a line with a slope of -5 passes through the point (3,6).?
813141
both are easy\n\nbecause slope is -5 so\n\neqution of line is\n\ny = -5x+c\nput y = 6 and x = 3 to get\n\n6= -15+c \nor c= 21\n\nequaton of line \n\ny = - 5x + 21\nor 5x+y - 21 = 0 as you got it\n\nNow for the area\n\nit cuts y axis at x =0\nso y = 21\n\nwhen it cuts x axis at y =0 or x = 4.2\n\nit is a right triangle\nso area = 21 *4.2/2 = 21*2.1 = 44.1
311234
-2/3\n\nTo find the slope, rewrite the equation in the form y = mx + b\n\n3y = -2x + 6\ny = (-2/3)x + 2
953419
The slope intercept form of the line is \n\ny = mx + b where m = slope and b = y-intercept\n\nSince (5, -3) lies on the line then\n\n-3 = 5m + b\n\nie 5m + b - 3 = o\n\nSo any line in which m = (3 - b)/5\n\neg b = -2, m = 1 So y = x - 2\nb = 8, m = -1 So y = 8 - x ....... and so forth\n\nGraphing m against b would give all the values of m, b that satisfy the given condition.
477914
slope= mx+b\nslope=1/2x\nequation= y=1/2x-3.5\ny-int: -3.5
834079
y=-1/3x + 6 is your line\ny=-1/3x+1 is a parallel line
910261
Slope = m\n= y2-y1&#92;x2-x1 \n= -4-6/-3-2\n = -10&#92;-5\n=2\n\nnote: / means divided by\n\nthe slope of y= 2&#92;3x-4 is 2&#92;3
890222
for (0,2),(4,0) the slope is -1/2\nfor (-2,0),(4,0) the slope is 0
920205
.y = 6x + 5\n(0, ) \nx=0\nso y=5\n\n\n(3, )\nx=3, so y= 6(3) +5 =23
130487
it's delta-y divided by delta-x\n\n(5 - (-1))/((-1) - (-3))
858135
m = (9-(-1))/(-2-3) = 10/-5 = -2\n\n-1 = -2(3) + b\nb = -1 + 6\nb = 5\n\n\nf(x)= -2x + 5\n\n\nYou got the slope sign reversed.  Be careful about that.  Otherwise, it's correct!
889514
slope of first line: (2-(-4))/ (5-3) = 6/2 =3\nslope of second line: ( -1 -5)/( -2-0) = -6/-2 = 3\nsince the slopes are equal that means that the 2 lines are parallel !
533058
Point slope form:\n\ny-(-3)=(1/2)(x-5)
961745
the slope of the line perpendicular to that line would be the inverse negative of the slope of x-3y = 8\n\n(x-8)/3 = y\n\nthe slope = x/3, so the perpendicular line has a slope of -3x\n\ny = mx + b\n\nm = -3\n\nx = 5\n\ny = -4\n\n-4 = -15 + b\n\nb = 11 \n\ny = -3x + 11
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.
232213
I'll do the fourth problem.\n\n-3x< -6  or x+5< -2\n\nfirst deal with one half- \n-3x< -6 (now divide by -3)\nx>2 (when you divide by a negative the sign is reversed)\n\nthen the second half- \nx+5<-2 (get x alone, so subtract five from each side)\nx< -7\n\nso you have:\n2 < x  or  x < -7\n\nto graph this there is a line with an arrow to the left of the first slash and another line to the right of the second slash\n\n<-----/-------0--/-->\n____-7______2
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!
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.