You use the well-known point-point formula . Here is it: suppose P1 = (x1,y1)
and P2 = (x2,y2) An equation for the line containing P1 and P2 is
y - y1 = [(y1-y2)/(x1-x2)] (x -x1)
Note that the quantity in brackets is the slope of the line. Note also that it does not matter which point is P1 and which is P2.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
the slope is the 'm' in y=mx+b so even if the points aren't given, if there is an equation, then you can find the slope. for example, if you have an equation like this: y=2x+5 the slope is 2 and the y-intercept is 5.
If you have two equations give AND one parametric equation why do you need to find yet another equation?
Select any three values of x in the domain of the equation. Solve the equation at these three points for the other variable, y. Then each (x, y) will be an ordered pair that is a solution of the equation.
First, you calculate the slope between the two points (difference of y / difference of x). Then you can use the equation, using one of the points (x1, y1): y - y1 = m(x - x1) Just replace x1 and y1 with the coordinates of the point, and m with with the slope.
Draw the graph of the equation. the solution is/are the points where the line cuts the x(horisontal) axis .
the slope is the 'm' in y=mx+b so even if the points aren't given, if there is an equation, then you can find the slope. for example, if you have an equation like this: y=2x+5 the slope is 2 and the y-intercept is 5.
Use the equation; y=mx+b where m is the slope Use your 2 points as y and b (intercept)
If you have two equations give AND one parametric equation why do you need to find yet another equation?
I suggest that the simplest way is as follows:Assume the equation is of the form y = ax2 + bx + c.Substitute the coordinates of the three points to obtain three equations in a, b and c.Solve these three equations to find the values of a, b and c.
In order to find the equation of a tangent line you must take the derivative of the original equation and then find the points that it passes through.
The least needed information can be given in different formats, which are equivalent: -- the slope of the line and its intercept on either axis -- the slope of the line and any one point on it -- any two points on the line
It is always easier to use an equation to find points since all you would have to do is substitute values into the equation to find the final unknown value that will tell the point. To get the equation, however, you would usually need to have some points at the start to help derive the equation in the end.
The equation is -x -16 equals y. You find this by using the equation for a line mx plus b equals y, where 'm' is the slope and 'b' is the y-intercept. From the information given, you have two points which are 0.-16 ans -16,0. You can find 'm' the slope with the equation y2-y1/x2-x1, or -16-0/0- -16. This is -16/16 or -1 for m and the y-intercept is given as -16. So, substitute into the line equation these values to get the answer given.
Select any three values of x in the domain of the equation. Solve the equation at these three points for the other variable, y. Then each (x, y) will be an ordered pair that is a solution of the equation.
First, you calculate the slope between the two points (difference of y / difference of x). Then you can use the equation, using one of the points (x1, y1): y - y1 = m(x - x1) Just replace x1 and y1 with the coordinates of the point, and m with with the slope.
No, a calculator is useless, unelss you are dealing with values for x and y which require some difficult working out. Use the general form of a linear equation using two points on the line: y - y1 = (y1 - y2)/(x1 - x2)(x - x1), where the points given are (x1, y1) and (x2, y2).