what is the equation of the line that has a slope  of two-thirds and passes the origin?
341751
Using pt. slope equation:\n\nY=mX+B where m=slope = 2/3, b=intercept, in this case = 0,\n\nY = 2/3 * x + 0\n\ny=2x/3.\n\n3y=2x\n3y-2x = 0.\n2x-3y=0
610276
points (3,2) and (-3,-10) belong to f and the equation for f is\ny-y1=[(y2-y1)/(x2-x1)]*[x-x1]\nwhere y1=2,y2=-10,x1=3,x2=-3\n\ny-2=[(-10-2)/(-3-3)]*[x-3]\n\ny-2=[-12/-6]{x-3]\n\ny-2=2[x-3]\n\ny=2+2x-6\n\ny=2x-4\n\nso the slope of f is 2\n\nsince g is perpendicular to f the slope is -1/2\n\nand equation would be\n\ny-9=-1/2(x+4)\n\ny=9-[1/2]x-2\n\ny=-(1/2)x+7\n\nso g(x)=-(1/2)+7
953083
let your given point be P(a,b,c...) and your slope be a direction given by a vector V(v,w,x,....).  I'm assuming you want a vector equation of a line?  You can get to any point on the line by starting at P and moving some distance m (maybe negative) in the direction V.  So any point R(r,s,t,...) on the line is given by\n\nR = P + mV, where m is some real number.\n\nFor example, in 2 dimensions, the line y = (2/3)x + 1 has a y-intercept at (0,1).  Let that be P.  The slope 2/3 is the vector (3,2).  The point (2, 7/3) is on the line, and the equation\n\n(2, 7/3) = (0,1) + m(3,2) works when m = 2/3.  m/√13 is the distance you go along the line from P to R.
820512
Y=MX+B\nso identify which number is next to the X, that will be your slope.  The other number is the Y-intercept.\n\nFirst problem:\nY=-2X+3\nthe -2 is next to the X so it is the slope (M)\nthe 3 is by itself so it is the Y-intercept (B)\nthe other two problems are the same way
128403
general equation of a line with slope m and y-intercept c is \n\ny = mx + c ....... (1)\n\nsay this line is parallel to 2x+3y+4=0 which means the slopes of two lines are equal\n\nso, m = -2/3  [-(x-coefficient/y-coefficient)]\n\nand c = 6 (given) \n\nsubstituting m and c in (1) \n\nrequired equation is y = -(2/3)x + 6\n\n=> 2x+3y-18=0
1075301
find the slope, m, of your line (-2/3 in this case). \n\nThe slope of any perpendicular line will be -1/m (3/2 in this case).\n\nSo for any real number b,\n. . . . . . . y = (-1/m)x + b = (3/2)x + b\nis the equation of a line perpendicular to the original line.
715413
To find the y-intercept to the line satisfying the equation y = (2/3)x + 3, we set x = 0 to obtain y; that is,\ny = (2/3)(0) + 3 = 3.\n\nSo, the y-intercept is 3. (i.e., The given equation of the line passes through the point (0, 3) in the Cartesian plane. )
743897
gradient of the new line is -1/4\nso y=- 1/4x-c\nso using (2,3) 3 = (-1/4*2)-c\n3 = -1/2 - c\n3 and1/2 = -c\nc = -3 and 1/2\n\nso equation of line is:\n\ny = -1/4x + 3.5\n\nco-ordinates of C:\n when y=0\n\nso 0 = -1/4x + 3.5\n-3.5 = -1/4x\n3.5 = 1/4x\n14=x\n\nso co-ordinates of C = (14,0)\n\nHope this helps
354769
y=1/3\nin fact, slope is zero because you don't have x\ny intercept is (0,1/3)\nthat's the line which passes through 1/3 on the y axis and is parallel to x axis
179699
The equation of a line is y=mx+c\nHere when x = 6, m = 4/3, y = -2\nTherefore c = y-mx = -2 - (4/3)*6 = -10\n\nHence y = (4/3)x-10\n\nHope this helps!
600555
y = mx + b\n\n"m" is the slope, which is the rise over the run.  y2-y1 / x2/1 = (1-3)/(3-2) = -2/1\nSo, m = -2\n\nPlug in a set of coordinates to find b...\n3 = -2(2) + b\nb = 7\n\nSo, the equation is y = -2x + 7\n\nIn yout case, you want it as Ax+By+C = 0\nThis would be 2x + y - 7 = 0
316654
The midpoint of that segment is (3, -2) - all you do is take the average of the x coordinates and the average of the y coordinates.\n\nThe slope of that segment is found by\n\nm = (y2 - y1)/(x2 - x1)\n= (1 - (-5))/(4 - 2)\n= 6/2\n= 3\n\nSince you're looking for a line perpendicular to this one, take the slope, invert it, and change the sign.  So the slope of the new line will be -1/3.\n\nNow you have a point and a slope for the new point: m = -1/3, and point (3, -2).  Use the point-slope formula to find the equation of the line:\n\ny - y1 = m(x - x1)\ny - (-2) = (-1/3)(x - 3)\ny + 2 = (-1/3)x + 1\ny = (-1/3)x - 1
32903
if you are given points (ex. (2,1) and (3,5) you can use this formula:\nm (slope) = y2-y1 / x2-x1\nso: 5 - 1 / 3-2 = 4/-1 or -4\n\nif you are given the equation of a line you can use this:\nx - ay = b\ny = -x + b / -a\nex.x-3y = 15\n-3y = -x + 15\ny = x/3 + 15\nso the slope is:\nx/3 = 1/3\nm(slope) = 1/3\nalways look for the coefficient of x and that is the slope.\n\nyou can also use this when you are given an angle:\nm (slope) = tan alpha
753590
Basically, its a straight line going through (1,3), with gradient, m=-3. we have y=mx+c. putting the point (1,3) into here, \n\n3=(-3)(1)+c, so c=6, which is the y-intercept. so the line equation is,\n\ny=-3x + 6, which is a straight line with y-intercept at y=6. the line will be sloping downwards since the gradient (m) is negative. if you plot a few values of x and y which satisfy the equation you can see the line. to check axis intercepts, put y=0 for the x-intercept and x=0 for the y-intercept.
63789
A line's equation is y=mx+b where m is the slope of the line and b is where it crosses the y-axis.  In this case, you need to simplify the equation down to make it into the y=mx+b format.  If you do this, you get y=10x-13.  So in this instance, the slope of the line is 10.
674651
put it in standard form :\n\ny < -1/2x + 3/2 \nand \ny > 2/3x + 2\n\nthen you know the first line passes thru the y axis at 3/2 and the second at 2\n\nthe slopes are -1/2 and 2/3 .... then shade below the line on the first one cuz y is less than .\n\nthen shade above the second line because y is greater than
1124994
you can use the y=mx+b formula for a line.  \n\ny is the y-coordinate of a point\n\nm is the slope, x is the x-coordinate of a point\n\nb is the y-intercept (where the line touches or intercepts the y-axes).  \n\nsince you know the y intercept of the line (passes through origin, so it's 0), the value of b would be 0 and  you also know the slope of the line, which is two, the equation for the line would look like this.\n\ny=2x + 0 or y=2x\n\na point on the line would have coordinates that would fit the equation ---- y has to be two times x; x has to be 1/2 of y.\n\nhere are some points on this line: (0,0), (1,2), (-1,-2), (2,4), (-2,-4).\n\nhope that helped :)