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Question Number 92301 by jagoll last updated on 06/May/20
given eq of line (1) [ x,y ] = [3,−2] + t [4,−5] (2) [x,y] = [1,1] + s [ 7,k ] find t and s if (1) ∥ (2) if (1) ⊥ (2)
$$\mathrm{given}\:\mathrm{eq}\:\mathrm{of}\:\mathrm{line}\: \\ $$$$\left(\mathrm{1}\right)\:\left[\:\mathrm{x},\mathrm{y}\:\right]\:=\:\left[\mathrm{3},−\mathrm{2}\right]\:+\:\mathrm{t}\:\left[\mathrm{4},−\mathrm{5}\right]\: \\ $$$$\left(\mathrm{2}\right)\:\left[\mathrm{x},\mathrm{y}\right]\:=\:\left[\mathrm{1},\mathrm{1}\right]\:+\:\mathrm{s}\:\left[\:\mathrm{7},\mathrm{k}\:\right]\: \\ $$$$\mathrm{find}\:\mathrm{t}\:\mathrm{and}\:\mathrm{s}\:\mathrm{if}\:\left(\mathrm{1}\right)\:\parallel\:\left(\mathrm{2}\right) \\ $$$$\mathrm{if}\:\left(\mathrm{1}\right)\:\bot\:\left(\mathrm{2}\right) \\ $$
Commented by jagoll last updated on 06/May/20
thank you both
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$
Commented by john santu last updated on 06/May/20
line (1) x=3+4t ⇒ t = ((x−3)/4) y = −2−5t ⇒t =((−2−y)/5) ⇒ ((x−3)/4) = ((−2−y)/5) 5x +4y −7 = 0 line (2) x = 1+7s ⇒s = ((x−1)/7) y = 1+sk ⇒ s = ((y−1)/k) ⇒ ((x−1)/7) = ((y−1)/k) kx −7y +7−k =0 if parallel ⇒ ((−5)/4) = ((−k)/(−7)) k = − ((35)/4) if perpendicular ⇒ (4/5) = (k/7) k = ((28)/5)
$$\mathrm{line}\:\left(\mathrm{1}\right)\:{x}=\mathrm{3}+\mathrm{4}{t}\:\Rightarrow\:{t}\:=\:\frac{{x}−\mathrm{3}}{\mathrm{4}} \\ $$$${y}\:=\:−\mathrm{2}−\mathrm{5}{t}\:\Rightarrow\mathrm{t}\:=\frac{−\mathrm{2}−{y}}{\mathrm{5}} \\ $$$$\Rightarrow\:\frac{\mathrm{x}−\mathrm{3}}{\mathrm{4}}\:=\:\frac{−\mathrm{2}−\mathrm{y}}{\mathrm{5}}\: \\ $$$$\mathrm{5x}\:+\mathrm{4y}\:−\mathrm{7}\:=\:\mathrm{0} \\ $$$$\mathrm{line}\:\left(\mathrm{2}\right)\:{x}\:=\:\mathrm{1}+\mathrm{7}{s}\:\Rightarrow\mathrm{s}\:=\:\frac{{x}−\mathrm{1}}{\mathrm{7}} \\ $$$${y}\:=\:\mathrm{1}+{sk}\:\Rightarrow\:{s}\:=\:\frac{{y}−\mathrm{1}}{{k}} \\ $$$$\Rightarrow\:\frac{{x}−\mathrm{1}}{\mathrm{7}}\:=\:\frac{{y}−\mathrm{1}}{{k}} \\ $$$${kx}\:−\mathrm{7}{y}\:+\mathrm{7}−{k}\:=\mathrm{0} \\ $$$$\mathrm{if}\:\mathrm{parallel}\:\Rightarrow\:\frac{−\mathrm{5}}{\mathrm{4}}\:=\:\frac{−{k}}{−\mathrm{7}}\: \\ $$$${k}\:=\:−\:\frac{\mathrm{35}}{\mathrm{4}} \\ $$$$\mathrm{if}\:\mathrm{perpendicular}\:\Rightarrow\:\frac{\mathrm{4}}{\mathrm{5}}\:=\:\frac{\mathrm{k}}{\mathrm{7}} \\ $$$${k}\:=\:\frac{\mathrm{28}}{\mathrm{5}}\: \\ $$
Answered by mr W last updated on 06/May/20
(1)//(2): ((−5)/4)=(k/7) ⇒k=−((35)/4) (1)⊥(2): ((−5)/4)×(k/7)=−1 ⇒k=((28)/5) please check question: t and s are parameters and are of any values. it is to find k=?
$$\left(\mathrm{1}\right)//\left(\mathrm{2}\right): \\ $$$$\frac{−\mathrm{5}}{\mathrm{4}}=\frac{{k}}{\mathrm{7}} \\ $$$$\Rightarrow{k}=−\frac{\mathrm{35}}{\mathrm{4}} \\ $$$$ \\ $$$$\left(\mathrm{1}\right)\bot\left(\mathrm{2}\right): \\ $$$$\frac{−\mathrm{5}}{\mathrm{4}}×\frac{{k}}{\mathrm{7}}=−\mathrm{1} \\ $$$$\Rightarrow{k}=\frac{\mathrm{28}}{\mathrm{5}} \\ $$$$ \\ $$$${please}\:{check}\:{question}:\:{t}\:{and}\:{s}\:{are} \\ $$$${parameters}\:{and}\:{are}\:{of}\:{any}\:{values}. \\ $$$${it}\:{is}\:{to}\:{find}\:{k}=? \\ $$
Commented by jagoll last updated on 06/May/20
yes, t & s are parameters sir
$$\mathrm{yes},\:\mathrm{t}\:\&\:\mathrm{s}\:\mathrm{are}\:\mathrm{parameters}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 06/May/20

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