Question 966881
you can calculate the slope using the variable of T.


let's see how you would do that>


your (x,y) point is (2T+3, 3-3T).


let (x1,y1) be equal to (2T1+3, 3-3T1).


let (x2,y2) be equal to (2T2+3, 3-3T2).


slope is equal to (y2 - y1) / (x2 - x1).


this would be equal to ( (3-3T2) - (3-3T1) ) divided by ( (2T2+3) - (2T1+3) )


simplify this equation to get:


slope = (3 - 3T2 - 3 + 3T1) divided by (2T2 + 3 - 2T1 - 3)


the 3 and the -3 in both expressions cancels out and you are left with:


slope = (-3T2 + 3T1) divided by (2T2 - 2T1)


you can factor out the common -3 in the numerator and the common 2 in the denominator to get:


slope = (-3 * (T2 - T1)) divided by (2 * (T2 - T1)).


the (T2 - T1) in the numerator and denominator cancel out and you are left with:


slope = -3/2.


that's a common slope which indicates that the graph of the equation of the points is a straight line.


to solve for the equation of the line, take 2 points and then use the point slope form of the equation to find the slope intercept form of the equation.\


the point slope form of the equation is y - y1 = m * (x - x1)


you already know the slope because you just solved for it.


the slope if -3/2.


the point slope form of the equation becomes y - y1 = -3/2 * (x - x1).


you only need one point to finalize the pokint slope form of the equation.


you can choose any point.


let t = 0, and your point would be (3,3).


that's ecause 2*0 + 3 = 3 for the x value, and 3 - 3*0 = 3 for the y value.


you make (x1,y1) equal to (3,3) and put it in the slope intercept form of the equation.


you get:


y - 3 = -3/2 * (x - 3)


you now want to convert this to slope intercept form of the equation.


add 3 to both sides of the equation to get y = -3/2 * (x - 3) + 3


simplify the right side of the equation to get y = -3/2 * x + 9/2 + 3.


simplify further to get y = -3/2 * x + 15/2.


that's your equation.


the graph of that eqution is shown below:


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the point (3,3) is when T = 0.


the point (0,7.5) is when T = -3/2.


the point (5,0) is when T = 1.


if the algebra for finding the slope was too intense for you, you could have found it easiwer by simply plotting several points using various random values of T and then calculating the slope several time.


you would have found that the slope was always -3/2 which indicates that you have a straight line.


the rest would then be the same to find the equation for the straight line.