Find the value of K if the points (-2,3),(1,0) and (k,2) lie on the same line.
we need the slope of the line the slope of the line between \((-2,3)\) and \((1,0)\) is \(m=\frac{3-0}{-2-1}=\frac{3}{-3}=-1\)
yes
this means for each unit increase in \(x\) there is a unit decrease in \(y\)
we see that \((-2,3)\) is on the line one decrease in \(y\) gives \((k,2)\) and so \(x\) must have been increased by 1 unit from -2 therefore it is \((-1,2)\)
so you mean k is the answer from the slope?
we can also use algebra if that is too much thinking\frac{36m^4n^3}{24m^2n^5} slope is \(-1\) and so we get the equation \[\frac{3-2}{-2-k}=-1\] \[\frac{1}{-2-k}=-1\] \[1=2+k\] \[k=-1\]
k we can find from the slope since \(m=\frac{y_2-y_1}{x_2-x_1}=-1\) use the points \((-2,3)\) and \((k,2)\) compute the slope, the result \(=-1\) and solve for \(k\)
i dont understand
i thought k=-1
yes, \(k=-1\) and also the slope is \(-1\) but that is a coincidence
thanks
yw
why you use -2,3 why not 1,0?
try it and you will see that it makes no difference at all
good question though! lets check
thanks
this is my review for my finals... I really appreciate your help
\[\frac{2-0}{k-1}=-1\] \[\frac{2}{k-1}=-1\] \[2=-1(k-1)\] \[2=-k+1\] \[1=-k\] \[k=-1\] makes no difference which point you pick
no problem, glad to help
thanks a lot
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