88244
The ratio in which the YZ-plane divides the line joining \((2,4,5)\) and \((3,5,-4)\) is
1 \(2: 3\) internally
2 \(3: 2\) internally
3 \(3: 2\) externally
4 \(2: 3\) externally
Explanation:
(D): Let the \(\mathrm{yz}-\) plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(m: 1\) We know that on \(y z-\) plane the co-ordinate of \(x\) is 0 \(\frac{m \times 3+1 \times 2}{m+1}=0\) \(3 \mathrm{~m}+2=0\) \(\mathrm{~m}=-\frac{2}{3}\) Hence, yz- plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(2: 3\) externally,
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88247
Given that the points \(P(3,2,-4), Q(5,4,-6)\) and \(R(9,8,-10)\) are collinear, the ratio in which \(Q\) divides \(P R\) externally is
1 \(1: 2\)
2 \(2: 1\)
3 \(1: 1\)
4 \(2: 2\)
Explanation:
(A) : Given, The point \(\mathrm{P}(3,2,-4)\) and \(\mathrm{Q}(5,4,-6)\) and \(\mathrm{R}(9,8,-10)\) divides PR internally. Let \(\mathrm{Q}\) divided \(\mathrm{PR}\) in the ratio \(\mathrm{k}\) : 1 \(\mathrm{P} \Rightarrow \mathrm{R}\) \(\mathrm{Q}=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \((5,4,-6)=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}=5 \Rightarrow 9 \mathrm{k}+3=5 \mathrm{k}+5\) \(\mathrm{k}=1 / 2\) Thus, point \(\mathrm{Q}\) divides \(\mathrm{PR}\) in the ratio \(1: 2\)
Co-Ordinate system
88248
The harmonic conjugate of \((2,3,4)\) with respect to the points \((3,-2,2)\) and \((6,-17,-4)\) is
(D) : Given, \(\mathrm{P}(2,3,4), \mathrm{A}(3,-2,2) \& \mathrm{~B}(6,-17,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(=\mathrm{k}: 1\) then, \(\quad 2=\frac{6 \mathrm{k}+3}{\mathrm{k}+1}\) \(2 \mathrm{k}+2=6 \mathrm{k}+3 \Rightarrow \mathrm{k}=-1 / 4\) harmonic conjugate \(\mathrm{Q}\) divided to the ratio \(=-\mathrm{k}: 1\) Hence, coordinates of \(\mathrm{Q}\), \(=1 / 4: 1\) \(\left(\frac{1 / 4 \times 6+1 \times 3}{1 / 4+1}, \frac{1 / 4 \times(-17)+1 \times-2}{1 / 4+1}, \frac{1 / 4-4+1 \times 2}{1 / 4+1}\right)\) \(\left(\frac{18}{5}, \frac{-25}{5}, \frac{4}{5}\right),\left(\frac{18}{5}, \frac{-5}{1}, \frac{4}{5}\right)\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88249
If \((a, 8)\) is a point on the join of \((2,5)\) and \((4,-\) 1) then
1 \(\mathrm{a}=\frac{8}{3}\)
2 \(a=\frac{3}{8}\)
3 \(a=1\)
4 \(a=-1\)
Explanation:
(C) : Given, The coordinate of points are \((2,5)\) and \((4,-1)\) We know that, The equation of straight lines is \(y-y_{1} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \(\therefore \quad y-5 =\frac{-1-5}{4-2}(x-2)\) \(y-5 =\frac{-6}{2}(x-2)\) \(y-5 =-3(x-2) \tag{i}\) Now, point \((a, 8)\) will satisfy the equation (i) \(\therefore \quad 8-5=-3(a-2)\) \(3=-3(a-2)\) \(a-2=-1\) \(a=1\)
AP EAMCET-18.09.2020
Co-Ordinate system
88250
If the length of the intercept made on the line \(y\) \(=a x\) by the lines \(y=2\) and \(y=6\) is less than 5
4 \(\mathrm{a}\lt \frac{-4}{3}\) or \(\mathrm{a}>\frac{4}{3}\)
Explanation:
(D) : Given, The equation of line is \(y=a x\) We know that, The coordinate of point A is \(\left(\frac{2}{a}, 2\right)\) Similarly, the coordinate of point B is \(\left(\frac{6}{a}, 6\right)\) Now, the distance between \(A B\) is \(\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+(6-2)^{2}\lt 5}\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+4^{2}\lt 25\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}\lt 9,\left(\frac{4}{a}\right)^{2}\lt 9 .\) \(\frac{16}{a^{2}}\lt 9\) \(a^{2}>\frac{16}{9}\) \(\therefore \quad a> \pm \sqrt{\frac{16}{9}}\) \(\therefore \quad \text { a }> \pm \frac{4}{3}\) \(\therefore \quad \text { a } 4 / 3 \text { or } a\lt -4 / 3\)
88244
The ratio in which the YZ-plane divides the line joining \((2,4,5)\) and \((3,5,-4)\) is
1 \(2: 3\) internally
2 \(3: 2\) internally
3 \(3: 2\) externally
4 \(2: 3\) externally
Explanation:
(D): Let the \(\mathrm{yz}-\) plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(m: 1\) We know that on \(y z-\) plane the co-ordinate of \(x\) is 0 \(\frac{m \times 3+1 \times 2}{m+1}=0\) \(3 \mathrm{~m}+2=0\) \(\mathrm{~m}=-\frac{2}{3}\) Hence, yz- plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(2: 3\) externally,
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88247
Given that the points \(P(3,2,-4), Q(5,4,-6)\) and \(R(9,8,-10)\) are collinear, the ratio in which \(Q\) divides \(P R\) externally is
1 \(1: 2\)
2 \(2: 1\)
3 \(1: 1\)
4 \(2: 2\)
Explanation:
(A) : Given, The point \(\mathrm{P}(3,2,-4)\) and \(\mathrm{Q}(5,4,-6)\) and \(\mathrm{R}(9,8,-10)\) divides PR internally. Let \(\mathrm{Q}\) divided \(\mathrm{PR}\) in the ratio \(\mathrm{k}\) : 1 \(\mathrm{P} \Rightarrow \mathrm{R}\) \(\mathrm{Q}=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \((5,4,-6)=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}=5 \Rightarrow 9 \mathrm{k}+3=5 \mathrm{k}+5\) \(\mathrm{k}=1 / 2\) Thus, point \(\mathrm{Q}\) divides \(\mathrm{PR}\) in the ratio \(1: 2\)
Co-Ordinate system
88248
The harmonic conjugate of \((2,3,4)\) with respect to the points \((3,-2,2)\) and \((6,-17,-4)\) is
(D) : Given, \(\mathrm{P}(2,3,4), \mathrm{A}(3,-2,2) \& \mathrm{~B}(6,-17,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(=\mathrm{k}: 1\) then, \(\quad 2=\frac{6 \mathrm{k}+3}{\mathrm{k}+1}\) \(2 \mathrm{k}+2=6 \mathrm{k}+3 \Rightarrow \mathrm{k}=-1 / 4\) harmonic conjugate \(\mathrm{Q}\) divided to the ratio \(=-\mathrm{k}: 1\) Hence, coordinates of \(\mathrm{Q}\), \(=1 / 4: 1\) \(\left(\frac{1 / 4 \times 6+1 \times 3}{1 / 4+1}, \frac{1 / 4 \times(-17)+1 \times-2}{1 / 4+1}, \frac{1 / 4-4+1 \times 2}{1 / 4+1}\right)\) \(\left(\frac{18}{5}, \frac{-25}{5}, \frac{4}{5}\right),\left(\frac{18}{5}, \frac{-5}{1}, \frac{4}{5}\right)\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88249
If \((a, 8)\) is a point on the join of \((2,5)\) and \((4,-\) 1) then
1 \(\mathrm{a}=\frac{8}{3}\)
2 \(a=\frac{3}{8}\)
3 \(a=1\)
4 \(a=-1\)
Explanation:
(C) : Given, The coordinate of points are \((2,5)\) and \((4,-1)\) We know that, The equation of straight lines is \(y-y_{1} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \(\therefore \quad y-5 =\frac{-1-5}{4-2}(x-2)\) \(y-5 =\frac{-6}{2}(x-2)\) \(y-5 =-3(x-2) \tag{i}\) Now, point \((a, 8)\) will satisfy the equation (i) \(\therefore \quad 8-5=-3(a-2)\) \(3=-3(a-2)\) \(a-2=-1\) \(a=1\)
AP EAMCET-18.09.2020
Co-Ordinate system
88250
If the length of the intercept made on the line \(y\) \(=a x\) by the lines \(y=2\) and \(y=6\) is less than 5
4 \(\mathrm{a}\lt \frac{-4}{3}\) or \(\mathrm{a}>\frac{4}{3}\)
Explanation:
(D) : Given, The equation of line is \(y=a x\) We know that, The coordinate of point A is \(\left(\frac{2}{a}, 2\right)\) Similarly, the coordinate of point B is \(\left(\frac{6}{a}, 6\right)\) Now, the distance between \(A B\) is \(\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+(6-2)^{2}\lt 5}\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+4^{2}\lt 25\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}\lt 9,\left(\frac{4}{a}\right)^{2}\lt 9 .\) \(\frac{16}{a^{2}}\lt 9\) \(a^{2}>\frac{16}{9}\) \(\therefore \quad a> \pm \sqrt{\frac{16}{9}}\) \(\therefore \quad \text { a }> \pm \frac{4}{3}\) \(\therefore \quad \text { a } 4 / 3 \text { or } a\lt -4 / 3\)
88244
The ratio in which the YZ-plane divides the line joining \((2,4,5)\) and \((3,5,-4)\) is
1 \(2: 3\) internally
2 \(3: 2\) internally
3 \(3: 2\) externally
4 \(2: 3\) externally
Explanation:
(D): Let the \(\mathrm{yz}-\) plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(m: 1\) We know that on \(y z-\) plane the co-ordinate of \(x\) is 0 \(\frac{m \times 3+1 \times 2}{m+1}=0\) \(3 \mathrm{~m}+2=0\) \(\mathrm{~m}=-\frac{2}{3}\) Hence, yz- plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(2: 3\) externally,
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88247
Given that the points \(P(3,2,-4), Q(5,4,-6)\) and \(R(9,8,-10)\) are collinear, the ratio in which \(Q\) divides \(P R\) externally is
1 \(1: 2\)
2 \(2: 1\)
3 \(1: 1\)
4 \(2: 2\)
Explanation:
(A) : Given, The point \(\mathrm{P}(3,2,-4)\) and \(\mathrm{Q}(5,4,-6)\) and \(\mathrm{R}(9,8,-10)\) divides PR internally. Let \(\mathrm{Q}\) divided \(\mathrm{PR}\) in the ratio \(\mathrm{k}\) : 1 \(\mathrm{P} \Rightarrow \mathrm{R}\) \(\mathrm{Q}=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \((5,4,-6)=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}=5 \Rightarrow 9 \mathrm{k}+3=5 \mathrm{k}+5\) \(\mathrm{k}=1 / 2\) Thus, point \(\mathrm{Q}\) divides \(\mathrm{PR}\) in the ratio \(1: 2\)
Co-Ordinate system
88248
The harmonic conjugate of \((2,3,4)\) with respect to the points \((3,-2,2)\) and \((6,-17,-4)\) is
(D) : Given, \(\mathrm{P}(2,3,4), \mathrm{A}(3,-2,2) \& \mathrm{~B}(6,-17,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(=\mathrm{k}: 1\) then, \(\quad 2=\frac{6 \mathrm{k}+3}{\mathrm{k}+1}\) \(2 \mathrm{k}+2=6 \mathrm{k}+3 \Rightarrow \mathrm{k}=-1 / 4\) harmonic conjugate \(\mathrm{Q}\) divided to the ratio \(=-\mathrm{k}: 1\) Hence, coordinates of \(\mathrm{Q}\), \(=1 / 4: 1\) \(\left(\frac{1 / 4 \times 6+1 \times 3}{1 / 4+1}, \frac{1 / 4 \times(-17)+1 \times-2}{1 / 4+1}, \frac{1 / 4-4+1 \times 2}{1 / 4+1}\right)\) \(\left(\frac{18}{5}, \frac{-25}{5}, \frac{4}{5}\right),\left(\frac{18}{5}, \frac{-5}{1}, \frac{4}{5}\right)\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88249
If \((a, 8)\) is a point on the join of \((2,5)\) and \((4,-\) 1) then
1 \(\mathrm{a}=\frac{8}{3}\)
2 \(a=\frac{3}{8}\)
3 \(a=1\)
4 \(a=-1\)
Explanation:
(C) : Given, The coordinate of points are \((2,5)\) and \((4,-1)\) We know that, The equation of straight lines is \(y-y_{1} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \(\therefore \quad y-5 =\frac{-1-5}{4-2}(x-2)\) \(y-5 =\frac{-6}{2}(x-2)\) \(y-5 =-3(x-2) \tag{i}\) Now, point \((a, 8)\) will satisfy the equation (i) \(\therefore \quad 8-5=-3(a-2)\) \(3=-3(a-2)\) \(a-2=-1\) \(a=1\)
AP EAMCET-18.09.2020
Co-Ordinate system
88250
If the length of the intercept made on the line \(y\) \(=a x\) by the lines \(y=2\) and \(y=6\) is less than 5
4 \(\mathrm{a}\lt \frac{-4}{3}\) or \(\mathrm{a}>\frac{4}{3}\)
Explanation:
(D) : Given, The equation of line is \(y=a x\) We know that, The coordinate of point A is \(\left(\frac{2}{a}, 2\right)\) Similarly, the coordinate of point B is \(\left(\frac{6}{a}, 6\right)\) Now, the distance between \(A B\) is \(\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+(6-2)^{2}\lt 5}\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+4^{2}\lt 25\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}\lt 9,\left(\frac{4}{a}\right)^{2}\lt 9 .\) \(\frac{16}{a^{2}}\lt 9\) \(a^{2}>\frac{16}{9}\) \(\therefore \quad a> \pm \sqrt{\frac{16}{9}}\) \(\therefore \quad \text { a }> \pm \frac{4}{3}\) \(\therefore \quad \text { a } 4 / 3 \text { or } a\lt -4 / 3\)
88244
The ratio in which the YZ-plane divides the line joining \((2,4,5)\) and \((3,5,-4)\) is
1 \(2: 3\) internally
2 \(3: 2\) internally
3 \(3: 2\) externally
4 \(2: 3\) externally
Explanation:
(D): Let the \(\mathrm{yz}-\) plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(m: 1\) We know that on \(y z-\) plane the co-ordinate of \(x\) is 0 \(\frac{m \times 3+1 \times 2}{m+1}=0\) \(3 \mathrm{~m}+2=0\) \(\mathrm{~m}=-\frac{2}{3}\) Hence, yz- plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(2: 3\) externally,
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88247
Given that the points \(P(3,2,-4), Q(5,4,-6)\) and \(R(9,8,-10)\) are collinear, the ratio in which \(Q\) divides \(P R\) externally is
1 \(1: 2\)
2 \(2: 1\)
3 \(1: 1\)
4 \(2: 2\)
Explanation:
(A) : Given, The point \(\mathrm{P}(3,2,-4)\) and \(\mathrm{Q}(5,4,-6)\) and \(\mathrm{R}(9,8,-10)\) divides PR internally. Let \(\mathrm{Q}\) divided \(\mathrm{PR}\) in the ratio \(\mathrm{k}\) : 1 \(\mathrm{P} \Rightarrow \mathrm{R}\) \(\mathrm{Q}=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \((5,4,-6)=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}=5 \Rightarrow 9 \mathrm{k}+3=5 \mathrm{k}+5\) \(\mathrm{k}=1 / 2\) Thus, point \(\mathrm{Q}\) divides \(\mathrm{PR}\) in the ratio \(1: 2\)
Co-Ordinate system
88248
The harmonic conjugate of \((2,3,4)\) with respect to the points \((3,-2,2)\) and \((6,-17,-4)\) is
(D) : Given, \(\mathrm{P}(2,3,4), \mathrm{A}(3,-2,2) \& \mathrm{~B}(6,-17,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(=\mathrm{k}: 1\) then, \(\quad 2=\frac{6 \mathrm{k}+3}{\mathrm{k}+1}\) \(2 \mathrm{k}+2=6 \mathrm{k}+3 \Rightarrow \mathrm{k}=-1 / 4\) harmonic conjugate \(\mathrm{Q}\) divided to the ratio \(=-\mathrm{k}: 1\) Hence, coordinates of \(\mathrm{Q}\), \(=1 / 4: 1\) \(\left(\frac{1 / 4 \times 6+1 \times 3}{1 / 4+1}, \frac{1 / 4 \times(-17)+1 \times-2}{1 / 4+1}, \frac{1 / 4-4+1 \times 2}{1 / 4+1}\right)\) \(\left(\frac{18}{5}, \frac{-25}{5}, \frac{4}{5}\right),\left(\frac{18}{5}, \frac{-5}{1}, \frac{4}{5}\right)\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88249
If \((a, 8)\) is a point on the join of \((2,5)\) and \((4,-\) 1) then
1 \(\mathrm{a}=\frac{8}{3}\)
2 \(a=\frac{3}{8}\)
3 \(a=1\)
4 \(a=-1\)
Explanation:
(C) : Given, The coordinate of points are \((2,5)\) and \((4,-1)\) We know that, The equation of straight lines is \(y-y_{1} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \(\therefore \quad y-5 =\frac{-1-5}{4-2}(x-2)\) \(y-5 =\frac{-6}{2}(x-2)\) \(y-5 =-3(x-2) \tag{i}\) Now, point \((a, 8)\) will satisfy the equation (i) \(\therefore \quad 8-5=-3(a-2)\) \(3=-3(a-2)\) \(a-2=-1\) \(a=1\)
AP EAMCET-18.09.2020
Co-Ordinate system
88250
If the length of the intercept made on the line \(y\) \(=a x\) by the lines \(y=2\) and \(y=6\) is less than 5
4 \(\mathrm{a}\lt \frac{-4}{3}\) or \(\mathrm{a}>\frac{4}{3}\)
Explanation:
(D) : Given, The equation of line is \(y=a x\) We know that, The coordinate of point A is \(\left(\frac{2}{a}, 2\right)\) Similarly, the coordinate of point B is \(\left(\frac{6}{a}, 6\right)\) Now, the distance between \(A B\) is \(\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+(6-2)^{2}\lt 5}\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+4^{2}\lt 25\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}\lt 9,\left(\frac{4}{a}\right)^{2}\lt 9 .\) \(\frac{16}{a^{2}}\lt 9\) \(a^{2}>\frac{16}{9}\) \(\therefore \quad a> \pm \sqrt{\frac{16}{9}}\) \(\therefore \quad \text { a }> \pm \frac{4}{3}\) \(\therefore \quad \text { a } 4 / 3 \text { or } a\lt -4 / 3\)
88244
The ratio in which the YZ-plane divides the line joining \((2,4,5)\) and \((3,5,-4)\) is
1 \(2: 3\) internally
2 \(3: 2\) internally
3 \(3: 2\) externally
4 \(2: 3\) externally
Explanation:
(D): Let the \(\mathrm{yz}-\) plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(m: 1\) We know that on \(y z-\) plane the co-ordinate of \(x\) is 0 \(\frac{m \times 3+1 \times 2}{m+1}=0\) \(3 \mathrm{~m}+2=0\) \(\mathrm{~m}=-\frac{2}{3}\) Hence, yz- plane divide the line segment joining the points \((2,4,5)\) and \((3,-5,4)\) in \(2: 3\) externally,
AP EAMCET-2021-19.08.2021
Co-Ordinate system
88247
Given that the points \(P(3,2,-4), Q(5,4,-6)\) and \(R(9,8,-10)\) are collinear, the ratio in which \(Q\) divides \(P R\) externally is
1 \(1: 2\)
2 \(2: 1\)
3 \(1: 1\)
4 \(2: 2\)
Explanation:
(A) : Given, The point \(\mathrm{P}(3,2,-4)\) and \(\mathrm{Q}(5,4,-6)\) and \(\mathrm{R}(9,8,-10)\) divides PR internally. Let \(\mathrm{Q}\) divided \(\mathrm{PR}\) in the ratio \(\mathrm{k}\) : 1 \(\mathrm{P} \Rightarrow \mathrm{R}\) \(\mathrm{Q}=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \((5,4,-6)=\left(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}, \frac{8 \mathrm{k}+2}{\mathrm{k}+1}, \frac{-10 \mathrm{k}-4}{\mathrm{k}+1}\right)\) \(\frac{9 \mathrm{k}+3}{\mathrm{k}+1}=5 \Rightarrow 9 \mathrm{k}+3=5 \mathrm{k}+5\) \(\mathrm{k}=1 / 2\) Thus, point \(\mathrm{Q}\) divides \(\mathrm{PR}\) in the ratio \(1: 2\)
Co-Ordinate system
88248
The harmonic conjugate of \((2,3,4)\) with respect to the points \((3,-2,2)\) and \((6,-17,-4)\) is
(D) : Given, \(\mathrm{P}(2,3,4), \mathrm{A}(3,-2,2) \& \mathrm{~B}(6,-17,-4)\) Let \(\mathrm{P}\) divides \(\mathrm{AB}\) in the ratio \(=\mathrm{k}: 1\) then, \(\quad 2=\frac{6 \mathrm{k}+3}{\mathrm{k}+1}\) \(2 \mathrm{k}+2=6 \mathrm{k}+3 \Rightarrow \mathrm{k}=-1 / 4\) harmonic conjugate \(\mathrm{Q}\) divided to the ratio \(=-\mathrm{k}: 1\) Hence, coordinates of \(\mathrm{Q}\), \(=1 / 4: 1\) \(\left(\frac{1 / 4 \times 6+1 \times 3}{1 / 4+1}, \frac{1 / 4 \times(-17)+1 \times-2}{1 / 4+1}, \frac{1 / 4-4+1 \times 2}{1 / 4+1}\right)\) \(\left(\frac{18}{5}, \frac{-25}{5}, \frac{4}{5}\right),\left(\frac{18}{5}, \frac{-5}{1}, \frac{4}{5}\right)\)
AP EAMCET-2020-22.09.2020
Co-Ordinate system
88249
If \((a, 8)\) is a point on the join of \((2,5)\) and \((4,-\) 1) then
1 \(\mathrm{a}=\frac{8}{3}\)
2 \(a=\frac{3}{8}\)
3 \(a=1\)
4 \(a=-1\)
Explanation:
(C) : Given, The coordinate of points are \((2,5)\) and \((4,-1)\) We know that, The equation of straight lines is \(y-y_{1} =\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)\) \(\therefore \quad y-5 =\frac{-1-5}{4-2}(x-2)\) \(y-5 =\frac{-6}{2}(x-2)\) \(y-5 =-3(x-2) \tag{i}\) Now, point \((a, 8)\) will satisfy the equation (i) \(\therefore \quad 8-5=-3(a-2)\) \(3=-3(a-2)\) \(a-2=-1\) \(a=1\)
AP EAMCET-18.09.2020
Co-Ordinate system
88250
If the length of the intercept made on the line \(y\) \(=a x\) by the lines \(y=2\) and \(y=6\) is less than 5
4 \(\mathrm{a}\lt \frac{-4}{3}\) or \(\mathrm{a}>\frac{4}{3}\)
Explanation:
(D) : Given, The equation of line is \(y=a x\) We know that, The coordinate of point A is \(\left(\frac{2}{a}, 2\right)\) Similarly, the coordinate of point B is \(\left(\frac{6}{a}, 6\right)\) Now, the distance between \(A B\) is \(\sqrt{\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+(6-2)^{2}\lt 5}\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}+4^{2}\lt 25\) \(\left(\frac{6}{a}-\frac{2}{a}\right)^{2}\lt 9,\left(\frac{4}{a}\right)^{2}\lt 9 .\) \(\frac{16}{a^{2}}\lt 9\) \(a^{2}>\frac{16}{9}\) \(\therefore \quad a> \pm \sqrt{\frac{16}{9}}\) \(\therefore \quad \text { a }> \pm \frac{4}{3}\) \(\therefore \quad \text { a } 4 / 3 \text { or } a\lt -4 / 3\)
