88756
The distance between the lines \(3 x+4 y=9,6 x+\) \(8 \mathrm{y}=15\) is
1 \(\frac{3}{10}\)
2 \(\frac{7}{10}\)
3 \(\frac{3}{2}\)
4 \(\frac{2}{3}\)
Explanation:
(A) : The distance between the parallel line \(\mathrm{d}=\frac{\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\)
\(\Rightarrow 3 x+4 y=9\) and \(6 x+8 y=15\)
ie \(\rightarrow 3 x+4 y=\frac{15}{2}\)
\(c_{1}=9\) and \(c_{2}=\frac{15}{2}\)
\(\mathrm{d}=\frac{\left|9-\frac{15}{2}\right|}{\sqrt{3^{2}+4^{2}}} \Rightarrow \frac{\frac{3}{2}}{\sqrt{25}}\)
\(\mathrm{d}=\frac{3}{2 \times 5}=\frac{3}{10}\)
Straight Line
88757
Let \(d_{1}\) and \(d_{2}\) be the lengths of the perpendiculars drawn from any point of the line \(7 x-9 y+10=0\) upon the lines \(3 x+4 y=5\) and \(12 x+5 y=7\), respectively. Then,
1 \(\mathrm{d}_{1}>\mathrm{d}_{2}\)
2 \(\mathrm{d}_{1}=\mathrm{d}_{2}\)
3 \(\mathrm{d}_{1}\lt \mathrm{d}_{2}\)
4 \(\mathrm{d}_{1}=2 \mathrm{~d}_{2}\)
Explanation:
(B) : let \((\mathrm{h}, \mathrm{k})\) be any point on the line \(7 \mathrm{x}-9 \mathrm{y}+\) \(10=0\), then \(7 \mathrm{~h}-9 \mathrm{k}+10=0\)
\(\Rightarrow \quad 7 \mathrm{~h}=9 \mathrm{k}-10\)
\(\Rightarrow \mathrm{h}=\frac{9 \mathrm{k}-10}{7}\)
Now, perpendicular distance from point \((h, k)\) to the line \(3 x+4 y=5\) is \(d\),
\(\mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{\sqrt{3^{2}+4^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}\)
And perpendicular distance from \((\mathrm{h}, \mathrm{k})\) to the line \(12 \mathrm{x}+\) \(5 y=7\) is \(d_{2}\)
\(\therefore \quad \mathrm{d}_{2}=\frac{12 h+5 \mathrm{k}-7}{\sqrt{12^{2}+5^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{2}=\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\text { Now, } \mathrm{d}_{1}-\mathrm{d}_{2}=\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}-\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}\)
\(=\frac{13(3 \mathrm{~h}+4 \mathrm{k}-5)-5(12 \mathrm{~h}+5 \mathrm{k}-7)}{65}\)
\(=\frac{39 \mathrm{~h}+52 \mathrm{k}-65-60 \mathrm{~h}-25 \mathrm{k}+35}{65}=\frac{-21 \mathrm{~h}+27 \mathrm{k}-30}{65}\)
\(\Rightarrow \frac{-21\left(\frac{9 \mathrm{k}-10}{7}\right)+27 \mathrm{k}-30}{65}=\frac{-27+30+27 \mathrm{k}-30}{65}=0\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}=0 \Rightarrow \mathrm{d}_{1}=\mathrm{d}_{2}\)
WB JEE-2017
Straight Line
88758
The motion of a particle along a straight line is described by the function \(x=(2 t-3)^{2}\), where \(x\) is in metre and \(t\) is in second. Then, the velocity of the particle at origin is
1 0
2 1
3 2
4 None of these
Explanation:
(A) : Given that, \(\mathrm{x}=(2 \mathrm{t}-3)^{2}\)
On differentiating w.r.r.x, we get
\(\frac{\mathrm{dx}}{\mathrm{dt}}=2(2 \mathrm{t}-3)(2)\)
At origin, \(\mathrm{x}=0\)
\(\therefore \quad 2 \mathrm{t}-3=0 \Rightarrow \mathrm{t}=\frac{3}{2}\)
\(\therefore \quad\) Velocity \(=\frac{\mathrm{dx}}{\mathrm{dt}}=4\left(2 \times \frac{3}{2}-3\right)=0\)
Straight Line
88759
The ratio in which the line \(3 x+4 y+2=0\) divides the distance between \(3 x+4 y+5=0\) and \(3 x+4 y-5=0\), is
1 \(7: 3\)
2 \(3: 7\)
3 \(2: 3\)
4 None of above
Explanation:
(B) : The lines \(3 x+47+5\) and \(3 x+47+2=0\)
lies on the same side of origin the distance between these lines \(=\left|\mathrm{C}_{1}-\mathrm{C}_{2}\right|\)
\(=5-2=3\)
similarly lines \(3 x+47+2=0\) and \(3 x+47-5=0\)
lies opposite side w.r. to origin then distance between these lines \(=5+2=7\)
Hence required ratio is \(3: 7\)
88756
The distance between the lines \(3 x+4 y=9,6 x+\) \(8 \mathrm{y}=15\) is
1 \(\frac{3}{10}\)
2 \(\frac{7}{10}\)
3 \(\frac{3}{2}\)
4 \(\frac{2}{3}\)
Explanation:
(A) : The distance between the parallel line \(\mathrm{d}=\frac{\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\)
\(\Rightarrow 3 x+4 y=9\) and \(6 x+8 y=15\)
ie \(\rightarrow 3 x+4 y=\frac{15}{2}\)
\(c_{1}=9\) and \(c_{2}=\frac{15}{2}\)
\(\mathrm{d}=\frac{\left|9-\frac{15}{2}\right|}{\sqrt{3^{2}+4^{2}}} \Rightarrow \frac{\frac{3}{2}}{\sqrt{25}}\)
\(\mathrm{d}=\frac{3}{2 \times 5}=\frac{3}{10}\)
Straight Line
88757
Let \(d_{1}\) and \(d_{2}\) be the lengths of the perpendiculars drawn from any point of the line \(7 x-9 y+10=0\) upon the lines \(3 x+4 y=5\) and \(12 x+5 y=7\), respectively. Then,
1 \(\mathrm{d}_{1}>\mathrm{d}_{2}\)
2 \(\mathrm{d}_{1}=\mathrm{d}_{2}\)
3 \(\mathrm{d}_{1}\lt \mathrm{d}_{2}\)
4 \(\mathrm{d}_{1}=2 \mathrm{~d}_{2}\)
Explanation:
(B) : let \((\mathrm{h}, \mathrm{k})\) be any point on the line \(7 \mathrm{x}-9 \mathrm{y}+\) \(10=0\), then \(7 \mathrm{~h}-9 \mathrm{k}+10=0\)
\(\Rightarrow \quad 7 \mathrm{~h}=9 \mathrm{k}-10\)
\(\Rightarrow \mathrm{h}=\frac{9 \mathrm{k}-10}{7}\)
Now, perpendicular distance from point \((h, k)\) to the line \(3 x+4 y=5\) is \(d\),
\(\mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{\sqrt{3^{2}+4^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}\)
And perpendicular distance from \((\mathrm{h}, \mathrm{k})\) to the line \(12 \mathrm{x}+\) \(5 y=7\) is \(d_{2}\)
\(\therefore \quad \mathrm{d}_{2}=\frac{12 h+5 \mathrm{k}-7}{\sqrt{12^{2}+5^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{2}=\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\text { Now, } \mathrm{d}_{1}-\mathrm{d}_{2}=\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}-\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}\)
\(=\frac{13(3 \mathrm{~h}+4 \mathrm{k}-5)-5(12 \mathrm{~h}+5 \mathrm{k}-7)}{65}\)
\(=\frac{39 \mathrm{~h}+52 \mathrm{k}-65-60 \mathrm{~h}-25 \mathrm{k}+35}{65}=\frac{-21 \mathrm{~h}+27 \mathrm{k}-30}{65}\)
\(\Rightarrow \frac{-21\left(\frac{9 \mathrm{k}-10}{7}\right)+27 \mathrm{k}-30}{65}=\frac{-27+30+27 \mathrm{k}-30}{65}=0\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}=0 \Rightarrow \mathrm{d}_{1}=\mathrm{d}_{2}\)
WB JEE-2017
Straight Line
88758
The motion of a particle along a straight line is described by the function \(x=(2 t-3)^{2}\), where \(x\) is in metre and \(t\) is in second. Then, the velocity of the particle at origin is
1 0
2 1
3 2
4 None of these
Explanation:
(A) : Given that, \(\mathrm{x}=(2 \mathrm{t}-3)^{2}\)
On differentiating w.r.r.x, we get
\(\frac{\mathrm{dx}}{\mathrm{dt}}=2(2 \mathrm{t}-3)(2)\)
At origin, \(\mathrm{x}=0\)
\(\therefore \quad 2 \mathrm{t}-3=0 \Rightarrow \mathrm{t}=\frac{3}{2}\)
\(\therefore \quad\) Velocity \(=\frac{\mathrm{dx}}{\mathrm{dt}}=4\left(2 \times \frac{3}{2}-3\right)=0\)
Straight Line
88759
The ratio in which the line \(3 x+4 y+2=0\) divides the distance between \(3 x+4 y+5=0\) and \(3 x+4 y-5=0\), is
1 \(7: 3\)
2 \(3: 7\)
3 \(2: 3\)
4 None of above
Explanation:
(B) : The lines \(3 x+47+5\) and \(3 x+47+2=0\)
lies on the same side of origin the distance between these lines \(=\left|\mathrm{C}_{1}-\mathrm{C}_{2}\right|\)
\(=5-2=3\)
similarly lines \(3 x+47+2=0\) and \(3 x+47-5=0\)
lies opposite side w.r. to origin then distance between these lines \(=5+2=7\)
Hence required ratio is \(3: 7\)
88756
The distance between the lines \(3 x+4 y=9,6 x+\) \(8 \mathrm{y}=15\) is
1 \(\frac{3}{10}\)
2 \(\frac{7}{10}\)
3 \(\frac{3}{2}\)
4 \(\frac{2}{3}\)
Explanation:
(A) : The distance between the parallel line \(\mathrm{d}=\frac{\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\)
\(\Rightarrow 3 x+4 y=9\) and \(6 x+8 y=15\)
ie \(\rightarrow 3 x+4 y=\frac{15}{2}\)
\(c_{1}=9\) and \(c_{2}=\frac{15}{2}\)
\(\mathrm{d}=\frac{\left|9-\frac{15}{2}\right|}{\sqrt{3^{2}+4^{2}}} \Rightarrow \frac{\frac{3}{2}}{\sqrt{25}}\)
\(\mathrm{d}=\frac{3}{2 \times 5}=\frac{3}{10}\)
Straight Line
88757
Let \(d_{1}\) and \(d_{2}\) be the lengths of the perpendiculars drawn from any point of the line \(7 x-9 y+10=0\) upon the lines \(3 x+4 y=5\) and \(12 x+5 y=7\), respectively. Then,
1 \(\mathrm{d}_{1}>\mathrm{d}_{2}\)
2 \(\mathrm{d}_{1}=\mathrm{d}_{2}\)
3 \(\mathrm{d}_{1}\lt \mathrm{d}_{2}\)
4 \(\mathrm{d}_{1}=2 \mathrm{~d}_{2}\)
Explanation:
(B) : let \((\mathrm{h}, \mathrm{k})\) be any point on the line \(7 \mathrm{x}-9 \mathrm{y}+\) \(10=0\), then \(7 \mathrm{~h}-9 \mathrm{k}+10=0\)
\(\Rightarrow \quad 7 \mathrm{~h}=9 \mathrm{k}-10\)
\(\Rightarrow \mathrm{h}=\frac{9 \mathrm{k}-10}{7}\)
Now, perpendicular distance from point \((h, k)\) to the line \(3 x+4 y=5\) is \(d\),
\(\mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{\sqrt{3^{2}+4^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}\)
And perpendicular distance from \((\mathrm{h}, \mathrm{k})\) to the line \(12 \mathrm{x}+\) \(5 y=7\) is \(d_{2}\)
\(\therefore \quad \mathrm{d}_{2}=\frac{12 h+5 \mathrm{k}-7}{\sqrt{12^{2}+5^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{2}=\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\text { Now, } \mathrm{d}_{1}-\mathrm{d}_{2}=\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}-\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}\)
\(=\frac{13(3 \mathrm{~h}+4 \mathrm{k}-5)-5(12 \mathrm{~h}+5 \mathrm{k}-7)}{65}\)
\(=\frac{39 \mathrm{~h}+52 \mathrm{k}-65-60 \mathrm{~h}-25 \mathrm{k}+35}{65}=\frac{-21 \mathrm{~h}+27 \mathrm{k}-30}{65}\)
\(\Rightarrow \frac{-21\left(\frac{9 \mathrm{k}-10}{7}\right)+27 \mathrm{k}-30}{65}=\frac{-27+30+27 \mathrm{k}-30}{65}=0\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}=0 \Rightarrow \mathrm{d}_{1}=\mathrm{d}_{2}\)
WB JEE-2017
Straight Line
88758
The motion of a particle along a straight line is described by the function \(x=(2 t-3)^{2}\), where \(x\) is in metre and \(t\) is in second. Then, the velocity of the particle at origin is
1 0
2 1
3 2
4 None of these
Explanation:
(A) : Given that, \(\mathrm{x}=(2 \mathrm{t}-3)^{2}\)
On differentiating w.r.r.x, we get
\(\frac{\mathrm{dx}}{\mathrm{dt}}=2(2 \mathrm{t}-3)(2)\)
At origin, \(\mathrm{x}=0\)
\(\therefore \quad 2 \mathrm{t}-3=0 \Rightarrow \mathrm{t}=\frac{3}{2}\)
\(\therefore \quad\) Velocity \(=\frac{\mathrm{dx}}{\mathrm{dt}}=4\left(2 \times \frac{3}{2}-3\right)=0\)
Straight Line
88759
The ratio in which the line \(3 x+4 y+2=0\) divides the distance between \(3 x+4 y+5=0\) and \(3 x+4 y-5=0\), is
1 \(7: 3\)
2 \(3: 7\)
3 \(2: 3\)
4 None of above
Explanation:
(B) : The lines \(3 x+47+5\) and \(3 x+47+2=0\)
lies on the same side of origin the distance between these lines \(=\left|\mathrm{C}_{1}-\mathrm{C}_{2}\right|\)
\(=5-2=3\)
similarly lines \(3 x+47+2=0\) and \(3 x+47-5=0\)
lies opposite side w.r. to origin then distance between these lines \(=5+2=7\)
Hence required ratio is \(3: 7\)
NEET Test Series from KOTA - 10 Papers In MS WORD
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Straight Line
88756
The distance between the lines \(3 x+4 y=9,6 x+\) \(8 \mathrm{y}=15\) is
1 \(\frac{3}{10}\)
2 \(\frac{7}{10}\)
3 \(\frac{3}{2}\)
4 \(\frac{2}{3}\)
Explanation:
(A) : The distance between the parallel line \(\mathrm{d}=\frac{\left|\mathrm{c}_{1}-\mathrm{c}_{2}\right|}{\sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}}\)
\(\Rightarrow 3 x+4 y=9\) and \(6 x+8 y=15\)
ie \(\rightarrow 3 x+4 y=\frac{15}{2}\)
\(c_{1}=9\) and \(c_{2}=\frac{15}{2}\)
\(\mathrm{d}=\frac{\left|9-\frac{15}{2}\right|}{\sqrt{3^{2}+4^{2}}} \Rightarrow \frac{\frac{3}{2}}{\sqrt{25}}\)
\(\mathrm{d}=\frac{3}{2 \times 5}=\frac{3}{10}\)
Straight Line
88757
Let \(d_{1}\) and \(d_{2}\) be the lengths of the perpendiculars drawn from any point of the line \(7 x-9 y+10=0\) upon the lines \(3 x+4 y=5\) and \(12 x+5 y=7\), respectively. Then,
1 \(\mathrm{d}_{1}>\mathrm{d}_{2}\)
2 \(\mathrm{d}_{1}=\mathrm{d}_{2}\)
3 \(\mathrm{d}_{1}\lt \mathrm{d}_{2}\)
4 \(\mathrm{d}_{1}=2 \mathrm{~d}_{2}\)
Explanation:
(B) : let \((\mathrm{h}, \mathrm{k})\) be any point on the line \(7 \mathrm{x}-9 \mathrm{y}+\) \(10=0\), then \(7 \mathrm{~h}-9 \mathrm{k}+10=0\)
\(\Rightarrow \quad 7 \mathrm{~h}=9 \mathrm{k}-10\)
\(\Rightarrow \mathrm{h}=\frac{9 \mathrm{k}-10}{7}\)
Now, perpendicular distance from point \((h, k)\) to the line \(3 x+4 y=5\) is \(d\),
\(\mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{\sqrt{3^{2}+4^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{1} =\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}\)
And perpendicular distance from \((\mathrm{h}, \mathrm{k})\) to the line \(12 \mathrm{x}+\) \(5 y=7\) is \(d_{2}\)
\(\therefore \quad \mathrm{d}_{2}=\frac{12 h+5 \mathrm{k}-7}{\sqrt{12^{2}+5^{2}}}\)
\(\Rightarrow \quad \mathrm{d}_{2}=\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\text { Now, } \mathrm{d}_{1}-\mathrm{d}_{2}=\frac{3 \mathrm{~h}+4 \mathrm{k}-5}{5}-\frac{12 \mathrm{~h}+5 \mathrm{k}-7}{13}\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}\)
\(=\frac{13(3 \mathrm{~h}+4 \mathrm{k}-5)-5(12 \mathrm{~h}+5 \mathrm{k}-7)}{65}\)
\(=\frac{39 \mathrm{~h}+52 \mathrm{k}-65-60 \mathrm{~h}-25 \mathrm{k}+35}{65}=\frac{-21 \mathrm{~h}+27 \mathrm{k}-30}{65}\)
\(\Rightarrow \frac{-21\left(\frac{9 \mathrm{k}-10}{7}\right)+27 \mathrm{k}-30}{65}=\frac{-27+30+27 \mathrm{k}-30}{65}=0\)
\(\Rightarrow \mathrm{d}_{1}-\mathrm{d}_{2}=0 \Rightarrow \mathrm{d}_{1}=\mathrm{d}_{2}\)
WB JEE-2017
Straight Line
88758
The motion of a particle along a straight line is described by the function \(x=(2 t-3)^{2}\), where \(x\) is in metre and \(t\) is in second. Then, the velocity of the particle at origin is
1 0
2 1
3 2
4 None of these
Explanation:
(A) : Given that, \(\mathrm{x}=(2 \mathrm{t}-3)^{2}\)
On differentiating w.r.r.x, we get
\(\frac{\mathrm{dx}}{\mathrm{dt}}=2(2 \mathrm{t}-3)(2)\)
At origin, \(\mathrm{x}=0\)
\(\therefore \quad 2 \mathrm{t}-3=0 \Rightarrow \mathrm{t}=\frac{3}{2}\)
\(\therefore \quad\) Velocity \(=\frac{\mathrm{dx}}{\mathrm{dt}}=4\left(2 \times \frac{3}{2}-3\right)=0\)
Straight Line
88759
The ratio in which the line \(3 x+4 y+2=0\) divides the distance between \(3 x+4 y+5=0\) and \(3 x+4 y-5=0\), is
1 \(7: 3\)
2 \(3: 7\)
3 \(2: 3\)
4 None of above
Explanation:
(B) : The lines \(3 x+47+5\) and \(3 x+47+2=0\)
lies on the same side of origin the distance between these lines \(=\left|\mathrm{C}_{1}-\mathrm{C}_{2}\right|\)
\(=5-2=3\)
similarly lines \(3 x+47+2=0\) and \(3 x+47-5=0\)
lies opposite side w.r. to origin then distance between these lines \(=5+2=7\)
Hence required ratio is \(3: 7\)
