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What is the slope of the line passing through the points -1,3 and 4,-7?

What is the slope of the line passing through the points -1 + 3 + 4 + -7 is it too is it three quarters is at -4/3 is it -2

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How use the slope formula and find the slope of a line ...

Interactive lesson with video explanation of how to find the slope of a line given two points ... line through the points (3, 4) and (5, 1) ... 4. A line passes ...

What is the slope of the line passing through the points (–1 ...

What is the slope of the line passing through the points (–1, 3) and (4, ... B. 3/4 C. -4/3 D. -2 . weegy; Answer; Search; More; Help; Account; Feed; Signup; Log In ...

What is the slope of the line passing through the... - OpenStudy

What is the slope of the line passing through the points (–1, 3) and (4, –7)? ... Slope = $\Large\frac{ Rise }{ Run }$ So i Graphed the points: ...

What is the slope of the line that passes through the points ...

What is the equation of the line that passes through the point (8,8) and has a slope of 22?

find the slope of the line passing through the points (-1,3 ...

find the slope of the line passing through the points (-1,3) and (4,-7) - 2509613 ... Slope = (3-(-7))/(-1-4) = 10/(-5) = -2 The slope is -2. Comments; Report; 0 ...

What is the slope of the line that passes through the points ...

What is the slope of the line that passes through the ... slope of the line that passes through the points (1, 4 ... a line passing through two points (x 1,y 1) ...

What is the slope of the line passing through (1,-1); (4,7 ...

8/3 Slope m of a line passing through ... 4-1)=(7+1)/3=8/3 implies slope of the line passing through the ... of the line passing through the given points ...

What is the slope of the line passing through the... - OpenStudy

What is the slope of the line passing through the points (–3, 4) and (2, –1)? ... The slope of the line through points (x1,y1) and (x2,y2) is given by : ...

What is the slope of the line that passes through the points ...

What is the slope of the line that passes through the points (1, 7) and (–4, –8)? ... (1, 7) and (–4, –8)? A. –3 B. Slope C. Irrational Numbers D. 3 1. Ask ...

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.

graph the line with slope 2 passing through the point (-4, -4)

Problem: graph the line with slope 2 passing through the point (-4, -4) how do i graph a slope with line 2 passing thru (-4,-4) Plot the point (-4, -4) Find another point by counting to the right one space for every two spaces you count up. Draw a line between the two points and extend the line in both directions.

What is the slope of the line passing through the points -1,3 and 4,-7?

Problem: What is the slope of the line passing through the points -1,3 and 4,-7? What is the slope of the line passing through the points -1 + 3 + 4 + -7 is it too is it three quarters is at -4/3 is it -2 The slope is m = (y2 - y1) / (x2 - x1) (-1, 3) and (4, -7) m = (-7 - 3) / (4 - (-1)) m = -10 / (4 + 1) m = -10 / 5 m = -2

what is the y intercept of the line whose slope is 3 passing points (1,-2)

Problem: what is the y intercept of the line whose slope is 3 passing points (1,-2) i need help finding the y intercept of the line whose slope is 3 passing through points (1,-2) The general equation is y = mx + b. Re-work that equation to find b. b = y - mx b = -2 - 3(1) b = -2 - 3 b = -5 Answer: the y-intercept is -5.

Find the equation of the altitude of the triangle

A(1,-5),B(2,2) and C(-2,4) are the vertices of triangle ABC . Find the equation of the altitude of the triangle through B. The answer given at the back of my book is 3y=x+4. But how is it coming. The altitude of the point B is a line perpinficular to the base line AC, and passing through the point B(2,2). Find the slope of the base line AC Hence find the slope of any line perpindicular to it. This line will now pass through the point B(2,2) You have a line with a slope that passes through a fixed point. Hence you can find the equation of this line.

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.

whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and (-2,-4)?

-2 + 6y = 13x + 2 6y = 13x + 4 y = (13/6)x + 2/3 slope = 13/6 perpendicular slope = -6/13 we want the equation of a line with slope = -6/13, passing through point (-2,-4) point-slope form y - y1 = m (x - x1) y - -4 = (-6/13) (x - -2) y + 4 = (-6/13) (x + 2) y + 4 = (-6/13)x - 12/13 y = (-6/13)x - 12/13 -4 y = (-6/13)x -64/13

whats the equation of line perpendicular to the given line, passes through -2+6y=13x+2 and (-2,-4)?

-2 + 6y = 13x + 2 6y = 13x + 4 y = (13/6)x + 2/3 slope = 13/6 perpendicular slope = -6/13 we want the equation of a line with slope = -6/13, passing through point (-2,-4) point-slope form y - y1 = m (x - x1) y - -4 = (-6/13) (x - -2) y + 4 = (-6/13) (x + 2) y + 4 = (-6/13)x - 12/13 y = (-6/13)x - 12/13 -4 y = (-6/13)x -64/13

how do I graph the line if the slope is 1/2 and it runs through the points (4,-3)

How do I graph the line if the slope is 1/2 and it passes through the points (4,-3) on the graph paper start at the point (4, -3)  then go one point over and two points up.  You have two points now you can graph the line. (y - (-3)) = 1/2x y = 1/2 x - 3  is the line

Find the equation of the line with a slope of 4

According to the slope formula, if you have a line with slope a and which passes through the point (b, c), the equation of the line will be (y - c) / (x - b) = a. Since the slope is 4 and the point which the line passes through is (-1, 12), the equation of the line is: (y - 12) / (x - (-1)) = 4 (y - 12) / (x + 1) = 4 y - 12 = 4(x + 1) y - 12 = 4x + 4 y = 4x + 4 + 12 y = 4x + 16, which is the equation you need.