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Does an equation allow you to find the x and y coordinates of any point on the xy plane?
Yes if it is a straight line equation
What is the formula used to find slope?
The slope of a straight line equation is: y2-y1/x2-x1
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
How do you find the equation of the perpendicular line bisecting the line segment of -2 5 and -8 -3?
if u don't now then i don't nowImproved answer as follows:-First find the mid-point of (-2, 5) and (-8, -3) which is (-5, 1)Then find the slope or gradient of (-2, 5) and (-8, -5) which is 4/3The perpendicular slope is the negative reciprocal of 4/3 which is -3/4So the perpendicular bisector passes through (-5, 1) and has a slope of -3/4Use y -y1 = m(x -x1)y -1 = -3/4(x- -5)y = -3/4x-11/4 which can expressed in the form of 3x+4y+11 = 0So the equation of the perpendicular bisector is: 3x+4y+11 = 0
Area of a trapezoid from vertices?
Use the coordinates of the vertices to establish which two sides are parallel.Find the lengths of the two parallel sides (X and Y).Find the equation of a perpendicular to one of these lines at a point P.Find the point where this perpendicular line meets the other parallel line (Q).Find the distance PQ = H.Area = 1/2*(X + Y)*H
