You have two variables which without further information can not tell you about the other. Without saying what they equal you can't even give possible solutions.
simply add the like terms: 4p -3p +2q -2q = p
4p+7q
16p + 6q - 3p - 8q = (16p - 3p) + (6q - 8q) = 13p -2q
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
3
simply add the like terms: 4p -3p +2q -2q = p
4p+7q
Let f(X)=2X2+6X+3 So f(-p)=f(2q) or 2p2-6p+3=8q2+12q+3 or p2-3p=4q2+6q or p2-4q2=3p+6q or (p+2q)(p-2q)=3(p+2q) so p-2q=3
16p + 6q - 3p - 8q = (16p - 3p) + (6q - 8q) = 13p -2q
Yes. 9p2 + 12pq + 4q2 = (3p + 2q)2
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
It is a string of algebraic terms.
3
3p
3p
-(b + c - p - 2q)(b + c + p + 2q)
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