for line segment PQ 1 endpoint is p(-2,-6) & midpoint is M(-4,1) find coordinates of endpoint Q explain please
371443
The midpoint of two points P(a,b) and Q(x,y) is\nM((a+x)/2,(b+y)/2). That is, you add the x coordinates together and divide by 2 and then add the y-coordinates together and divide by 2.\nUsing P(-2,-6), Q(x,y) and M(-4,1), you have the two equations\n(-2+x)/2 = -4 and (-6+y)/2 = 1\nSolving for x and y you get\nx = -6 and y = 8\nThus, the other point is (-6,8).
790996
if cuts the x axis when p(x) = 0\n\nthat is 2x-3 = or x= 3/2\n\n2nd point x-5 =0 or x = 5\n\n3rd point x+1 = 0 or x = -1
19244
(Y-y1) = m(X-x1)\non point (4,7) => Y-7 = m(X-4)  (I)\non point (2,5) => Y-5 = m(X-2)  (II)\n\n(I&II) ==> m = 1
876684
y = -x + 6 is in slope intercept form, y = mx + b, where m is the slope.\nSo, you know that the slope of this line is -1. A perpendicular line has the opposite converse slope, so since this slope is -1/1, the slope of the new line will be positive 1/1, or 1.\n\nNow that you have the slope of the new line, you can use that and the point-slope form of a line to find the equation. The general form of a point slope line is y - Y = m(x-X), where m is the slop, Y is the y coordinate of a point on the line, and X is the x coordinate of the same point.\nSo plug in the point and slope to this form and you have:\n\ny - (-4) = 1(x - (-2))\nor\ny + 4 = x + 2\n\nYou can leave the equation in this form, or you can rearrange it so that it is in slope intercept form again.\n\ny = x - 2\n\nI hope this helped!
883980
find the mid pointof any side by using the formula (x1+x2/2,y1+y2/2)\nusing that and the opposite vertex using the two point formula (y-y1/y2-y1)=(x-x1/x2-x1) find the equation\ni shall do one for you\nbid point of BC=(0,-2)\npoint A=(0,7)\naquation of median AD=(y+2)/(7+2)=(x-0)/(0-0)\n0(y+2)=9(x-0)\n9x=0 is th equation of median AD
726789
What are the co-ordinates of the other point? just use the midppint formula for a segment. All the best!\n\nLife's Lovely! Love & Live Life!
400503
since each one gives you a *-----* i would say all of them are line segments.\n\nHowever since D. is the only one with an indefinite beginning and end and not a possible end or beginning, i would definately have to go with \n\nD.) -5 <= x <= 5\n\nThat is the only one that has a defined beginning and end and not a undefined end and/or beginning.
415594
R u saying about the midpoint on x and y axis?\n\nWe know that the midpoint is the half of a line length. mathematically..\n               \n               note that 1/2 is the definition of half\n               midpoint= the length of the line/2\n\non x axis, the midpoint(mp) is\n                  mp=(x2-x1)/2\n             \n        x2-x1 is the length of a line in x axis\n\nthe same goes to y axis, we get\n                 mp=(y2-y1)/2\n\n\nCombine both equations,( since mp is a coordinate), we get\n             mp= ( mp of x axis , mp of y axis )\n             mp=( (x2-x1)/2   ,   (y2-y1)/2)   )
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Coordinate Geometry\nThe Distance Between two Points\nThe length of the line joining the points (x1, y1) and (x2, y2) is:\n\nsqure root of (x1-x2)^2+(y1-y2)^2\n\nExample\nFind the distance between the points (5, 3) and (1, 4).\n(So in this case, x2 = 1, x1 = 5, y2 = 4 and y1 = 3).\n\nDistance = Ö[(1 - 5)2 + (4 - 3)2]\n= Ö[16 + 1] = Ö17\n\nThe Midpoint of a Line Joining Two Points\nThe midpoint of the line joining the points (x1, y1) and (x2, y2) is:\n\n[½(x1 + x2), ½(y1 + y2)] \n \nExample\nFind the coordinates of the midpoint of the line joining (1, 2) and (3, 1).\n\nMidpoint = [½(3 + 1), ½(2 + 1)] = (2, 1.5)\n\n \nThe Gradient of a Line Joining Two Points\nThe gradient of the line joining the points (x1, y1) and (x2, y2) is:\ny2 - y1\nx2 - x1\n\nExample\nFind the gradient of the line joining the points (5, 3) and (1, 4).\nGradient = 4 - 3 =  1 =  -0.25\n               1 - 5    -4\n\n \nParallel and Perpendicular Lines\nIf two lines are parallel, then they have the same gradient.\n\nIf two lines are perpendicular, then the product of the gradients of the two lines is -1.\n\nExample\na) y = 2x + 1\nb) y = -½ x + 2\nc) ½y = x - 3\n\nThe gradients of the lines are 2, -½ and 2 respectively. Therefore (a) and (b) and perpendicular, (b) and (c) are perpendicular and (a) and (c) are parallel.\n\nThe Equation of a Line Using One Point and the Gradient\nThe equation of a line which has gradient m and which passes through the point (x1, y1) is:\n\ny - y1 = m(x - x1) \nExample\nFind the equation of the line with gradient 2 passing through (1, 4).\n\ny - 4 = 2(x - 1)\ny - 4 = 2x - 2\ny = 2x + 2\n\nSince m = y2 - y1/\n              x2 - x1\n\nThe equation of a line passing through (x1, y1) and (x2, y2) can be written as:\ny - y1 / y2 - y1=\nx - x1 /   x2 - x1\n\n\n--------------------------------------------------------------------------------
137113
Draw the two points (a,b) and (c,d) on a graph so that the segment connecting them is NOT horizontal or vertical. Then make a right triangle with one vertical leg and one horizontal leg where the connecting segment is the hypotenuse.  \n\nThe length of the horizontal leg is c-a\nThe length of the vertical leg is d-b\n\nSo the length of the hypotenuse, which is the distance from (a,b) to (c,d), is\n\nsquare root of (c-a)^2+(d-b)^2
859683
Find equation of the lines then solve them 2 get point of intersection.
745771
first find the midpoint of AB which is C(9,-2).\nNow obviously the one-fourth distance will be the midpoint of AC which is ...well I let you do this. This is simple knowing the formula for the midpoint .\nMathguy did well too but you are supposed to apply, I think, midpoint formula.
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considering both y values are -2 for A and C....\n\nwe know that a straight line exists between these two points, which means that it HAS to be 1 of the 4 sides.\n\ngiven that, the coordinates or K and L HAVE to be (3, infinity (EXCEPT -2)) and (-4, infinity (EXCEPT -2))
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http://www.citlink.net/~math/geometry/ch1/1-4_notes.pdf#search='segment%20addition%20postulate'
685032
plz tell me wat is meant by x.wat angle is it.as i understood ur question QT is a line outside the figure & perpendicular to PQ.hence the angle should be 180 degrees
838763
if y=0\n\n2x+4(0)=-8\n2x+0=-8\nx=-4\n\n\nif x=0\n\n2(0)+4y=-8\n0+4y=-8\ny=-2\n\nIf your doing co-ordinate geometery, two points on the line are (0,-2) and (-4,0)
819556
Hey, here's the setup for these types of problems.\nYou want the equation in the form of y = mx + b\nsince we know that x = 2, y = -7, and m = -1/5, we can plug these values into that equation to find the equation of the line that passes through the desired point. SO:\n-7 = (-1/5)(2) + b\n-7 = (-2/5) + b\nb = (-7)+(2/5)\nb = -33/5\nso the equation for the line is:\ny = (-1/5)x - (33/5)\n\nhope that helps!!