NEET Test Series from KOTA - 10 Papers In MS WORD
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Co-Ordinate system
88251
A straight line through the origin \(O\) meets the parallel lines \(4 x+2 y=9\) and \(2 x+y+6=0\) at \(\mathbf{P}\) and \(Q\) respectively. The point \(O\) divides the segment \(P Q\) in the ratio
1 \(1: 2\)
2 \(3: 4\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
(B) : Given, The equation of parallel lines are \(4 x+2 y=9\) and \(2 x+y+6=0\) We know that, The equation of line passing through origin is \(\mathrm{y}=\mathrm{mx} \tag{iii}\) Now, from equation (i) and (iii) \(4 x+2 y=9\) \(2 x+y=9 / 2\) \(2 x+m x=9 / 2\) \(x(2+m)=\frac{9}{2}\) \(x=\frac{9}{2(2+m)}\) Now, substitute the value of \(x\) in \(y\) of equation (i) \(y=\frac{9 m}{2(2+m)}\) The coordinate of point of intersection on line \(4 x+2 y=9 \text { is } P\left(\frac{9}{2(2+m)}, \frac{9 m}{2(2+m)}\right)\) Similarly, the coordinate of point of intersection on the line \(\quad 2 x+y+6=0\) is \(Q\left(\frac{-6}{2+m}, \frac{-6 m}{2+m}\right)\) Now, The section formula for the ratio \(\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\right) \quad\left[\begin{array}{l}\text { Where } \mathrm{m} \text { and } \mathrm{n} \mathrm{be} \\ \text { the ratios }\end{array}\right]\) Let the ratio is \(\lambda: 1\) \(\therefore \quad \left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}\right)=0\) \(\frac{\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9(1)}{2 \mathrm{~m}+4}}{\lambda+1}=0\) \(\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9}{2(2+\mathrm{m})}=0\) \(-12 \lambda=-9\) \(\lambda=3 / 4\) \(\therefore\) Ratio is \(\frac{3}{4}: 1=3: 4\)
WB JEE-2020
Co-Ordinate system
88252
The ratio in which the line joining points \(\mathbf{A}(-1\), \(-1)\) and \(B(2,1)\) divides the line joining \(C(3,4)\) D \((1, \mathbf{1})\)
1 7:5 Internally
2 7:5 Externally
3 7:11 Internally
4 7:11 Externally
Explanation:
(B): Given, A (-1, -1) , B (2, 1), C (3, 4) , D (1, 2) Let the ratio be \(\lambda: 1\) equation of line through \(A B\) \(y+1=\frac{2}{3}(x+1)\) \(3 y+3=2 x+2\) \(2 x-3 y-1=0\) Let, \(\mathrm{E}\) be the point where line intersect \(\mathrm{E}=\left(\frac{\lambda+3}{\lambda+1}, \frac{2 \lambda+4}{\lambda+1}\right)\) \(2\left(\frac{\lambda+3}{\lambda+1}\right)-3\left(\frac{2 \lambda+4}{\lambda+1}\right)-1=0\) \(2 \lambda+6-6 \lambda-12-\lambda-1=0\) \(-5 \lambda-7=0\) \(\lambda=\frac{-7}{5}\) 7: 5 Externally
AP EAMCET-2021-23.08.2021
Co-Ordinate system
88253
The vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(2,2), \mathrm{B}(-4,-4)\) and \(C(5,-8)\). Find the length of a median of a triangle, which is passing through the point \(C\).
1 \(\sqrt{65}\)
2 \(\sqrt{117}\)
3 \(\sqrt{85}\)
4 \(\sqrt{116}\)
Explanation:
(C) \(\mathrm{D} \text { is a mid-point of } \mathrm{AB}\) \(\mathrm{h}=\frac{-4+2}{2}=-1, \mathrm{k}=\frac{-4+2}{2}=-1\) \((\mathrm{~h}, \mathrm{k})=(-1,-1)\) Distance from \((5,-8)\) and \((-1,-1)\) is \(\mathrm{D}=\sqrt{(5+1)^{2}+(-8+1)^{2}}=\sqrt{36+49}=\sqrt{85}\)
GUJCET-2011
Co-Ordinate system
88254
A straight line passing through origin 0 intersects the lines \(10 x-8 y-10=0\) and \(\frac{x}{4}-\frac{y}{5}+1=0\) at right angles and at the points \(P\) and \(Q\) respectively. Then the ratio in which \(O\) divides the line segment \(P Q\) is
1 \(1: 2\)
2 \(1: 4\)
3 \(1: 1\)
4 \(3: 4\)
Explanation:
: Let, Equation of required line \(y=m x+c\) Given, it is normal to \(10 x-8 y=10\) and \(5 x-4 y+20=0\) So, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\mathrm{m} \times\left(\frac{5}{4}\right)=-1\) \(\mathrm{~m}=-\frac{4}{5}\) \(P\) will be the point of intersection of \(5 x-4 y=5\) \(4 x+5 y=0\) \(P:\left(\frac{25}{41}, \frac{-20}{41}\right)\) \(Q\) will be the point of intersection of \(5 x-4 y=-20\) \(4 x+5 y=0\) \(\mathrm{Q}:\left(\frac{-100}{41}, \frac{80}{41}\right)\) Let \(\mathrm{O}\) divides PQ in \(\lambda: 1\) \((0,0) =\left(\frac{-100 \lambda+25}{41(\lambda+1)}, \frac{80 \lambda-20}{41(\lambda+1)}\right)\) \(\lambda =\frac{1}{4}\) \(\lambda: 1=1: 4\)
88251
A straight line through the origin \(O\) meets the parallel lines \(4 x+2 y=9\) and \(2 x+y+6=0\) at \(\mathbf{P}\) and \(Q\) respectively. The point \(O\) divides the segment \(P Q\) in the ratio
1 \(1: 2\)
2 \(3: 4\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
(B) : Given, The equation of parallel lines are \(4 x+2 y=9\) and \(2 x+y+6=0\) We know that, The equation of line passing through origin is \(\mathrm{y}=\mathrm{mx} \tag{iii}\) Now, from equation (i) and (iii) \(4 x+2 y=9\) \(2 x+y=9 / 2\) \(2 x+m x=9 / 2\) \(x(2+m)=\frac{9}{2}\) \(x=\frac{9}{2(2+m)}\) Now, substitute the value of \(x\) in \(y\) of equation (i) \(y=\frac{9 m}{2(2+m)}\) The coordinate of point of intersection on line \(4 x+2 y=9 \text { is } P\left(\frac{9}{2(2+m)}, \frac{9 m}{2(2+m)}\right)\) Similarly, the coordinate of point of intersection on the line \(\quad 2 x+y+6=0\) is \(Q\left(\frac{-6}{2+m}, \frac{-6 m}{2+m}\right)\) Now, The section formula for the ratio \(\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\right) \quad\left[\begin{array}{l}\text { Where } \mathrm{m} \text { and } \mathrm{n} \mathrm{be} \\ \text { the ratios }\end{array}\right]\) Let the ratio is \(\lambda: 1\) \(\therefore \quad \left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}\right)=0\) \(\frac{\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9(1)}{2 \mathrm{~m}+4}}{\lambda+1}=0\) \(\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9}{2(2+\mathrm{m})}=0\) \(-12 \lambda=-9\) \(\lambda=3 / 4\) \(\therefore\) Ratio is \(\frac{3}{4}: 1=3: 4\)
WB JEE-2020
Co-Ordinate system
88252
The ratio in which the line joining points \(\mathbf{A}(-1\), \(-1)\) and \(B(2,1)\) divides the line joining \(C(3,4)\) D \((1, \mathbf{1})\)
1 7:5 Internally
2 7:5 Externally
3 7:11 Internally
4 7:11 Externally
Explanation:
(B): Given, A (-1, -1) , B (2, 1), C (3, 4) , D (1, 2) Let the ratio be \(\lambda: 1\) equation of line through \(A B\) \(y+1=\frac{2}{3}(x+1)\) \(3 y+3=2 x+2\) \(2 x-3 y-1=0\) Let, \(\mathrm{E}\) be the point where line intersect \(\mathrm{E}=\left(\frac{\lambda+3}{\lambda+1}, \frac{2 \lambda+4}{\lambda+1}\right)\) \(2\left(\frac{\lambda+3}{\lambda+1}\right)-3\left(\frac{2 \lambda+4}{\lambda+1}\right)-1=0\) \(2 \lambda+6-6 \lambda-12-\lambda-1=0\) \(-5 \lambda-7=0\) \(\lambda=\frac{-7}{5}\) 7: 5 Externally
AP EAMCET-2021-23.08.2021
Co-Ordinate system
88253
The vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(2,2), \mathrm{B}(-4,-4)\) and \(C(5,-8)\). Find the length of a median of a triangle, which is passing through the point \(C\).
1 \(\sqrt{65}\)
2 \(\sqrt{117}\)
3 \(\sqrt{85}\)
4 \(\sqrt{116}\)
Explanation:
(C) \(\mathrm{D} \text { is a mid-point of } \mathrm{AB}\) \(\mathrm{h}=\frac{-4+2}{2}=-1, \mathrm{k}=\frac{-4+2}{2}=-1\) \((\mathrm{~h}, \mathrm{k})=(-1,-1)\) Distance from \((5,-8)\) and \((-1,-1)\) is \(\mathrm{D}=\sqrt{(5+1)^{2}+(-8+1)^{2}}=\sqrt{36+49}=\sqrt{85}\)
GUJCET-2011
Co-Ordinate system
88254
A straight line passing through origin 0 intersects the lines \(10 x-8 y-10=0\) and \(\frac{x}{4}-\frac{y}{5}+1=0\) at right angles and at the points \(P\) and \(Q\) respectively. Then the ratio in which \(O\) divides the line segment \(P Q\) is
1 \(1: 2\)
2 \(1: 4\)
3 \(1: 1\)
4 \(3: 4\)
Explanation:
: Let, Equation of required line \(y=m x+c\) Given, it is normal to \(10 x-8 y=10\) and \(5 x-4 y+20=0\) So, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\mathrm{m} \times\left(\frac{5}{4}\right)=-1\) \(\mathrm{~m}=-\frac{4}{5}\) \(P\) will be the point of intersection of \(5 x-4 y=5\) \(4 x+5 y=0\) \(P:\left(\frac{25}{41}, \frac{-20}{41}\right)\) \(Q\) will be the point of intersection of \(5 x-4 y=-20\) \(4 x+5 y=0\) \(\mathrm{Q}:\left(\frac{-100}{41}, \frac{80}{41}\right)\) Let \(\mathrm{O}\) divides PQ in \(\lambda: 1\) \((0,0) =\left(\frac{-100 \lambda+25}{41(\lambda+1)}, \frac{80 \lambda-20}{41(\lambda+1)}\right)\) \(\lambda =\frac{1}{4}\) \(\lambda: 1=1: 4\)
88251
A straight line through the origin \(O\) meets the parallel lines \(4 x+2 y=9\) and \(2 x+y+6=0\) at \(\mathbf{P}\) and \(Q\) respectively. The point \(O\) divides the segment \(P Q\) in the ratio
1 \(1: 2\)
2 \(3: 4\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
(B) : Given, The equation of parallel lines are \(4 x+2 y=9\) and \(2 x+y+6=0\) We know that, The equation of line passing through origin is \(\mathrm{y}=\mathrm{mx} \tag{iii}\) Now, from equation (i) and (iii) \(4 x+2 y=9\) \(2 x+y=9 / 2\) \(2 x+m x=9 / 2\) \(x(2+m)=\frac{9}{2}\) \(x=\frac{9}{2(2+m)}\) Now, substitute the value of \(x\) in \(y\) of equation (i) \(y=\frac{9 m}{2(2+m)}\) The coordinate of point of intersection on line \(4 x+2 y=9 \text { is } P\left(\frac{9}{2(2+m)}, \frac{9 m}{2(2+m)}\right)\) Similarly, the coordinate of point of intersection on the line \(\quad 2 x+y+6=0\) is \(Q\left(\frac{-6}{2+m}, \frac{-6 m}{2+m}\right)\) Now, The section formula for the ratio \(\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\right) \quad\left[\begin{array}{l}\text { Where } \mathrm{m} \text { and } \mathrm{n} \mathrm{be} \\ \text { the ratios }\end{array}\right]\) Let the ratio is \(\lambda: 1\) \(\therefore \quad \left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}\right)=0\) \(\frac{\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9(1)}{2 \mathrm{~m}+4}}{\lambda+1}=0\) \(\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9}{2(2+\mathrm{m})}=0\) \(-12 \lambda=-9\) \(\lambda=3 / 4\) \(\therefore\) Ratio is \(\frac{3}{4}: 1=3: 4\)
WB JEE-2020
Co-Ordinate system
88252
The ratio in which the line joining points \(\mathbf{A}(-1\), \(-1)\) and \(B(2,1)\) divides the line joining \(C(3,4)\) D \((1, \mathbf{1})\)
1 7:5 Internally
2 7:5 Externally
3 7:11 Internally
4 7:11 Externally
Explanation:
(B): Given, A (-1, -1) , B (2, 1), C (3, 4) , D (1, 2) Let the ratio be \(\lambda: 1\) equation of line through \(A B\) \(y+1=\frac{2}{3}(x+1)\) \(3 y+3=2 x+2\) \(2 x-3 y-1=0\) Let, \(\mathrm{E}\) be the point where line intersect \(\mathrm{E}=\left(\frac{\lambda+3}{\lambda+1}, \frac{2 \lambda+4}{\lambda+1}\right)\) \(2\left(\frac{\lambda+3}{\lambda+1}\right)-3\left(\frac{2 \lambda+4}{\lambda+1}\right)-1=0\) \(2 \lambda+6-6 \lambda-12-\lambda-1=0\) \(-5 \lambda-7=0\) \(\lambda=\frac{-7}{5}\) 7: 5 Externally
AP EAMCET-2021-23.08.2021
Co-Ordinate system
88253
The vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(2,2), \mathrm{B}(-4,-4)\) and \(C(5,-8)\). Find the length of a median of a triangle, which is passing through the point \(C\).
1 \(\sqrt{65}\)
2 \(\sqrt{117}\)
3 \(\sqrt{85}\)
4 \(\sqrt{116}\)
Explanation:
(C) \(\mathrm{D} \text { is a mid-point of } \mathrm{AB}\) \(\mathrm{h}=\frac{-4+2}{2}=-1, \mathrm{k}=\frac{-4+2}{2}=-1\) \((\mathrm{~h}, \mathrm{k})=(-1,-1)\) Distance from \((5,-8)\) and \((-1,-1)\) is \(\mathrm{D}=\sqrt{(5+1)^{2}+(-8+1)^{2}}=\sqrt{36+49}=\sqrt{85}\)
GUJCET-2011
Co-Ordinate system
88254
A straight line passing through origin 0 intersects the lines \(10 x-8 y-10=0\) and \(\frac{x}{4}-\frac{y}{5}+1=0\) at right angles and at the points \(P\) and \(Q\) respectively. Then the ratio in which \(O\) divides the line segment \(P Q\) is
1 \(1: 2\)
2 \(1: 4\)
3 \(1: 1\)
4 \(3: 4\)
Explanation:
: Let, Equation of required line \(y=m x+c\) Given, it is normal to \(10 x-8 y=10\) and \(5 x-4 y+20=0\) So, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\mathrm{m} \times\left(\frac{5}{4}\right)=-1\) \(\mathrm{~m}=-\frac{4}{5}\) \(P\) will be the point of intersection of \(5 x-4 y=5\) \(4 x+5 y=0\) \(P:\left(\frac{25}{41}, \frac{-20}{41}\right)\) \(Q\) will be the point of intersection of \(5 x-4 y=-20\) \(4 x+5 y=0\) \(\mathrm{Q}:\left(\frac{-100}{41}, \frac{80}{41}\right)\) Let \(\mathrm{O}\) divides PQ in \(\lambda: 1\) \((0,0) =\left(\frac{-100 \lambda+25}{41(\lambda+1)}, \frac{80 \lambda-20}{41(\lambda+1)}\right)\) \(\lambda =\frac{1}{4}\) \(\lambda: 1=1: 4\)
88251
A straight line through the origin \(O\) meets the parallel lines \(4 x+2 y=9\) and \(2 x+y+6=0\) at \(\mathbf{P}\) and \(Q\) respectively. The point \(O\) divides the segment \(P Q\) in the ratio
1 \(1: 2\)
2 \(3: 4\)
3 \(2: 1\)
4 \(4: 3\)
Explanation:
(B) : Given, The equation of parallel lines are \(4 x+2 y=9\) and \(2 x+y+6=0\) We know that, The equation of line passing through origin is \(\mathrm{y}=\mathrm{mx} \tag{iii}\) Now, from equation (i) and (iii) \(4 x+2 y=9\) \(2 x+y=9 / 2\) \(2 x+m x=9 / 2\) \(x(2+m)=\frac{9}{2}\) \(x=\frac{9}{2(2+m)}\) Now, substitute the value of \(x\) in \(y\) of equation (i) \(y=\frac{9 m}{2(2+m)}\) The coordinate of point of intersection on line \(4 x+2 y=9 \text { is } P\left(\frac{9}{2(2+m)}, \frac{9 m}{2(2+m)}\right)\) Similarly, the coordinate of point of intersection on the line \(\quad 2 x+y+6=0\) is \(Q\left(\frac{-6}{2+m}, \frac{-6 m}{2+m}\right)\) Now, The section formula for the ratio \(\left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}, \frac{\mathrm{my}_{2}+\mathrm{ny}_{1}}{\mathrm{~m}+\mathrm{n}}\right) \quad\left[\begin{array}{l}\text { Where } \mathrm{m} \text { and } \mathrm{n} \mathrm{be} \\ \text { the ratios }\end{array}\right]\) Let the ratio is \(\lambda: 1\) \(\therefore \quad \left(\frac{\mathrm{mx}_{2}+\mathrm{nx}_{1}}{\mathrm{~m}+\mathrm{n}}\right)=0\) \(\frac{\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9(1)}{2 \mathrm{~m}+4}}{\lambda+1}=0\) \(\frac{-6 \lambda}{2+\mathrm{m}}+\frac{9}{2(2+\mathrm{m})}=0\) \(-12 \lambda=-9\) \(\lambda=3 / 4\) \(\therefore\) Ratio is \(\frac{3}{4}: 1=3: 4\)
WB JEE-2020
Co-Ordinate system
88252
The ratio in which the line joining points \(\mathbf{A}(-1\), \(-1)\) and \(B(2,1)\) divides the line joining \(C(3,4)\) D \((1, \mathbf{1})\)
1 7:5 Internally
2 7:5 Externally
3 7:11 Internally
4 7:11 Externally
Explanation:
(B): Given, A (-1, -1) , B (2, 1), C (3, 4) , D (1, 2) Let the ratio be \(\lambda: 1\) equation of line through \(A B\) \(y+1=\frac{2}{3}(x+1)\) \(3 y+3=2 x+2\) \(2 x-3 y-1=0\) Let, \(\mathrm{E}\) be the point where line intersect \(\mathrm{E}=\left(\frac{\lambda+3}{\lambda+1}, \frac{2 \lambda+4}{\lambda+1}\right)\) \(2\left(\frac{\lambda+3}{\lambda+1}\right)-3\left(\frac{2 \lambda+4}{\lambda+1}\right)-1=0\) \(2 \lambda+6-6 \lambda-12-\lambda-1=0\) \(-5 \lambda-7=0\) \(\lambda=\frac{-7}{5}\) 7: 5 Externally
AP EAMCET-2021-23.08.2021
Co-Ordinate system
88253
The vertices of \(\triangle \mathrm{ABC}\) are \(\mathrm{A}(2,2), \mathrm{B}(-4,-4)\) and \(C(5,-8)\). Find the length of a median of a triangle, which is passing through the point \(C\).
1 \(\sqrt{65}\)
2 \(\sqrt{117}\)
3 \(\sqrt{85}\)
4 \(\sqrt{116}\)
Explanation:
(C) \(\mathrm{D} \text { is a mid-point of } \mathrm{AB}\) \(\mathrm{h}=\frac{-4+2}{2}=-1, \mathrm{k}=\frac{-4+2}{2}=-1\) \((\mathrm{~h}, \mathrm{k})=(-1,-1)\) Distance from \((5,-8)\) and \((-1,-1)\) is \(\mathrm{D}=\sqrt{(5+1)^{2}+(-8+1)^{2}}=\sqrt{36+49}=\sqrt{85}\)
GUJCET-2011
Co-Ordinate system
88254
A straight line passing through origin 0 intersects the lines \(10 x-8 y-10=0\) and \(\frac{x}{4}-\frac{y}{5}+1=0\) at right angles and at the points \(P\) and \(Q\) respectively. Then the ratio in which \(O\) divides the line segment \(P Q\) is
1 \(1: 2\)
2 \(1: 4\)
3 \(1: 1\)
4 \(3: 4\)
Explanation:
: Let, Equation of required line \(y=m x+c\) Given, it is normal to \(10 x-8 y=10\) and \(5 x-4 y+20=0\) So, \(\mathrm{m}_{1} \mathrm{~m}_{2}=-1\) \(\mathrm{m} \times\left(\frac{5}{4}\right)=-1\) \(\mathrm{~m}=-\frac{4}{5}\) \(P\) will be the point of intersection of \(5 x-4 y=5\) \(4 x+5 y=0\) \(P:\left(\frac{25}{41}, \frac{-20}{41}\right)\) \(Q\) will be the point of intersection of \(5 x-4 y=-20\) \(4 x+5 y=0\) \(\mathrm{Q}:\left(\frac{-100}{41}, \frac{80}{41}\right)\) Let \(\mathrm{O}\) divides PQ in \(\lambda: 1\) \((0,0) =\left(\frac{-100 \lambda+25}{41(\lambda+1)}, \frac{80 \lambda-20}{41(\lambda+1)}\right)\) \(\lambda =\frac{1}{4}\) \(\lambda: 1=1: 4\)
