express the final equation in standard form. contains pts. (2,-1) and is perpendicular to the line 4x-5y=2?
342522
4x - 5y = 2\n5y = 4x - 2\ny = 4x/5 - 2/5\n\nThe slope of this is 4/5, so the slope of the perpendicular is -5/4 (just take the opposite of the reciprocal).\n\nFor the new line, m = -5/4, and you have a point (2, -1).  Using the point-slope formula,\n\ny - y1 = m(x - x1)\ny - (-1) = (-5/4)(x - 2)\ny + 1 = -5x/4 + 5/2\n5x/4 + y = 3/2\n5x + 4y = 6
708303
The perpendicular would have a slope 7/3 and any perpendicular line would have an equation like\n\ny = 7/3x + b        (b any number)
576742
ANSWER: y = 2; x = -1/5\n\nSOLUTION:\n\n1) First solve for y.  You can do this by adding the two equations together:\n\n5x - 2y = -5\ny - 5x = 3\n\nNotice that you have a positive 5x on the first equation, and a minus 5x on the second equation, so they both cancel out and you can re-write the equation as follows:\n\n- 2y = -5\n   y = 3\n\nAdd up these 2 equations and you get: \n\n- 1y = -2   or y =  2\n\n2) Then solve for x by replacing y with 2.  You can use any of the 2 equations from the problem.\n\n5x - 2(2) = -5\n\n5x - 4 = -5\n\n5x = -5 + 4\n\n5x = -1\n\nx = -1/5
720107
4x -12y =5...........(1)\n-x +3y = -1..........(2)\n\nMultiplying eq.(2) by 4 and add with eq.(1)\n(4x -12y =5) + (-4x +12y = -4)=\n0 = 1\nwhich is invalid.
925629
polyeig(-5,-2,4,2) \n\nwill solve the equasion.  To learn about interpelations type:\n\nhelp interp1
957836
4x+(-5y)+(-6x)+3y\n4x+(-6x)+(-5y)+3y\n=-2x-2y
143646
LET US SEE:\n\n1. y = (1/2)x + 5. This is an equation in standard form.\nIt is called slope-intercept form. If you mean general form\nthen:\n\ny = (1/2)x + 5\n\nMultiply by 2:\n2y = x + 10\n\n0 = x - 2y + 10\n\nx - 2y + 10 = 0  ANSWER\n\n=======================\n2. Line parallel to y = (1/4)x + 2:\n\nAny line parallel to y = (1/4)x + 2 has the equation:\ny = mx + b  ===> Equation of line\nwhere:\nm = slope = same as slope of given parallel line = 1/4\n\ny = (1/4)x + b ANSWER (equation of parallel lines)\n\nex:\na) y = (1/4)x + 0\nb) y = (1/4)x + 1\nc) y = (1/4)x + 3\nd) y = (1/4)x + 4\nd) y = (1/4)x - 1\ne) y = (1/4)x - 2\n.\n.\n.\n\n=======================\n\n3. y varies directly as x ; y = 2/3 and x = 1/2.\n\ny = kx (direct variation equation) where:\nk = constant of variation\n\nif y = 2/3 and x = 1/2:\n\ny = kx\n2/3 = k(1/2)\nk = (2/3)/(1/2)\nk = (2/3)(2/1)\nk = 4/3\n\nTherefore:\n\ny = (4/3)x ANSWER\n\n=======================\n4. Line passing thru (0,2) and perpendicular to \ny = (1/4)x + 2\n\nLet m = slope = -4 = negative reciprocal to slope\nof perpendicular line.\n\ny = mx + b  ==> slope-intercept form, where:\n\nm = slope\nb = y-intercept\n\nSolving for b, use point (0,2)\n\ny = mx + b\n2 = -4(0) + b\n2 = 0 + b\nb = 2\n\nTherefore, equation is:\n\ny = -4x + 2 ANSWER\n\n=======================\n\n5) Point(s) that lie on the graph of the equation \n\n3x + 6y = 27.\n\nThese are points that will make the left side of \nthe equation equal to 27.\n\nEX. ...;(-5,7)(-3,6)(-1,5);(1,4);(3,3);(5,2);(7,1);...\n\n=============================\n6. Equation of the line that contains (2,4) and undefined slope.\n\nA line that has an undefined slope is a vertical line. So,\na vertical line that contains (2,4) has the equation:\n\nx = 2  same as the x-coordinate of the given point.\n\n===========================\n\n7. Equation of the line that passes thru the x-axis \nat x = -1 and y-axis at y = 3.\n\nFrom the given intercepts, the two points are (-1,0)\nand (0,3)\n\nTherefore:\n\ny = mx + b ===> equation in slope intercept form.\nwhere:\nm = slope = (3-0)/(0--1) = 3/1 = 3\nb = y-intercept from point (0,3) = 3\n\nTherefore:\n\ny = mx + b\n\ny = 3x + 3 ANSWER.
525451
5y -3y = 14 -3\n2y = 11\ny = 5.5\n\n5y -3y = 14 - 9\n2y = 5\ny = 2.5
897842
The first thing you need to be able to figure out is what the question is asking for.  Slope-intercept form (which I prefer to standard form for any and all linear equations) is y = mx + b, where m is the slope and b is the y-intercept (the place where x=0).  To be perpendicular to a line, another line must have a slope which is the negative reciprocal of the slope of the first (e.g. 2 & -1/2).\n\nFirst, figure out the slope of the given line.  In standard form, which is ax + by = c, the slope is -a/b, so the slope of this line must be 5/2.  Take the negative reciprocal of this to find that your second line must have a slope of -2/5.\n\nNow set the equation up in slope-intercept form:\ny = -2/5x + b\nPlug in your point:\n-5 = (-2/5)(3) + b\nSolve for b\n-5 = -6/5 + b\n-19/5 = b\n\ny = -2/5x -19/5
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!
250429
the answer for the line eqwuation  is \n\nY= m(x)+c;\ny=(y2-y1)/(x2-x1)(x)+c;\n\nx1,y1 be the first coordinate and \nx2,y2 be the second coordinate.\n\nm is slope of the Equation.\nc is the Constant \n\nx,y  is  first cordinate value s or seconde cordinate values..\n\n\n\nBaalu..?
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2x-3y=6 is the same as y=2/3x-2\n\nso in order to be perpendicular to this line, you must have slope -3/2\n\ntherefore L has slope -3/2 and has y-intercept of (0,2).  so plug that into slope-intercept form\n\ny=(-3/2)x+2
284371
To solve this, you need to know how to write the equation of a line:\n\ny = mx + b\n\nYour math book should describe in detail how to derive this equation and what it means. Basically, m is the slope of the line and b is the y-intercept (the x coordinate where the line crosses the y-axis).\n\nYou can take the equation of the line given, y = 4x + 3, and figure these things out.\n\nm = 4 (the slope)\nb = 3 (the y-intercept)\nThis y-intercept can also be written as (0,3) being a point on the line.\n\nThe second part is to figure out the equation for a perpendicular line. First, the slope will always be -1/m. Therefore, the slope of your new line will be -1/4. What about the y intercept? We don't yet know this, but we do know a point on the line. What we can do is use the equation above, fill in the slope and the (x,y) coordinates, and solve for b.\n\ny = mx + b\nWe know: point (x,y) is (-1,3) and m = -1/4\n3 = 1/4 * (-1) + b\n3 = -1/4 + b\nb = 2 3/4\n\nHope that helps explain how to do these problems.\n\nEdit: Sorry, I had the point written backwards as (3,-1). I corrected the mistake. That's why it helps to write everything out!
704941
The second equation can be rewritten as:\n3x + 2y = -1\nMultiply both sides by 2\n6x + 4y = -2\n6x + 5y = -7 (this is the first equation)\nSubtract the first equation from the second one.\ny = -5\nNow, plug the y = -5 into the second equation:\n3x +2(-5) = -1\n3x - 10 = -1\n3x = 9\nx = 3\nSo your solution is x = 3 and y = -5
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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
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y = ax^2 + bx + c, a not equal to 0\n\nvertex = [(-b/2a), f(-b/2a)]\n\n-b/2a = -4/2 = -2\n\n-2^2 + (4 x -2) + 5 = 4 - 2 + 5 = 7\n\nthe vertex of the graph of the equation is -2, 7\n\nTherefore the axis of symmetry is x = -2
541010
1.  Find the line that goes through d that is perpendicular to line l.  The slope of the perpendicular line is -1/m, where m is the slope from line l.  Since you have the slope for that line, then just plug in d to get the intercept, which I will call c.\n\n2.  Find the point (x,y) where the two lines meet by solving the system of equations.\n\ny = mx + b\ny = (-1/m)x + c\n\nI will call this point e.\n\n3.  Find the distance between d and e.  If d = (x1,y1) and e = (x2,y2) then the distance between d and e is\n\nsqrt((x1-x2)^2 + (y1-y2)^2)