2q²Additional Information:-Final answer in its general form: 3px+qy-12p^2-2q^2 = 0
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