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Question Number 32834 by Rio Mike last updated on 03/Apr/18
find  the value of k for which the   length of the line segment joining  (k^� ,2) and (−1,4) is 2(√2) units.      show full working...
$${find}\:\:{the}\:{value}\:{of}\:{k}\:{for}\:{which}\:{the}\: \\ $$$${length}\:{of}\:{the}\:{line}\:{segment}\:{joining} \\ $$$$\left(\bar {{k}},\mathrm{2}\right)\:{and}\:\left(−\mathrm{1},\mathrm{4}\right)\:{is}\:\mathrm{2}\sqrt{\mathrm{2}}\:{units}. \\ $$$$\:\:\:\:{show}\:{full}\:{working}… \\ $$
Answered by MJS last updated on 03/Apr/18
(√((k+1)^2 +(2−4)^2 ))=2(√2)  (√(k^2 +2k+5))=2(√2)  k^2 +2k+5=8  k^2 +2k−3=0  k=−1±(√(1+3))=−1±2  k_1 =−3  k_2 =1
$$\sqrt{\left({k}+\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}−\mathrm{4}\right)^{\mathrm{2}} }=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$$\sqrt{{k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{5}}=\mathrm{2}\sqrt{\mathrm{2}} \\ $$$${k}^{\mathrm{2}} +\mathrm{2}{k}+\mathrm{5}=\mathrm{8} \\ $$$${k}^{\mathrm{2}} +\mathrm{2}{k}−\mathrm{3}=\mathrm{0} \\ $$$${k}=−\mathrm{1}\pm\sqrt{\mathrm{1}+\mathrm{3}}=−\mathrm{1}\pm\mathrm{2} \\ $$$${k}_{\mathrm{1}} =−\mathrm{3} \\ $$$${k}_{\mathrm{2}} =\mathrm{1} \\ $$

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