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the line passes through (7,-2) and is parallel to the line whose equation y=4x+2 is

type the slope interception form of the equation of the line

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passing through (4,4) and parallel to the line whose equation ...


passing through (4,4) and parallel to the line whose equation is 2x+y-7=0. 0 votes. ... what line passes through 1,2 and is parallel to the graph of the line y=3x+8.
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What is the equation of the line that passes through (1,2 ...


What is the equation of the line that passes through #(1,2)# and is parallel to the line whose equation is #4x+y-1=0#?
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What's the equation of the line through (3,7) and parallel to ...


What's the equation of the line through (3,7) ... an equation for a line parallel to y=4x+2 and that is ... parallel to the line y − 1 = 4 (x + 3) and passes ...
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What is the equation of the line that passes through (1, 2 ...


What is the equation of the line that passes through (1, 2) and is parallel to the line whose equation is 4x + y - 1 = 0? algebra 1 , functions . 7/10/2013 ...
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What is the equation of the line that passes through (2, -2 ...


What is the equation of the line that passes through (2 ... of the lines 4x+y-1=0 and ... the equation of a line parallel to y-3x+2=0 passing ...
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the line passing through (2,-2) and is parallel to the line ...


the line passing through (2,-2) and is parallel to the line whose equation is y=3x + 3. ... (-3,0) and (a) the line parallel that passes through the origin (b) ...
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Write Equation of Line parallel to a given line and point-- a ...


Write the Equation of a Line Parallel to a line and through a ... line that is parallel to another line and that passes ... equation of the line y = 3x + b. Step 2.
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Suggested Questions And Answer :


equation for line

Problem: equation for line Write and equation for passing through (4,-1) and parallel to the line whose equation is y=-2x+3 First, we notice that the slope of the given line is -2. That will be the slope of the line we are trying to determine. Our equation: y = mx + b Using the co-ordinates we were given: -1 = -2(4) + b From this, we determine the y-intercept. -1 = -2(4) + b -1 = -8 + b -1 + 8 = -8 + b + 8 7 = b Now for our equation: y = -2x + 7  
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Passing through (-4,4) and parallel to the line whose equation is y=3x+2

strate line thru point=(-4,4) with slope=3 y=3x+konst projekt line from x=-4 tu x=0...deltax=+4 deltay=slope*deltax=3*4=12 y=4+12+16 y=3x+16
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how do i solve this?

y = mx + b is the equation of a line.  m is the slope.  If you have two parallel lines, they will have the same slope, so m will be the same on both of them. In your case, you have y = -3x + 5, so m = -3 This means the parallel line will also have a slope of -3, so it's equation is y = -3x + b.  You still need to solve for b.  You have a point, (-2, -6), so you can just plug that in to solve for b. y = -3x + b -6 = -3 * -2 + b -6 = 6 + b Subtract 6 from both sides to get b by itself -12 = b So the equation of the line you are looking for is y = -3x - 12
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Find the equation of the cylinder having its base the circle

When z=1, x^2+y^2=4, which is a circle, centre at the origin and radius 2. x-2y=1 is a plane intercepting the x-y plane with y intercept=-1/2 and x intercept=1. The equation x^2+y^2+(z-1)^2=4 is a sphere, centre (0,0,1) and radius 2. The plane cuts through the sphere when (1+2y)^2+y^2=4; 1+4y+4y^2+y^2=4; 5y^2+4y-3=0, y=(-4±sqrt(16+60)/10=0.47178 ((-2+sqrt(19))/5) and -1.27178 ((-2-sqrt(19))/5). These values give x=1.94356 ((1+2sqrt(19))/5) and -1.54356 ((1-2sqrt(19))/5 respectively, by using the equation x=1+2y. So the points of intersection are (1.94356,0.47178) and (-1.54356,-1.27178). From these we can find the diameter of the cylinder: sqrt(76/25+304/25)=sqrt(380)/5=2sqrt(95)/5=3.9 approx. The radius is sqrt(95)/5. The sides of the cylinder are perpendicular to the diameter.  The slope of the diameter is the same as the slope of the plane x-2y=1, which is 1/2, so the slope of the perpendicular is -2. The equations of the sides of the cylinder where z=1 in the x-y plane are y=-2x+c where c is found by plugging in the intersection points: (-2+sqrt(19))/5=-2(1+2sqrt(19))/5+c; -2+sqrt(19)=-2-4sqrt(19)+5c. So c=sqrt(19) and y=sqrt(19)-2x as one side. Similarly, -2-sqrt(19)=-2+4sqrt(19)+5c, c=-sqrt(19) and y=-(sqrt(19)+2x) for the other side. Consider the view looking along the z axis. The circular cross-section of the cylinder will appear edge on, while the sides will have the slope -2 and will be spaced apart according to the value of z. When z=1±radius of cylinder=1±sqrt(95), the sides will appear to be as one line passing through the x-y plane's origin, the equation of the line being y=-2x when z=1±sqrt(95). The picture shows the view from z=1 looking at the x-y plane. The circle is the cross-section of the sphere and the line passing through (1,0) and (0,-1/2) is the edge of the plane x-2y=1. The diameter of the cylinder is constant and is shown by the two lines perpendicular to each end of the chord where the plane cuts the sphere. These lines represent the sides of the cylinder as they would appear at z=1. Parallel to them and passing through (0,0) is the single line that appears when z is at the extreme limits of the diameter of the cylinder. This line is also the central axis of the cylinder. (The vertical line at y=-2 is just a marker to show the the leftmost limit of the diameter of the sphere.) The general equation of a cylinder is the same as the 2-dimensional equation of a circle: x^2+y^2=a^2. This cylinder, radius a, has the z axis its central axis. x^2+z^2=a^2 is a cylinder with the y axis as its central axis. a=sqrt(95)/5 so a^2=95/25=19/5, making the equation of the cylinder 5x^2+5z^2=19. This is the same size as the cylinder in the problem, but with its central axis as the y axis. 5x^2+5(z-1)^2=19 is the equation of the cylinder with central axis passing through the centre of the sphere. If the cylinder is tilted so that its base coincides with the circle produced by the plane cutting through the original sphere, more calculations need to be made  to transform the coordinates. The picture shows the axial tilt of the cylinder. The central axis of the cylinder, y=-2x, bisects the chord on the line x-2y=1 when x-2(-2x)=1; 5x=1, x=1/5. So y=(x-1)/2=-2/5. If x-2y=1 represents the horizontal axis (we'll call X) and y=-2x represents the vertical axis (Y) we can see that, relative to these X-Y coordinates, the cylinder is upright. The  z value is unaffected by rotation. Take a point P(x,y) in the x-y plane. What are its coordinates in the X-Y plane? To find out we use geometry and trigonometry. The axial tilt of the X-Y axes is angle ø where tanø=1/2, so sinø=1/sqrt(5) and cosø=2/sqrt(5).  X=SP=QPcosø=2(x+y/2)/sqrt(5), Y=PRcosø=2(y-(x-1)/2)/sqrt(5). Z=z In the X-Z plane, 5X^2+5Z^2=19. This transforms to 4(x+y/2)^2+5z^2=19=(2x+y)^2+5z^2. In the X-Y plane the sides of the cylinder are the lines X=-a and X=a, where a^2=19/5. So X^2=19/5, which transforms to 4(x+y/2)^2 or (2x+y)^2=19/5; 2x+y=±sqrt(3.8), corresponding to the equation of two parallel, sloping line forming the sides of the cylinder.  
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what is the slope of the line whose equation is y-5=3(x-2)

y-5 =3(x-2) y=3x -6 +5 y=3x-1 slope=3
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the line passes through (7,-2) and is parallel to the line whose equation y=4x+2 is

start: point=(7,-2) & slope=4 need tu get y-intersept...x=0 deltax=7, deltay=slope*deltax=28 y-intersept=-2 -28=-30 line: y=4x-30
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equation of the line Through (4, 8); parallel to the line passing through (5, 6) and (1, 2)

Find an equation of the line that satisfies the given conditions. Through (4, 8);  parallel to the line passing through  (5, 6) and (1, 2)    
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find the equation of a line parallel to 6x-7y+14 passing through (-4, 8)

Finding such an equation is very easy, so study the method I use and learn it. Don't just take the answer. The best way to start is rewriting the given function in slope-intercept form. 6x-7y+14 = 0 (I'm assuming it's set equal to 0; you didn't specify!) --> 6x - 7y = -14   --> -7y = -6x - 14     --> y = (6/7)x + 2 (6/7) is the slope of the line. Parallel lines have the same slope, so we know the equation of the line you want will follow this format: y = (6/7)x + b Using the point given, we can solve for b, the y-intercept. (-4 , 8) 8 = (6/7)*-4 + b Solve for b and replace it in y = (6/7)x + b and you have the equation of the line.
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I need to see how you work this Wrote the equation of the line passing throught (4,1) and parallel to the line y=3x-2

parallel to y = 3x - 2 and passing through (4,1): Parallel lines have the same slope, so the slope is 3: using the equation (y - y1) = m(x - x1): y - 1 = 3(x - 4) y - 1 = 3x - 12 in slope-intercept form: y = 3x - 11 in standard form: 3x - y = 11
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How do I write an equation of the line that passes through the given point (-2,5), 2y=4x-6

When two lines are parallel they have the same slope but different y intercept. 2y=4x-6 can also be written y=2x-3 because 2 is a common factor to both sides. The slope of this line is the number before the x term=2. Now we can write the equation of the parallel line: y=2x+b, where we have to find b by plugging in the given point: 5=-4+b, so b=9 and the equation is y=2x+9.
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