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# What is the slope of a line that is perpendicular to (4, -1) and (-3, 5)?

what is the slope of a line that is perpendicular to (4, -1) and (-3, 5)

## Research, Knowledge and Information :

### Slopes and Equations of Lines - University of Missouri–St ...

Slopes and Equations of Lines Some ... the slope of a line: The slopes of perpendicular lines are ... Example 4 What is the equation of a line with slope 3/4 and y ...

### Algebra Examples | Linear Equations | Finding the Slope of a ...

Algebra Examples. Step-by-Step ... Linear Equations. Find the Slope of a Perpendicular Line. Rewrite in ... American Express security codes are 4 digits located on ...

### Horizontal and Vertical Lines | Purplemath

The second line's equation was y = –2 x + 3, and the line's slope was m = –2. ... From the line's graph, I'll use the (arbitrary) points (4, 5) and (4, –3 ...

### Slope of a line

Slope of a line explained,videos,practice quizzes.The slope is a measure of how the line ... Perpendicular are two lines that intersect and create right 90 degree ...

### The slope of a line is 1/3 . What is the slope of a line ...

The slope of a line is 1/3 . What is the slope of a line ... perpendicular is opposite and reciprocal slope of a line is 1/3 so slope of a line perpendicular is ...

### Slope of Parallel and Perpendicular Lines - AlgebraLAB

Slope of Parallel and Perpendicular Lines: ... The parallel line will have the same slope which is 5. The perpendicular line will have a slope of which is the ...

### Finding Parallel and Perpendicular Lines - Maths Resources

Parallel and Perpendicular Lines. How to use Algebra to find parallel and perpendicular lines. ... So the perpendicular line will have a slope of 1/4: y − y 1 = ...

### in slope intercept form how would you find perpendicular line ...

in slope intercept form how would you find perpendicular line to y=3/4x+1 and goes ... and m is the slope of the perpendicular line (which is -4/3); plug in these ...

### Perpendicular Lines 1 | Coolmath.com

Perpendicular Lines. ... These lines are perpendicular. (They form a angle.) So, what's ... Finding the Slope of a Line from the Graph.

## Suggested Questions And Answer :

### Are the two lines parallel, perpendicular, or neither?

a) The slopes (-1 and 1) have turned over fractions (1/1 vs 1/1) and their + - signs are switched, so these lines are perpendicular. b) y = 0x - 5 4x - 3y = 12 -3y = -4x + 12 y = (4/3)x - 4 0 and 4/3 are different, so the lines aren't parallel. 0 and 4/3 aren't turned over fractions of each other, so the lines aren't perpendicular. Answer:  Neither. c) (2,2) to (4,8) slope = (8 - 2) / (4 / 2) slope = 6/2 slope = 3 12x - 2y = 42 -2y = -12x + 42 y = 6x - 21 The slopes (3 and 6) aren't the same, so the lines aren't parallel. The slopes aren't fractions turned over of each other, so the lines aren't perpendicular. Answer:  Neither.

### 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.

### how do you find the slope of a perpendicular line?

how do you find the slope of a perpendicular line? Presumably you know the slope of the other line, the one this line is perpendicular to. Given that the slope of that line is m, the slope of the perpendicular line is -1/m.

### Parallel and Perpendicular Lines matching exercise

starte line...x-3y=9 bekum 3y=x-9, or y=(x/3)-3... slope=(1/3)... perpendikular line hav slope=-3

### Give the slope-intercept form of the equation of the line that is perpendicular to and contains (-1, 9)

Start by solving for y to get this in y=mx+b form subtract 5x from both sides 4y=-5x+4 divide by 4 y=4/5x+4/4 simplify to y=4/5x+1 Now you know that the slope of the first line is 4/5 The slope of the second line is -4/5 so the second equation is y=-4/5x+? since the line has to cross the point -1,9 you can plot that point on the graph and then use the slope to find another point. from the point -1,9 you can use rise over run to go down the graph 4 spots to 5 and to the right on the graph 5 spots to 4 your second point would be (4,5) y=-4/5x+41/5 is the equation for the second line connect those points to get your line.  The two lines look like this:

### 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.

### Find the equation of the: a) tangent to the circle with center (-1,2) at the point (3,1) b) perpendicular bisector bisector of (AB) for A (2,6) and B(5,-2).

Problem: Find the equation of the: a) tangent to the circle with center (-1,2) at the point (3,1) b) perpendicular bisector bisector of (AB) for A (2,6) and B(5,-2). I  need help. a) Any line that is tangent to the circle has a slope that is    the negative inverse of the slope of the radius at the tangent    point.    For the radius: m = (y1 - y2)/(x1 - x2)    m = (2 - 1)/(-1 - 3)    m = 1/-4    m = -1/4    The slope of the tangent line is m = 4    The equation is found by using the slope and one point on    the line. We are given that one point: (3, 1)    y = mx + b    b = y - mx    b = 1 - 4(3)    b = 1 - 12    b = -11    Our equation is y = 4x - 11 b) As with the first part, we need the slope of the line we    were given. m = (y1 - y2)/(x1 - x2)    m = (6 - (-2))/(2 - 5)    m = (6 + 2)/(-3)    m = 8/-3    m = -8/3    The slope of any line perpendicular to this line is m = 3/8    The midpoint of the given line is ((x1 + x2)/2, (y1 + y2)/2)    (x1 + x2)/2 = (2 + 5)/2 = 7/2 = 3.5    (y1 + y2)/2) = (6 + (-2))/2 = 4/2 = 2    The midpoint is (3.5, 2)    We find the y-intercept with the formula b = y - mx    b = 2 - (3/8)3.5    b = 2 - 1.3125    b = 0.6875    Our equation is y = 3/8 x + 0.6875

### What is the equation of a line that passes through (2, -1) and is perpendicular to the line 2x-y=1

first step:  find slope of line 2x - y = 1 2x = y + 1 y = 2x - 1 slope of line:  2 perpendicular to slope of line:  - 1/2 now we want the equation of the line passing through (2, -1) with slope = -1/2 y - y1 = m (x - x1)  (point-slope form) y - -1 = (-1/2) (x -2) y + 1 = (-1/2)x + 1 y = (-1/2)x

### What is the slope of a line that is perpendicular to (4, -1) and (-3, 5)?

slope of the line with points (4,-1) and (-3,5) slope = (5 - -1)/(-3 -4) slope = 6/-7 slope = -6/7 since the product of gradients of perpendicular lines is -1 . then m x -6/7 = -1 where m is the gradient of the required lined therefore m = 7/6 therefore the slope is 7/6

### if two straight lines are perpendicular to the same line prove that they are parallel to each other

First line's slope = m Middle line (perpendicular to first line) has slope = -1/m  (change sign and flip fraction) Second line is perpendicular to middle line, so we take the slope of the middle line (-1/m) and change the sign and flip the equation, so it becomes +m/1 or just m. The first line and the second line have the same slope (m), so they are parallel.