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cartesian equation question!!!

please help... im finding it difficult to eliminate t from these:
Q. show that the straight line given parametrically by the equations
x= (2-t)/(1+2t) and y=(3+t)/(1+2t) passes through the points (6,7) and (-2,-1). Find the values of t corresponding to these points
Original post by Carokelly123
please help... im finding it difficult to eliminate t from these:
Q. show that the straight line given parametrically by the equations
x= (2-t)/(1+2t) and y=(3+t)/(1+2t) passes through the points (6,7) and (-2,-1). Find the values of t corresponding to these points


You don't need to eliminate t for this question. Substitute x = 6 and work out the value of t. Check that substituting the value of t into y gives 7. Similarly for the other point.
Original post by Carokelly123
please help... im finding it difficult to eliminate t from these:
Q. show that the straight line given parametrically by the equations
x= (2-t)/(1+2t) and y=(3+t)/(1+2t) passes through the points (6,7) and (-2,-1). Find the values of t corresponding to these points


It's a straight line so you can find the gradient using y−y1x−x1 \frac{y-y_1}{x-x_1}
dydx=dydt÷dxdt\frac{dy}{dx} = \frac{dy}{dt} \div \frac{dx}{dt} so you can equate the gradients and find t.

Or do what tinyhobbit said as that's much easier.
Reply 3
Avatar for dnr_23
dnr_23
9
x=(2-t)/(1+2t)
cross multiply
x+2tx=2-t
get t on one side
2tx+t=2-x
factorise t
t(2x+1)=2-x
divide both sides by (2x+1) to leave t alone
t=(2-x)/(2x+1)

you can sub into y
y=[3+(2-x/2x+1)]/[1+2(2-x/2x+1)]

to prove it passes through points, sub in x=6 to make sure your value for y=7
do the same for x=-2 giving y=-1

to find corresponding values of t, equate x in terms of t with 6, then equate y in terms of t with 7, you should get the same value for t. (i think its -4/13)

i.e. (2-t)/(1+2t) = 6
(3+t)/(1+2t)=7
t=?

(2-t)/(1+2t) = -2
(3+t)/(1+2t)=7=-1
t=?

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