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