in 1-6, find the following for each pair of points: (a) the distance between the points | studysoup
in problems 1-6, find the following for each pair of points: (a) the distance between the points (b) the midpoint of the line segment connecting the points (c) the slope of the line containing the points (d) interpret the slope found in part (c) (0, 0); ( -4, 6)
general systems thursday, september 29, 2016 3:50 pm a fruit store sell two types of fruit, apples and oranges. apples are $2 and oranges are $3.today 50 fruit was sold and the store made $110. 110 = 2▯ + 3▯ 50 = ▯ + ▯ ▯ = 40 ▯ = 10 general ▯▯ ▯+ ▯ ▯▯ ▯⋯+ ▯ ▯ = ▯ ▯ a linear equation is an equation in the form of: • x's are variables and a's are constants. • solution set for a linear equation is the set of all solutions. • set of linear equations where a commonsolution is desired is called a system of linear equations. graphs • 1 variables is a point • 2 variables is a line • 3 variables is a plane • 4+ variables is a hyperplane examples of systems 2 equations, 2 unknowns • no solutions • one
you seem to be missing some information about e.if e is the midpoint between d and f, thene is (1,1) take half of (4-2) = 1 and half of (5-3) = 1e alphabetically does come between d and f.just average the x values and average the y values of the two points d and for plot them graphically, carefully, maybe on some graph paper, and you'llvisually see the midpoint is at (1,1) 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 the straight distance between two place on a map can be measured by a strip of paper and divider.use a ruler to measure the distance between the two places. if the line is quite curved, use a string to determine the distance, and then measure the string. find the scale for the map you're going to use. it might be a ruler bar scale or a written scale, in words or numbers. 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` 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. several ways to do this, here's one:

to go from (-6,4) to (-2,5) you need to go 4 in the x direction and 1 in the y direction. if you move again 4 in the x direction and 1 in the y direction you end up at (2, 6). that's p2 (i) co-ordinates of point p are `((1xx17+2xx(-1))/(1+2),(1xx10+2xx1)/(1+2))` `=((17-8)/3,(10+2)/3)` `=(9/3,12/3)` `=(3,4)` (ii) `op=sqrt((0-3)^2+(0-4)^2)` `op=sqrt(9+16)` `op=sqrt25` `op=5` units (iii) let ab be divided by the point p(0, y) lying on y-axis in the ratio `therefore (0,y)=((kxx17+1xx(-4))/(k+1),(kxx10+1xx1)/(k+1))` `=>(0,y)=((17k-4)/(k+1),(10k+1)/(k+1))` `=>0=(17k-4)/(k+1)` `=>17k-4=0` `=>k=4/17` thus, the ratio in which the y-axis divide the line ab is 4: 17. the given equations arex2 + y2 = 16 …(i)and y2 = 6x …(ii)on solving the equation (i) and (ii),or x2 + 6x = 16or x2 + 6x – 16 = 0or (x + 8)(x – 2) = 0or x = 2 or – 8 is not possible solution∴ when x = 2, y = ± = ± = ±equation (i) is a circle with centre (0, 0) and meets axes at (±4,0), (0,±4)equation (ii) represents a parabola with axis as x - axis.points of intersection are a () and c (2,)these are shown in the graph below:area bounded by the circle and parabola= 2[area (oado) + area (adba)]area of circle = π(r) 2= π(4)2= 16π sq. unitsthus, required areathe area of the circle x2 + y2 = 16 which is exterior the parabola y2 = 6x is given : let a , b and c be three students whose chances of solving a problem is given i.e , p(a) = , p(b) = and p(c) = . p() = , p() = and p()= to find: the probability that the problem is solved .here , p(the problem is solved) = 1 - p(the problem is not solved)= 1- p( )= 1 – [p( ) p( ) p( )]= 1- []= 1 - = therefore , the probability that the problem is solved is . 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.