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What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?
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
What is 7p more than 3p?
10
What is 4p plus 15 equals 7p-3?
15 + 3 = 7p - 4p ie 18 = 3p so p = 6
What is the answer to this question 7p plus 2q equals 46 and 5p plus 3q equals 36?
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)
What is 3p-8 equals 13-4p p equals?
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.
What is the perpendicular bisector equation of the line joined by the points of p q and 7p 3q on the Cartesian plane?
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
What is the perpendicular bisector equation of a line segment with endpoints of p q and 7p 3q?
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
How much is 7p more than 3p?
To find out how much 7p is more than 3p, subtract 3p from 7p. This gives you 7p - 3p = 4p. Therefore, 7p is 4p more than 3p.
What is the answer to 7p-12 equals 3p?
3p +12= 7p 7p-3p=12 12?3=4 p=4
-7p plus 5r-6p plus 3r?
-7p + 5r - 6p + 3r = -7p - 6p + 5r + 3r = -(7 + 6)p + (5 + 3)r = -13p + 8r
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
What is p and q when 2p add 4q is 16 and 7p add 12q is 52?
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
What is the perpendicular bisector equation of the line segment of p q and 7p 3q?
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
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
What is 5p-15p plus 7p in algebra?
-3p
What is 5p 15p plus 7p in algebra?
5p + 15p + 7p = 27p 5p - 15p + 7p = -3p 5p x 15p + 7p = 75p2 + 7p 5p/15p + 7p = 7p + 1/3
What is 7p more than 3p?
10
