88255
Let \(D\) be the foot of the perpendicular drawn from the point \(A(2,0,3)\) to the line joining the points \(B(0,4,1)\) and \(C(-2,0,4)\). Then, the ratio in which \(D\) divides \(B C\) is
1 \(3: 2\)
2 \(2 \sqrt{6}: \sqrt{17}\)
3 \(18: 11\)
4 \(16: 9\)
Explanation:
(C) : Equation of line \(\mathrm{BC}\) is- \(\frac{\mathrm{x}-0}{-2}=\frac{\mathrm{y}-4}{-4}=\frac{\mathrm{z}-1}{3}=\mathrm{K}\) \(\therefore\) Any point on the line \(\mathrm{BC}\) i.e. \(\mathrm{D}\) is \(=(-2 \mathrm{~K},-4 \mathrm{~K}+4\), \(3 \mathrm{~K}+1)\) Direction ratio's of line \(\mathrm{AD}\) is- \((-2 \mathrm{~K}-2,-4 \mathrm{~K}+4-0,3 \mathrm{~K}+1-3)\) i.e. \(\quad(-2 K-2),(-4 K+4)\) and \((3 K-2)\) Direction ratio's of line \(\mathrm{BC}=(-2,-4,3)\) Since line \(A D\) and \(B C\) are perpendicular to each other \(-2(-2 K-2)-4(-4 K+4)+3(3 K-2)=0\) \((4 K+16 K+9 K)+(4-16-6)=0\) \(29 K=18=K=\frac{18}{29}\) \(\therefore\) Co-ordinates of \(\mathrm{D}=\left(\frac{-36}{29}, \frac{44}{29}, \frac{83}{29}\right)\) Now, \(\frac{-2 \mathrm{~m}}{\mathrm{~m}+\mathrm{n}}=\frac{-36}{29}\) \(\frac{4 \mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{44}{29}\) \(\frac{4 \mathrm{~m}+\mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{83}{29}\) \(\text { Solving the equation (i) we get }\) \(-58 \mathrm{~m}=-36 \mathrm{~m}-36 \mathrm{n}\) \(22 \mathrm{~m}-36 \mathrm{n}=0\) \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{18}{11}\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88256
Let the line \(L\) drawn perpendicular to the lines \(2 x-3 y+4=0\) and \(6 x-9 y+7=0\) meet them at \(A\) and \(B\), respectively. If \(P(1,1)\) is a point on \(L\), then the ratio in which \(P\) divides \(A B\) is
1 \(9: 4\) internally
2 \(9: 4\) externally
3 \(4: 9\) internally
4 \(4: 9\) externally
Explanation:
(B) : Since two lines are parallel to each other. \(\therefore \quad \mathrm{PA}=\frac{2-3+4}{\sqrt{13}}=\frac{3}{\sqrt{13}}\) \(\mathrm{PB}=\frac{6-9+7}{\sqrt{36+81}}=\frac{4}{3 \sqrt{13}}\) Now, \(\quad \frac{\mathrm{AP}}{\mathrm{BP}}=\frac{3}{\sqrt{13}} \times \frac{3 \sqrt{13}}{4}=\frac{9}{4}\) or AP : BP :: \(9: 4\) (externally)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88257
If the line \(2 x-y-4=0\) divides the line segment joining the points \((2,-1)\) and \((1,-4)\) at the point \((a, b)\) in the ratio \(m: n\), then \(4\left(\mathbf{a}-\mathbf{b}\left(\frac{\mathbf{m}}{\mathbf{n}}\right)^{2}\right)=\)
1 -5
2 14
3 11
4 10
Explanation:
(D) : Equation of line through \((2,-1)\) and \((1,-4)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\Rightarrow \quad \mathrm{y}-(-1)=\frac{-4+1}{1-2}(\mathrm{x}-2)\) \(\mathrm{y}+1=3(\mathrm{x}-2)\) \(\text { or } 3 \mathrm{x}-\mathrm{y}=7\) \(2 \mathrm{x}-\mathrm{y}-4=0 \tag{ii}\) Solving equation (i) and (ii), we get the point of intersection \((a, b)\) i.e. \(\mathrm{a}=3, \mathrm{~b}=2\) Also, \(\quad a=\frac{m+2 n}{m+n}\) and \(b=\frac{-4 m-n}{m+n}\) \(\text { or } \quad a=\frac{\frac{m}{n}+2}{\frac{m}{m}+1} \text { and } b=\frac{\frac{-4 m}{n}-1}{\frac{m}{n}+1}\) \(\Rightarrow \quad \frac{a}{b}=\frac{\frac{m}{n}+2}{\frac{-4 m}{n}-1}=\frac{3}{2}\) Which gives- Thus, \(\frac{\mathrm{m}}{\mathrm{n}}=-\frac{1}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=4\left[3-2\left(-\frac{1}{2}\right)^{2}\right]=4\left(3-\frac{1}{2}\right)=4 \times \frac{5}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=10\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88258
The quadrilateral formed by the points \(\mathbf{A}(1,2\), \(5), B(-1,6,1), C(3,4,-3)\) and \(D(5,0,1)\) is a
1 Parallelogram
2 Rectangle
3 Square
4 Rhombus
Explanation:
(C) : Given, Vertices of quadrilateral \(\mathrm{A}(1,2,5), \mathrm{B}(-1,6,1), \mathrm{C}(3,4,-3), \mathrm{D}(5,0,1)\) \(\therefore \quad \mathrm{AB}=\sqrt{4+16+16}=6\) \(\mathrm{BC}=\sqrt{16+4+16}=6\) \(\mathrm{CD}=\sqrt{4+16+16}=6\) \(\mathrm{AD}=\sqrt{16+4+16}=6\) \(\mathrm{AC}=\sqrt{4+4+64}=\sqrt{72}\) \(\mathrm{BD}=\sqrt{36+36+0}=\sqrt{72}\) Here, \(\quad \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}\) and \(\mathrm{AC}=\mathrm{BD}\) \(\therefore \mathrm{ABCD}\) forms a square.
88255
Let \(D\) be the foot of the perpendicular drawn from the point \(A(2,0,3)\) to the line joining the points \(B(0,4,1)\) and \(C(-2,0,4)\). Then, the ratio in which \(D\) divides \(B C\) is
1 \(3: 2\)
2 \(2 \sqrt{6}: \sqrt{17}\)
3 \(18: 11\)
4 \(16: 9\)
Explanation:
(C) : Equation of line \(\mathrm{BC}\) is- \(\frac{\mathrm{x}-0}{-2}=\frac{\mathrm{y}-4}{-4}=\frac{\mathrm{z}-1}{3}=\mathrm{K}\) \(\therefore\) Any point on the line \(\mathrm{BC}\) i.e. \(\mathrm{D}\) is \(=(-2 \mathrm{~K},-4 \mathrm{~K}+4\), \(3 \mathrm{~K}+1)\) Direction ratio's of line \(\mathrm{AD}\) is- \((-2 \mathrm{~K}-2,-4 \mathrm{~K}+4-0,3 \mathrm{~K}+1-3)\) i.e. \(\quad(-2 K-2),(-4 K+4)\) and \((3 K-2)\) Direction ratio's of line \(\mathrm{BC}=(-2,-4,3)\) Since line \(A D\) and \(B C\) are perpendicular to each other \(-2(-2 K-2)-4(-4 K+4)+3(3 K-2)=0\) \((4 K+16 K+9 K)+(4-16-6)=0\) \(29 K=18=K=\frac{18}{29}\) \(\therefore\) Co-ordinates of \(\mathrm{D}=\left(\frac{-36}{29}, \frac{44}{29}, \frac{83}{29}\right)\) Now, \(\frac{-2 \mathrm{~m}}{\mathrm{~m}+\mathrm{n}}=\frac{-36}{29}\) \(\frac{4 \mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{44}{29}\) \(\frac{4 \mathrm{~m}+\mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{83}{29}\) \(\text { Solving the equation (i) we get }\) \(-58 \mathrm{~m}=-36 \mathrm{~m}-36 \mathrm{n}\) \(22 \mathrm{~m}-36 \mathrm{n}=0\) \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{18}{11}\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88256
Let the line \(L\) drawn perpendicular to the lines \(2 x-3 y+4=0\) and \(6 x-9 y+7=0\) meet them at \(A\) and \(B\), respectively. If \(P(1,1)\) is a point on \(L\), then the ratio in which \(P\) divides \(A B\) is
1 \(9: 4\) internally
2 \(9: 4\) externally
3 \(4: 9\) internally
4 \(4: 9\) externally
Explanation:
(B) : Since two lines are parallel to each other. \(\therefore \quad \mathrm{PA}=\frac{2-3+4}{\sqrt{13}}=\frac{3}{\sqrt{13}}\) \(\mathrm{PB}=\frac{6-9+7}{\sqrt{36+81}}=\frac{4}{3 \sqrt{13}}\) Now, \(\quad \frac{\mathrm{AP}}{\mathrm{BP}}=\frac{3}{\sqrt{13}} \times \frac{3 \sqrt{13}}{4}=\frac{9}{4}\) or AP : BP :: \(9: 4\) (externally)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88257
If the line \(2 x-y-4=0\) divides the line segment joining the points \((2,-1)\) and \((1,-4)\) at the point \((a, b)\) in the ratio \(m: n\), then \(4\left(\mathbf{a}-\mathbf{b}\left(\frac{\mathbf{m}}{\mathbf{n}}\right)^{2}\right)=\)
1 -5
2 14
3 11
4 10
Explanation:
(D) : Equation of line through \((2,-1)\) and \((1,-4)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\Rightarrow \quad \mathrm{y}-(-1)=\frac{-4+1}{1-2}(\mathrm{x}-2)\) \(\mathrm{y}+1=3(\mathrm{x}-2)\) \(\text { or } 3 \mathrm{x}-\mathrm{y}=7\) \(2 \mathrm{x}-\mathrm{y}-4=0 \tag{ii}\) Solving equation (i) and (ii), we get the point of intersection \((a, b)\) i.e. \(\mathrm{a}=3, \mathrm{~b}=2\) Also, \(\quad a=\frac{m+2 n}{m+n}\) and \(b=\frac{-4 m-n}{m+n}\) \(\text { or } \quad a=\frac{\frac{m}{n}+2}{\frac{m}{m}+1} \text { and } b=\frac{\frac{-4 m}{n}-1}{\frac{m}{n}+1}\) \(\Rightarrow \quad \frac{a}{b}=\frac{\frac{m}{n}+2}{\frac{-4 m}{n}-1}=\frac{3}{2}\) Which gives- Thus, \(\frac{\mathrm{m}}{\mathrm{n}}=-\frac{1}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=4\left[3-2\left(-\frac{1}{2}\right)^{2}\right]=4\left(3-\frac{1}{2}\right)=4 \times \frac{5}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=10\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88258
The quadrilateral formed by the points \(\mathbf{A}(1,2\), \(5), B(-1,6,1), C(3,4,-3)\) and \(D(5,0,1)\) is a
1 Parallelogram
2 Rectangle
3 Square
4 Rhombus
Explanation:
(C) : Given, Vertices of quadrilateral \(\mathrm{A}(1,2,5), \mathrm{B}(-1,6,1), \mathrm{C}(3,4,-3), \mathrm{D}(5,0,1)\) \(\therefore \quad \mathrm{AB}=\sqrt{4+16+16}=6\) \(\mathrm{BC}=\sqrt{16+4+16}=6\) \(\mathrm{CD}=\sqrt{4+16+16}=6\) \(\mathrm{AD}=\sqrt{16+4+16}=6\) \(\mathrm{AC}=\sqrt{4+4+64}=\sqrt{72}\) \(\mathrm{BD}=\sqrt{36+36+0}=\sqrt{72}\) Here, \(\quad \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}\) and \(\mathrm{AC}=\mathrm{BD}\) \(\therefore \mathrm{ABCD}\) forms a square.
88255
Let \(D\) be the foot of the perpendicular drawn from the point \(A(2,0,3)\) to the line joining the points \(B(0,4,1)\) and \(C(-2,0,4)\). Then, the ratio in which \(D\) divides \(B C\) is
1 \(3: 2\)
2 \(2 \sqrt{6}: \sqrt{17}\)
3 \(18: 11\)
4 \(16: 9\)
Explanation:
(C) : Equation of line \(\mathrm{BC}\) is- \(\frac{\mathrm{x}-0}{-2}=\frac{\mathrm{y}-4}{-4}=\frac{\mathrm{z}-1}{3}=\mathrm{K}\) \(\therefore\) Any point on the line \(\mathrm{BC}\) i.e. \(\mathrm{D}\) is \(=(-2 \mathrm{~K},-4 \mathrm{~K}+4\), \(3 \mathrm{~K}+1)\) Direction ratio's of line \(\mathrm{AD}\) is- \((-2 \mathrm{~K}-2,-4 \mathrm{~K}+4-0,3 \mathrm{~K}+1-3)\) i.e. \(\quad(-2 K-2),(-4 K+4)\) and \((3 K-2)\) Direction ratio's of line \(\mathrm{BC}=(-2,-4,3)\) Since line \(A D\) and \(B C\) are perpendicular to each other \(-2(-2 K-2)-4(-4 K+4)+3(3 K-2)=0\) \((4 K+16 K+9 K)+(4-16-6)=0\) \(29 K=18=K=\frac{18}{29}\) \(\therefore\) Co-ordinates of \(\mathrm{D}=\left(\frac{-36}{29}, \frac{44}{29}, \frac{83}{29}\right)\) Now, \(\frac{-2 \mathrm{~m}}{\mathrm{~m}+\mathrm{n}}=\frac{-36}{29}\) \(\frac{4 \mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{44}{29}\) \(\frac{4 \mathrm{~m}+\mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{83}{29}\) \(\text { Solving the equation (i) we get }\) \(-58 \mathrm{~m}=-36 \mathrm{~m}-36 \mathrm{n}\) \(22 \mathrm{~m}-36 \mathrm{n}=0\) \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{18}{11}\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88256
Let the line \(L\) drawn perpendicular to the lines \(2 x-3 y+4=0\) and \(6 x-9 y+7=0\) meet them at \(A\) and \(B\), respectively. If \(P(1,1)\) is a point on \(L\), then the ratio in which \(P\) divides \(A B\) is
1 \(9: 4\) internally
2 \(9: 4\) externally
3 \(4: 9\) internally
4 \(4: 9\) externally
Explanation:
(B) : Since two lines are parallel to each other. \(\therefore \quad \mathrm{PA}=\frac{2-3+4}{\sqrt{13}}=\frac{3}{\sqrt{13}}\) \(\mathrm{PB}=\frac{6-9+7}{\sqrt{36+81}}=\frac{4}{3 \sqrt{13}}\) Now, \(\quad \frac{\mathrm{AP}}{\mathrm{BP}}=\frac{3}{\sqrt{13}} \times \frac{3 \sqrt{13}}{4}=\frac{9}{4}\) or AP : BP :: \(9: 4\) (externally)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88257
If the line \(2 x-y-4=0\) divides the line segment joining the points \((2,-1)\) and \((1,-4)\) at the point \((a, b)\) in the ratio \(m: n\), then \(4\left(\mathbf{a}-\mathbf{b}\left(\frac{\mathbf{m}}{\mathbf{n}}\right)^{2}\right)=\)
1 -5
2 14
3 11
4 10
Explanation:
(D) : Equation of line through \((2,-1)\) and \((1,-4)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\Rightarrow \quad \mathrm{y}-(-1)=\frac{-4+1}{1-2}(\mathrm{x}-2)\) \(\mathrm{y}+1=3(\mathrm{x}-2)\) \(\text { or } 3 \mathrm{x}-\mathrm{y}=7\) \(2 \mathrm{x}-\mathrm{y}-4=0 \tag{ii}\) Solving equation (i) and (ii), we get the point of intersection \((a, b)\) i.e. \(\mathrm{a}=3, \mathrm{~b}=2\) Also, \(\quad a=\frac{m+2 n}{m+n}\) and \(b=\frac{-4 m-n}{m+n}\) \(\text { or } \quad a=\frac{\frac{m}{n}+2}{\frac{m}{m}+1} \text { and } b=\frac{\frac{-4 m}{n}-1}{\frac{m}{n}+1}\) \(\Rightarrow \quad \frac{a}{b}=\frac{\frac{m}{n}+2}{\frac{-4 m}{n}-1}=\frac{3}{2}\) Which gives- Thus, \(\frac{\mathrm{m}}{\mathrm{n}}=-\frac{1}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=4\left[3-2\left(-\frac{1}{2}\right)^{2}\right]=4\left(3-\frac{1}{2}\right)=4 \times \frac{5}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=10\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88258
The quadrilateral formed by the points \(\mathbf{A}(1,2\), \(5), B(-1,6,1), C(3,4,-3)\) and \(D(5,0,1)\) is a
1 Parallelogram
2 Rectangle
3 Square
4 Rhombus
Explanation:
(C) : Given, Vertices of quadrilateral \(\mathrm{A}(1,2,5), \mathrm{B}(-1,6,1), \mathrm{C}(3,4,-3), \mathrm{D}(5,0,1)\) \(\therefore \quad \mathrm{AB}=\sqrt{4+16+16}=6\) \(\mathrm{BC}=\sqrt{16+4+16}=6\) \(\mathrm{CD}=\sqrt{4+16+16}=6\) \(\mathrm{AD}=\sqrt{16+4+16}=6\) \(\mathrm{AC}=\sqrt{4+4+64}=\sqrt{72}\) \(\mathrm{BD}=\sqrt{36+36+0}=\sqrt{72}\) Here, \(\quad \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}\) and \(\mathrm{AC}=\mathrm{BD}\) \(\therefore \mathrm{ABCD}\) forms a square.
88255
Let \(D\) be the foot of the perpendicular drawn from the point \(A(2,0,3)\) to the line joining the points \(B(0,4,1)\) and \(C(-2,0,4)\). Then, the ratio in which \(D\) divides \(B C\) is
1 \(3: 2\)
2 \(2 \sqrt{6}: \sqrt{17}\)
3 \(18: 11\)
4 \(16: 9\)
Explanation:
(C) : Equation of line \(\mathrm{BC}\) is- \(\frac{\mathrm{x}-0}{-2}=\frac{\mathrm{y}-4}{-4}=\frac{\mathrm{z}-1}{3}=\mathrm{K}\) \(\therefore\) Any point on the line \(\mathrm{BC}\) i.e. \(\mathrm{D}\) is \(=(-2 \mathrm{~K},-4 \mathrm{~K}+4\), \(3 \mathrm{~K}+1)\) Direction ratio's of line \(\mathrm{AD}\) is- \((-2 \mathrm{~K}-2,-4 \mathrm{~K}+4-0,3 \mathrm{~K}+1-3)\) i.e. \(\quad(-2 K-2),(-4 K+4)\) and \((3 K-2)\) Direction ratio's of line \(\mathrm{BC}=(-2,-4,3)\) Since line \(A D\) and \(B C\) are perpendicular to each other \(-2(-2 K-2)-4(-4 K+4)+3(3 K-2)=0\) \((4 K+16 K+9 K)+(4-16-6)=0\) \(29 K=18=K=\frac{18}{29}\) \(\therefore\) Co-ordinates of \(\mathrm{D}=\left(\frac{-36}{29}, \frac{44}{29}, \frac{83}{29}\right)\) Now, \(\frac{-2 \mathrm{~m}}{\mathrm{~m}+\mathrm{n}}=\frac{-36}{29}\) \(\frac{4 \mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{44}{29}\) \(\frac{4 \mathrm{~m}+\mathrm{n}}{\mathrm{m}+\mathrm{n}}=\frac{83}{29}\) \(\text { Solving the equation (i) we get }\) \(-58 \mathrm{~m}=-36 \mathrm{~m}-36 \mathrm{n}\) \(22 \mathrm{~m}-36 \mathrm{n}=0\) \(\frac{\mathrm{m}}{\mathrm{n}}=\frac{18}{11}\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88256
Let the line \(L\) drawn perpendicular to the lines \(2 x-3 y+4=0\) and \(6 x-9 y+7=0\) meet them at \(A\) and \(B\), respectively. If \(P(1,1)\) is a point on \(L\), then the ratio in which \(P\) divides \(A B\) is
1 \(9: 4\) internally
2 \(9: 4\) externally
3 \(4: 9\) internally
4 \(4: 9\) externally
Explanation:
(B) : Since two lines are parallel to each other. \(\therefore \quad \mathrm{PA}=\frac{2-3+4}{\sqrt{13}}=\frac{3}{\sqrt{13}}\) \(\mathrm{PB}=\frac{6-9+7}{\sqrt{36+81}}=\frac{4}{3 \sqrt{13}}\) Now, \(\quad \frac{\mathrm{AP}}{\mathrm{BP}}=\frac{3}{\sqrt{13}} \times \frac{3 \sqrt{13}}{4}=\frac{9}{4}\) or AP : BP :: \(9: 4\) (externally)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88257
If the line \(2 x-y-4=0\) divides the line segment joining the points \((2,-1)\) and \((1,-4)\) at the point \((a, b)\) in the ratio \(m: n\), then \(4\left(\mathbf{a}-\mathbf{b}\left(\frac{\mathbf{m}}{\mathbf{n}}\right)^{2}\right)=\)
1 -5
2 14
3 11
4 10
Explanation:
(D) : Equation of line through \((2,-1)\) and \((1,-4)\) is- \(\mathrm{y}-\mathrm{y}_{1}=\frac{\mathrm{y}_{2}-\mathrm{y}_{1}}{\mathrm{x}_{2}-\mathrm{x}_{1}}\left(\mathrm{x}-\mathrm{x}_{1}\right)\) \(\Rightarrow \quad \mathrm{y}-(-1)=\frac{-4+1}{1-2}(\mathrm{x}-2)\) \(\mathrm{y}+1=3(\mathrm{x}-2)\) \(\text { or } 3 \mathrm{x}-\mathrm{y}=7\) \(2 \mathrm{x}-\mathrm{y}-4=0 \tag{ii}\) Solving equation (i) and (ii), we get the point of intersection \((a, b)\) i.e. \(\mathrm{a}=3, \mathrm{~b}=2\) Also, \(\quad a=\frac{m+2 n}{m+n}\) and \(b=\frac{-4 m-n}{m+n}\) \(\text { or } \quad a=\frac{\frac{m}{n}+2}{\frac{m}{m}+1} \text { and } b=\frac{\frac{-4 m}{n}-1}{\frac{m}{n}+1}\) \(\Rightarrow \quad \frac{a}{b}=\frac{\frac{m}{n}+2}{\frac{-4 m}{n}-1}=\frac{3}{2}\) Which gives- Thus, \(\frac{\mathrm{m}}{\mathrm{n}}=-\frac{1}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=4\left[3-2\left(-\frac{1}{2}\right)^{2}\right]=4\left(3-\frac{1}{2}\right)=4 \times \frac{5}{2}\) \(4\left[\mathrm{a}-\mathrm{b}\left(\frac{\mathrm{m}}{\mathrm{n}}\right)^{2}\right]=10\)
TS EAMCET-2022-19.07.2022
Co-Ordinate system
88258
The quadrilateral formed by the points \(\mathbf{A}(1,2\), \(5), B(-1,6,1), C(3,4,-3)\) and \(D(5,0,1)\) is a
1 Parallelogram
2 Rectangle
3 Square
4 Rhombus
Explanation:
(C) : Given, Vertices of quadrilateral \(\mathrm{A}(1,2,5), \mathrm{B}(-1,6,1), \mathrm{C}(3,4,-3), \mathrm{D}(5,0,1)\) \(\therefore \quad \mathrm{AB}=\sqrt{4+16+16}=6\) \(\mathrm{BC}=\sqrt{16+4+16}=6\) \(\mathrm{CD}=\sqrt{4+16+16}=6\) \(\mathrm{AD}=\sqrt{16+4+16}=6\) \(\mathrm{AC}=\sqrt{4+4+64}=\sqrt{72}\) \(\mathrm{BD}=\sqrt{36+36+0}=\sqrt{72}\) Here, \(\quad \mathrm{AB}=\mathrm{BC}=\mathrm{CD}=\mathrm{AD}\) and \(\mathrm{AC}=\mathrm{BD}\) \(\therefore \mathrm{ABCD}\) forms a square.
