find the equation of the line that passes through | ch 2 - 55 | studysoup
find the equation of the line that passes through the given point and also satises the additional piece of information. perpendicular to the line
media effects (chapter 15)  media effects is the field of research that attempts to understand, predict, and explain the effects of mass media on individuals and society o scientific method: understand, predict, and explain o individuals and society: media does not only effect the individual but society as well  not all media researchers are studying "effects" or aiming for prediction o cultural studies researchers try to understand how media constructs and is constructed by questions of gender, race, social class, nationality, etc. o rhetorical studies researchers try to understand how media functions in terms of persuasion, meaning, political contestation, ideology, identity, etc. media effects theories  the hypodermic needle model o an early model for television effects on audiences o the idea that powerful media shoot their messages directly into vulnerable audiences' brains o also known as the "magic bullet" or "direct effects" model o fails to account for the fact that people are critical and skeptical of media messages o widely discredited by subsequent research  the minimal effects model o media effects are less powerful, and are mostly indirect. o also known as the "limited model" o people engage in selective exposure:  we tend to expose ourselves only to messages that we are already comfortable with o people engage in selective retention:  we tend to retain messages that confirm the values and ideas that we already hold o this model still assumes that audiences are passive in their consumption of media  uses and gratifications 
 o studies how and why media are used, rather than what their effects are o the surveillance function suggests that people use the media to monitor the world around them o similarly, the socialization function suggests that people use the media to help fit in with others o while the textbook says this model was never widely adopted, it has been incorporated into rhetorical and cultural studies research. what effect does the media have?  the most significant demonstrated effect of media is agenda- setting: o not telling people what to think, but telling them what to think about  many of the other effects listed in the textbook are less well proven but still worth discussing  the cultivation effect is the idea that heavy tv viewing causes is to believe that tvis giving us an accurate portrayal of the real world o one common version of this is the mean world syndrome: we believe the world is more dangerous than it really is because tv shows so much crime  the spiral of silence theory is the idea that the media silences controversial opinions by making those who have them fear social invasion o this can even be true of opinions that are not really controversial, since the media edits out very widely held opinions o some studies indicate that we are even willing to go along with wrong answers if "everyone else" is giving those answers too
you can find the radius in the following way. 
write the equation of the line that is perpendicular to the given tangent line. this perpendicular goes through the stated center, and intersects the tangent; the perpendicular's slope is the negative reciprocal of the stated tangent line.
find the intersection (point t) of the tangent and radius line--then use the distance formula to find the radius, which is a segment from the stated center to the point t. intersecting lines – two lines are intersecting if they have a common point. the common point is called point of intersection. slope of the line ⇒ slope of the lineequation of line :⇒ 3(y - 5) = 2(x - 2)⇒ 2x - 3y + 11 = 0 …equation (i)putting x = x + 3 and y = y + 2 in the above equation we get2(x + 3) - 3(y + 2) + 11 = 0⇒ 2x - 3y + 11 = 0since it again gives the same equation as (i) so it is a point on this line.hence proved in this we are asked to divide it in 2:3 ratioso, 3+2 = 55, number congruent of line segment is formed.i. firstly, we have to draw a line segment ab of length 7 cm. draw a line with an acute angle from point aii. point the compass on a, and cut the new line with some radiusplace the compass along the line and make 5 arcs. since we need 5 congruent parts, and name the last one as “g”iii. take length of bg as radius draw an arc from a just below it.again, take the length of ag as radius and draw an arc which cuts the previous arc from b, name the intersection point as hconnect b and hiv. taking the same length that we used in line ag, draw arcs along bh and name the points.v. connect the points along ag and bh. name the intercepts on the line ab as m, n, o & p.vi. thus we can divide a line segment as ao:ob = 6:4= 3:2 the graph of this equation is a line.  we know that because each term is either a constant (1.5) or the product of a constant and the first power of a variable (d, 0.8t).
since we have two variables (d and t), the line sits in a two-dimensional plane, with a vertical axis and a horizontal axis.  d stands alone in the equation, so its value depends on the value oft.  therefore, the vertical--or dependent--axis will represent values of d, and the horizontal--or independent--axis will represent values of t.
at this point, then, we can set up a coordinate plane with a vertical axis labeled d and a horizontal axis labeled t.  all of the points on our plane have a coordinate (t, d).
to graph a line, we need two points.  since the question asks for the axial intercepts anyway, we can use those.
we'll start with the d-intercept, where the line crosses the d-axis.  all of the points on the d-axis have a t-coordinate of zero, so to find where the line crosses the axis, we substitute zero for t in our equation and solve.
d = 1.5 + 0.8 × t
d = 1.5 + 0.8 × (0)
d = 1.5 + 0
d = 1.5
so our d-intercept has the coordinates (0, 1.5), and we have our first point and our first intercept.
similarly, to find the t-intercept, we note that all of the points on the t-axis have a d-coordinate of zero, so we set d equal to zero and solve.
d = 1.5 + 0.8 × t
0 = 1.5 + 0.8 × t
0 - 1.5 = 1.5 - 1.5 + 0.8 × t
-1.5 = 0.8 × t
-1.5 ÷ 0.8 = (0.8 ÷ 0.8) × t
-1.875 = 1 × t
-1.875 = t
and the d-intercept is (-1.875, 0).
now we have both axial intercepts, so that part of the question is complete.  we can plot those two points and draw the line that passing through them (extending on, at either side, to infinity).
all that remains is to find the turning points.  a turning point, also called a local maximum or local minimum, is where the value of a function goes from decreasing to increasing or from increasing to decreasing.  (the graph shown at http://en.wikipedia.org/wiki/file:parabola\_with\_focus\_and\_directrix.svg has a minimum at the origin, for example.)
a line has a constant slope.  to say that differently, the angle between any two points on a line is the same as the angle between any two other points on the line.  if a line is increasing in one location (like our line), it is increasing everywhere, and it is increasing everywhere at the same rate.  therefore, a line has no turning points. answer: 15
solution: there are six straight lines `l\_(1),l\_(2),l\_(3),l\_(4),l\_(5),l\_(6)` in a plane.   
 since no two lines are parallel and no three of which pass through the same point, we have maximum points of intersection.   
 line `l\_(1)` intersect five lines, so there will be five points of intersection.   
 similarly, for each of the lines `l\_(2),l\_(3),l\_(4),l\_(5),l\_(6)`, there will be five points of intersection.   
 but there will be double counting .   
 so, total number of points of intersection is `(5xx6)/(2)=15` May as well promote my comment to an answer.


You need to add points to a `Curve2D` object, then adjust the control points of the curve, and then finally draw the curve. So here's some simple, almost pseudo code:



```
var array_of_line_points # This already has the vectors which describe our line
for point in array_of_line_points:
  # The "get_perpendicular_vector()" function returns a vector that's a copy of the point, yet has been slid along a line parallel to two neighboring points. The "distance" variable is how far the control point should be from the originating point.
  var control_point1 = get_perpendicular_vector(point, distance)
  var control_point2 = get_perpendicular_vector(point, -distance)
  curve.add_point(point, control_point1, control_point2)
# ...In the draw function
func _draw():
  draw_polyline(curve.get_baked_points(), red, 2.0)

```

I got the inspiration for this from [Rob Spencer's page on Spline Interpolation](http://scaledinnovation.com/analytics/splines/aboutSplines.html). I'd like to make this more thorough (i.e. defining the `get_perpendicular_vector()` function). But this is the best I can currently do. if , passing through (1, 2), satisfies the differential equation , then dear student

kindly refer to the link for the similar query :


https://www meritnation
com/ask-answer/question/q-23-and-24-q23-give-the-formula-of-each-of-the-follow/the-p-block-elements/11940305

regards check out the intersect function.
here is one approach:
% generate a rectangular boundry
rectx = [0.1 0.1 0.9 0.9];
recty = [0.1 0.5 0.5 0.1];
rect = polyshape(rectx,recty);

% initialize a figure for display purposes
figure(1);clf
titles = [...
 "all lines";...
 "partially within";...
 "exclusively within"];
for i = 1:3
 subplot(1,3,i)
 hold on
 title(titles(i))
 plot(rect)
 xlim([0 1])
 ylim([0 1])
end

% define the number of line segments to generate
nlines = 25;

% initialize a structure to store the lines (if you need the data later)
lines = struct();

% generate each line, calculate it's intersection conditions, and plot
for i=1:nlines
 
 % generate points for a random line segment
 lines(i).points = rand(2);
 
 % section the line into parts at intersection with the rectangle
 [in, out] = intersect(rect, lines(i).points);
 
 % plot all the lines that were generated
 subplot(1,3,1)
 line(lines(i).points(:,1),lines(i).points(:,2))
 
 % plot lines that are partially within the boundary
 if isempty(in)==0
 subplot(1,3,2)
 line(lines(i).points(:,1),lines(i).points(:,2))
 end
 
 % plot lines that are exclusively within the boundary
 if isempty(out)==1
 subplot(1,3,3)
 line(lines(i).points(:,1),lines(i).points(:,2))
 end
end
here is the output: