How do you find k so that the line through (k,k+1) and (2,3) will have a slope of -1?

Answer 1

I could not find #k#!

Consider the general relationship that gives you the slope #m# as: #m=(Deltay)/(Deltax)=(y_2-y_1)/(x_2-x_1)# (1) Let us use our data: #-1=(3-(k+1))/(2-k)# and solve for #k#: #k-2=3-k-1# #k+k=3+2-1# #2k=4# so #k=4/2=2# BUT if you try this solution into your slope (equation 1) it doesn't work. So I think it is not possible to have a line with this slope...

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Answer 2

Hazel Anglin

To find the value of ( k ) so that the line passing through ( (k, k+1) ) and ( (2,3) ) has a slope of -1, we can use the slope formula:

[ m = \frac{{y_2 - y_1}}{{x_2 - x_1}} ]

Given that the slope is -1 and the points ( (k, k+1) ) and ( (2,3) ) lie on the line, we have:

[ -1 = \frac{{3 - (k+1)}}{{2 - k}} ]

Now, we can solve this equation for ( k ):

[ -1 = \frac{{3 - k - 1}}{{2 - k}} ] [ -1 = \frac{{2 - k}}{{2 - k}} ] [ -1 = 1 ]

Since this equation does not hold true, there is no value of ( k ) that satisfies the condition for the line to have a slope of -1.

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.