if there are six straight lines in a plane, no two of which are parallel and no three of which pass through the same point, then find the number of points in which these lines intersect.
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`
intersecting lines – two lines are intersecting if they have a common point. the common point is called point of intersection. 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: 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. Straight lines (a)The regular octahedron has 3 symmetry planes;  
(b)The regular dodecahedron has 15 symmetry planes;  
(c)The regular icosahedron has 15 symmetry planes; hi [@banati.bodo](/u/banati.bodo)

the key here is to find which points are on the symmetry plane and keep them there with an onplane goal, then for the points ‘above’ the symmetry plane, use a transform goal to lock them to their mirror image below the symmetry plane.

[symmetry\_planarhex.gh](/uploads/short-url/wlxtpug6o3kghsxsd43l0qwa6jy.gh) (17.3 kb)

[symmetry] 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 let x denote the number of sixes. p(getting a six when a die is thrown) = p = `(1)/(6)` ∴ q = 1 – p = `1 - (1)/(6) = (5)/(6)` given, n = 6∴ x ∼ b`(6, 1/6)`the p.m.f. of x is given by p(x = x) = `""^6"c"\_x (1/6)^x (5/6)^(6 - x), x` = 0, 1,...,6p(getting at most 2 sixes) = p(x ≤ 2) = p(x = 0 or x = 1 or x = 2) = p(x = 0) + p(x = 1) + p(x = 2) = `""^6"c"\_0(1/6)^0 (5/6)^6 + ""^6"c"\_1(1/6)(5/6)5 + ""^6"c"\_2(1/6)^2 (5/4)^4` = `(5/6)^4 + 6(1/6)(5/6)^5 + (6!)/(2! xx 4!)(1/6)^2 (5/6)^4` = `(5/6)^4 [(5/6)^2 + (5/6) + (6 xx 5 xx 4!)/(2 xx 1 xx 4) xx (1/6)^2]` = `(5/6)^4 [25/6^2 + 30/6^ + 15/6^2]` = `(5/6)^4 (70/36)` = `(5/6)^4 (35/18)` = `(5/4)^4 ((5 xx 7)/(6 xx 3))` = `(7)/(3)(5/6)^5`. this is a type of problem called linear programming. you'll want to first define your variables:
 
x = number of motorcycles made
y = number of bikes made
 
 
 
then, write equations that bound your problem:
 
0 ≤ x ≤ 100    (you can't make more than 100 or fewer than 0 motorcycles)
0 ≤ y ≤ 250    (you can't make more than 250 or fewer than 0 bikes)
x + y ≤ 300    (you can't make more than 300 vehicles total)
 
 
 
now, plot those equations and find the region of intersection. the region should be some enclosed polygon where all the equations are true. so:
 
draw vertical lines at x = 0 and x = 100. the first equation is true between those lines.
 
draw horizontal lines at y = 0 and y = 250. the second equation is true between those lines.
 
draw a diagonal line at y = -x + 300. the third equation is true below that line.
 
the enclosed region is a pentagon, with one side formed by each of the five lines.
 
 
 
now, find the vertices of the region of intersection. since yours is a pentagon, you'll want to find five vertices. some are straightforward:
 
x = 0 and y = 0 intersect at (0, 0)
x = 0 and y = 250 intersect at (0, 250)
x = 100 and y = 0 intersect at (100, 0)
 
some are slightly less straightforward:
 
y = -x + 300 and x = 100 intersect where y = -100 + 300. that's (100, 200)
y = -x + 300 and y = 250 intersect where 250 = -x + 300, or -x = -50, or x = 50. that's (50, 250)
 
so, the five points of intersection are (0, 0), (0, 250), (100, 0), (100, 200), and (50, 250).
 
 
 
now, write an equation for profit, p:
 
p = 1220x + 380y
 
this equation will have its maximum value (and its minimum value) at one of those vertices. plug each vertex in to figure out which gives you the maximum profit:
 
p = 1220(0) + 380(0) = 0 + 0 = $0
p = 1220(0) + 380(250) = 0 + 95000 = $95,000
p = 1220(100) + 380(0) = 122000 + 0 = $122,000
p = 1220(100) + 380(200) = 122000 + 76000 = $198,000
p = 1220(50) + 380(250) = 61000 + 95000 = $156,000
 
your maximum profit is $198,000, at 100 motorcycles and 200 bikes. two poles of height 6 m and 11 m, respectively stand on a plane ground. if the distance between their feet is 12 m, then what is the distance between their tops?