Guide :

find and graph and equation parallel to 3x-2y=6 and passing through (-2,-1)

give the whole equation or steps how to solve it

Research, Knowledge and Information :


Find the equation of the line passing through $(-2,6)$ and ...


The question is: Find the equation of the line passing through the point $ ... the equation of the line parallel to $2y - 3x = 8$ that passes through the point $ ...
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Find the Equation of a Line Parallel or Perpendicular to ...


Find the Equation of a Line Parallel or Perpendicular to ... Find the equation of a line passing through the point (–6, 5) parallel to the line 3x – 5y = 9. Step ...
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Find The Equation Of A Line Parallel To 3x - chegg.com


Show transcribed image text Find the equation of a line parallel to 3x ... to 3x - 2y = 8 and passing through (-1, -2). ... Graph the equation. Find the equation ...
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1.Perpendicular to 7x-2y=-2 and passing through the point (7 ...


1.Perpendicular to 7x-2y=-2 and passing through the point (7,1); ... Above equation is slope intercept form of the equation of the required line 2.Parallel to 3y-4x=3 ...
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How would i find an equation of a line passing through point ...


How would i find an equation of a line passing through point B(-2,4) ... is 2 point lower on the graph in ... equation of a line parallel to 3x-2y=6 and ...
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Find the Equation of a Line Parallel or Perpendicular to ...


Find the Equation of a Line Parallel or Perpendicular to Another ... Find the equation of a line passing through ... passing through the point (–7, 2) parallel to ...
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Algebra 2-1 Unit 3 Semester Review Flashcards | Quizlet


Algebra 2-1 Unit 3 Semester Review. ... The y-intercept of the line whose equation is 3x - 2y = 6 is-3. ... Find the slope of the line passing through the points ...
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Standard Form, Parallel/Perpendicular, and X/Y Intercepts ...


Start studying Standard Form, Parallel/Perpendicular, ... 6y = -12 and passes through (-6, 2) ... 3x - 2y = 2. Write in Standard Form Passes
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Suggested Questions And Answer :


write an equation of the line that passes through the given point and is parallel to the give line (6,8), y=-5/2x+10

Since we know the lines are parallel they must have the same slope. That is the first place to start. We know that the slope intercept form of a line is y=mx+b where m is the slope and b is the y itnercept Since we know the slope or m value we can start by writing this as y=-5/2x+b now we need to find the y intercept or b value. The easiest way to do this is to graph the line by plotting the point and using the slope to find other points.  Here is a graph of the two lines:   The y intercept of the unknown line, or the point at which it crosses the y axis is 23. The equation then for the unknown line is y=-5/2x+23
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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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find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1)

Problem: find the equation of a parabola with axis parallel to y passing through (-1,-3), (1,-2), (2,1) Pre-calculus, getting the general equation of a parabola We are looking for an equation in the form of y = ax^2 + bx + c. Given three points, we can replace x and y with real values to create three equations in three unknowns: a, b and c. a(-1)^2 + b(-1) + c = -3 a(1)^2 + b(1) + c = -2 a(2)^2 + b(2) + c = 1 a(1) + b(-1) + c = -3 a(1) + b(1) + c = -2 a(4) + b(2) + c = 1 1) a - b + c = -3 2) a + b + c = -2 3) 4a + 2b + c = 1 We can use that system of equations to find the values for what are actually the constants in the general equation. Subtract equation 2 from equation 1.    a - b + c = -3 -(a + b + c = -2) ---------------------      -2b      = -1 4) -2b = -1 b = -1/-2 = 1/2          <<<<<<<<<<<<<< Subtract equation 3 from equation 2.      a +   b + c = -2 -(4a + 2b + c =  1) ------------------------  -3a -    b      = -3 5) -3a - b = -3 Multiply equation 5 by 2. 2(-3a - b) = -3 * 2 6) -6a -2b = -6 Subtract equation 6 from equation 4.          -2b = -1 -(-6a -2b = -6) -------------------    6a        =  5 6a = 5 a = 5/6          <<<<<<<<<<<<<< Use equation 3 to solve for c. 4a + 2b + c = 1 4(5/6) + 2(1/2) + c = 1 10/3 + 1 + c = 1 10/3 + 1 + c - 1 = 1 - 1 10/3 + c = 0 10/3 + c - 10/3 = 0 = 10/3 c = -10/3 c = -3 1/3          <<<<<<<<<<<<<< The equation is y = 5/6 x^2 + 1/2 x - 3 1/3  
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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 help on this Calculus problem. Thanks.

If f(x)=-4x^2/9+2x/3 as it was in your previous question, f'(x)=-8x/9+2/3, and is the slope of the tangent for all x values. This slope is parallel to y=-2x+4 when -8x/9+2/3=-2, because the line and the tangent line must have the same slope of -2 to be parallel. a) The equation that has to be solved is -8x/9+2/3=-2. b) The slope of the perpendicular is -1/(-2)=1/2, so -8x/9+2/3=1/2. This is the equation that has to be solved. Because the equations in a) and b) are linear, there is only one solution for x, and by plugging this value into f(x) you can find the exact point on the graph where the tangent is parallel or perpendicular.
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how do you find the slope of a line sagment?

how do you find the slope of a line sagment? you are given two points which define a line segment. Find the slope of the line sagment, and then use this slope to write the equation of a parallel line [and a] perpendicular line which pass through the new y-intercept 1. line segment (1,8) and (7,-4)  y-intercept of new lines: (0,-7) The slope is (y2 - y1) / (x2 - x1) m = (-4 - 8) / (7 - 1) = -12/6 = -2 This line segment is y = -2x + b. We don't know the y-intercept, but we are not concerned with that. A parallel line has the same slope but a different y-intercept. We are given the y-intercept, so the equation is y = -2x - 7. A line perpendicular to the given line segment has a slope that is the negative reciprocal: 1/2. The equation of that line is y = (1/2)x - 7.
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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 and graph and equation parallel to 3x-2y=6 and passing through (-2,-1)

1. 3x-2y=6...-2y=6-3x, 2y=3x-6, y=(3/2)x-3...this giv slope=3/2 2. thru (-2,-1) tu get y-intersept: deltax=+2 deltay=slope*deltax=2*3/2=3 yintersept=-1+3=2 line: y=(3/2)x+2
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what line passes through 1,2 and is parallel to the graph of the line y=3x+8

what line passes through 1,2 and is parallel to the graph of the line y=3x+8 y = 3x + 8     the slope is 3 y = mx + b b = y - mx b = 2 - 3(1) b = 2 - 3 b = -1 The equation is y = 3x - 1
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Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0.

Question: Find the pts on the graph of z=xy^3+8y^-1 where the tangent plane is parallel to 2x+7y+2z=0. The tangent plane to the surface z=f(x,y)=xy^3+8/y is given by 2x+7y+2z=D Where D is some constant. The equation of the tangent plane to the surface f(x,y), at the point (x_0 y_0 ) can be written as z=f(x_0,y_0 ) + f_x (x_0 y_0 )∙(x-x_0 ) + f_(y ) (x_0,y_0 )∙(y-y_0 ) Where (x_0,y_0) is some point on that plane andf_x (x_0 y_0 ) is the slope of the tangent line to the surface at that point in the x-direction and similarly for f_(y ) (x_0,y_0 ). We have f(x,y)=xy^3+8/y. f_x=y^3,  f_y=3xy^2-8/y^2 . So, f_x (x_0 y_0 )=y_0^3, f_(y ) (x_0,y_0 )=3x_0 y_0^2-8/(y_0^2 ), f(x_0,y_0 )=x_0 y_0^3+8/y_0 . Substituting for the above into the equation of the tangent plane, z = x_0 y_0^3+8/y_0 + y_0^3∙(x-x_0 ) + (3x_0 y_0^2 - 8/(y_0^2 ))∙(y-y_0 ) z = x_0 y_0^3+8/y_0 + y_0^3 x - x_0 y_0^3 + (3x_0 y_0^2 - 8/(y_0^2 ))y - 3x_0 y_0^3 + 8/y_0 z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {x_0 y_0^3 + 8/y_0 - x_0 y_0^3 - 3x_0 y_0^3 + 8/y_0 } z = y_0^3 x + (3x_0 y_0^2 - 8/(y_0^2 ))y + {16/y_0 - 3x_0 y_0^3 } -2y_0^3 x - 2(3x_0 y_0^2 - 8/(y_0^2 ))y + 2z = 32/y_0 - 6x_0 y_0^3 We have just written down the equation of the tangent plane, which has also been written down as 2x+7y+2z=D Comparing the two forms of these equations, -2y_0^3 = 2     ----------------------------------- (1) 2(3x_0 y_0^2 - 8/(y_0^2 )) = 7     ---------- (2) 32/y_0 - 6x_0 y_0^3 = D   ------------------- (3) From (1), y_0 = -1 Substituting for y_0=-1 into (2), we get x_0 = 3/2. Substituting for x_0 and y_0 into (3), we get D = -23, and z = f(x_0,y_0 ) = (3/2) (-1)^3 + 8/(-1) = (-9.5) The equation of the tangent plane is thus 2x+7y+2z=-23 And this plane is tangential to the surface f(x,y) at one point only, which is: (1.5,-1,-9.5)  
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