88259
Let OABC be a parallelogram. The equation of one diagonal \(A C\) is \(x+y-1=0\) and the combined equation of the sides \(O A, O C\) is \(2 x^{2}-\) \(y^{2}=0\). If \(G\) is centroid of the triangle \(O A C\), then \(\mathrm{BG}=\)
1 \(2 \sqrt{5}\)
2 \(\frac{4}{3} \sqrt{5}\)
3 \(\frac{2}{3} \sqrt{15}\)
4 \(\frac{4}{9} \sqrt{5}\)
Explanation:
(B) : Equation of diagonal of \(\mathrm{AC}\) of a parallelogram \(\mathrm{OABC}\) is \(\mathrm{x}+\mathrm{y}-1=0\) Equation of \(\mathrm{OA}\) and \(\mathrm{OC}\) is \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=0\) \(\because \quad \mathrm{y}= \pm \sqrt{2} \mathrm{x}\) Coordinate of \(\mathrm{A}(\sqrt{2}-1,2-\sqrt{2})\) \(\mathrm{C}=(-\sqrt{2}-1,2+\sqrt{2})\) \(\mathrm{D}\) is mid-point of \(\mathrm{AC}\) \(\mathrm{D}=\left(\frac{\sqrt{2}-1-\sqrt{2}-1}{2}, \frac{2-\sqrt{2}+2+\sqrt{2}}{2}\right)=(-1,2)\) \(\mathrm{D}\) is also mid-point of \(\mathrm{OB}\) \(\because \mathrm{B}(-2,4)\) Centroid \(G=\left(\frac{\sqrt{2}-1-\sqrt{2}-1+0}{3}, \frac{2-\sqrt{2}+2+\sqrt{2}+0}{3}\right)\) \(\mathrm{G}=\left(\frac{-2}{3}, \frac{4}{3}\right)\) \(\mathrm{BG}=\sqrt{\left(\frac{-2}{3}+2\right)^{2}+\left(\frac{4}{3}-4\right)^{2}}=\sqrt{\frac{16+64}{9}}=\frac{4 \sqrt{5}}{3}\)
TS EAMCET-2020-14.09.2020
Co-Ordinate system
88260
The point on the line \(4 x-y-2=0\) which is equidistant from the points \((-5,6)\) and \((3,2)\) is
1 \((2,6)\)
2 \((4,14)\)
3 \((1,2)\)
4 \((3,10)\)
Explanation:
(B): Let, the point on line, \(4 x-y-2=0 \text { be } P(x, y)\) Let, \(\quad \mathrm{A}=(-5,6)\) and \(\mathrm{B}=(3,2)\) \(4 x-y-2=0 \tag{i}\) Point, \(\mathrm{P}\) is equivalent from point \(\mathrm{A}\) and \(\mathrm{B}\) \(\therefore \mathrm{AP}=\mathrm{PB}\) By distance formula \((x+5)^{2}+(y-6)^{2}=(x-3)^{2}+(y-2)^{2}\) \(x^{2}+25+10 x+y^{2}+36-12 y=x^{2}+9-6 x+y^{2}+4-4 y\) \(16 x-8 y+48=0\) \(4 x-2 y+12=0 \tag{ii}\) \(\text { subtracting eq }{ }^{n} .(\text { ii })-\text { eq }^{n} .(\text { (i) we get -........ (ii) }\) \(y=14\) \(x=4\) So, point on the line is \((4,14)\)
TS EAMCET-2021-04.08.2021
Co-Ordinate system
88261
The distance of the point \((3,5)\) from \(2 x+3 y\) \(-14=0\) measured parallel to \(x-2 y=1\) is
1 \(\frac{7}{\sqrt{5}}\)
2 \(\frac{7}{\sqrt{13}}\)
3 \(\sqrt{5}\)
4 \(\sqrt{13}\)
Explanation:
(C) : Let the equation of the line parallel to \(x-2 y=\) 1 is \(x-2 y+\lambda=0\) Since, it passes through \((3,5)\) \(3-10+\lambda=0 \Rightarrow \lambda=7\) \(\therefore\) The line is \(\mathrm{x}-2 \mathrm{y}+7=0\) The point of intersection of \(x-2 y+7=0\) and \(2 \mathrm{x}+3 \mathrm{y}-14=0\) is \((1,4)\) \(\therefore\) The distance between \((3,5)\) and \((1,4)\) \(=\sqrt{(3-1)^{2}+(5-4)^{2}}=\sqrt{4+1}=\sqrt{5}\)
88259
Let OABC be a parallelogram. The equation of one diagonal \(A C\) is \(x+y-1=0\) and the combined equation of the sides \(O A, O C\) is \(2 x^{2}-\) \(y^{2}=0\). If \(G\) is centroid of the triangle \(O A C\), then \(\mathrm{BG}=\)
1 \(2 \sqrt{5}\)
2 \(\frac{4}{3} \sqrt{5}\)
3 \(\frac{2}{3} \sqrt{15}\)
4 \(\frac{4}{9} \sqrt{5}\)
Explanation:
(B) : Equation of diagonal of \(\mathrm{AC}\) of a parallelogram \(\mathrm{OABC}\) is \(\mathrm{x}+\mathrm{y}-1=0\) Equation of \(\mathrm{OA}\) and \(\mathrm{OC}\) is \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=0\) \(\because \quad \mathrm{y}= \pm \sqrt{2} \mathrm{x}\) Coordinate of \(\mathrm{A}(\sqrt{2}-1,2-\sqrt{2})\) \(\mathrm{C}=(-\sqrt{2}-1,2+\sqrt{2})\) \(\mathrm{D}\) is mid-point of \(\mathrm{AC}\) \(\mathrm{D}=\left(\frac{\sqrt{2}-1-\sqrt{2}-1}{2}, \frac{2-\sqrt{2}+2+\sqrt{2}}{2}\right)=(-1,2)\) \(\mathrm{D}\) is also mid-point of \(\mathrm{OB}\) \(\because \mathrm{B}(-2,4)\) Centroid \(G=\left(\frac{\sqrt{2}-1-\sqrt{2}-1+0}{3}, \frac{2-\sqrt{2}+2+\sqrt{2}+0}{3}\right)\) \(\mathrm{G}=\left(\frac{-2}{3}, \frac{4}{3}\right)\) \(\mathrm{BG}=\sqrt{\left(\frac{-2}{3}+2\right)^{2}+\left(\frac{4}{3}-4\right)^{2}}=\sqrt{\frac{16+64}{9}}=\frac{4 \sqrt{5}}{3}\)
TS EAMCET-2020-14.09.2020
Co-Ordinate system
88260
The point on the line \(4 x-y-2=0\) which is equidistant from the points \((-5,6)\) and \((3,2)\) is
1 \((2,6)\)
2 \((4,14)\)
3 \((1,2)\)
4 \((3,10)\)
Explanation:
(B): Let, the point on line, \(4 x-y-2=0 \text { be } P(x, y)\) Let, \(\quad \mathrm{A}=(-5,6)\) and \(\mathrm{B}=(3,2)\) \(4 x-y-2=0 \tag{i}\) Point, \(\mathrm{P}\) is equivalent from point \(\mathrm{A}\) and \(\mathrm{B}\) \(\therefore \mathrm{AP}=\mathrm{PB}\) By distance formula \((x+5)^{2}+(y-6)^{2}=(x-3)^{2}+(y-2)^{2}\) \(x^{2}+25+10 x+y^{2}+36-12 y=x^{2}+9-6 x+y^{2}+4-4 y\) \(16 x-8 y+48=0\) \(4 x-2 y+12=0 \tag{ii}\) \(\text { subtracting eq }{ }^{n} .(\text { ii })-\text { eq }^{n} .(\text { (i) we get -........ (ii) }\) \(y=14\) \(x=4\) So, point on the line is \((4,14)\)
TS EAMCET-2021-04.08.2021
Co-Ordinate system
88261
The distance of the point \((3,5)\) from \(2 x+3 y\) \(-14=0\) measured parallel to \(x-2 y=1\) is
1 \(\frac{7}{\sqrt{5}}\)
2 \(\frac{7}{\sqrt{13}}\)
3 \(\sqrt{5}\)
4 \(\sqrt{13}\)
Explanation:
(C) : Let the equation of the line parallel to \(x-2 y=\) 1 is \(x-2 y+\lambda=0\) Since, it passes through \((3,5)\) \(3-10+\lambda=0 \Rightarrow \lambda=7\) \(\therefore\) The line is \(\mathrm{x}-2 \mathrm{y}+7=0\) The point of intersection of \(x-2 y+7=0\) and \(2 \mathrm{x}+3 \mathrm{y}-14=0\) is \((1,4)\) \(\therefore\) The distance between \((3,5)\) and \((1,4)\) \(=\sqrt{(3-1)^{2}+(5-4)^{2}}=\sqrt{4+1}=\sqrt{5}\)
88259
Let OABC be a parallelogram. The equation of one diagonal \(A C\) is \(x+y-1=0\) and the combined equation of the sides \(O A, O C\) is \(2 x^{2}-\) \(y^{2}=0\). If \(G\) is centroid of the triangle \(O A C\), then \(\mathrm{BG}=\)
1 \(2 \sqrt{5}\)
2 \(\frac{4}{3} \sqrt{5}\)
3 \(\frac{2}{3} \sqrt{15}\)
4 \(\frac{4}{9} \sqrt{5}\)
Explanation:
(B) : Equation of diagonal of \(\mathrm{AC}\) of a parallelogram \(\mathrm{OABC}\) is \(\mathrm{x}+\mathrm{y}-1=0\) Equation of \(\mathrm{OA}\) and \(\mathrm{OC}\) is \(2 \mathrm{x}^{2}-\mathrm{y}^{2}=0\) \(\because \quad \mathrm{y}= \pm \sqrt{2} \mathrm{x}\) Coordinate of \(\mathrm{A}(\sqrt{2}-1,2-\sqrt{2})\) \(\mathrm{C}=(-\sqrt{2}-1,2+\sqrt{2})\) \(\mathrm{D}\) is mid-point of \(\mathrm{AC}\) \(\mathrm{D}=\left(\frac{\sqrt{2}-1-\sqrt{2}-1}{2}, \frac{2-\sqrt{2}+2+\sqrt{2}}{2}\right)=(-1,2)\) \(\mathrm{D}\) is also mid-point of \(\mathrm{OB}\) \(\because \mathrm{B}(-2,4)\) Centroid \(G=\left(\frac{\sqrt{2}-1-\sqrt{2}-1+0}{3}, \frac{2-\sqrt{2}+2+\sqrt{2}+0}{3}\right)\) \(\mathrm{G}=\left(\frac{-2}{3}, \frac{4}{3}\right)\) \(\mathrm{BG}=\sqrt{\left(\frac{-2}{3}+2\right)^{2}+\left(\frac{4}{3}-4\right)^{2}}=\sqrt{\frac{16+64}{9}}=\frac{4 \sqrt{5}}{3}\)
TS EAMCET-2020-14.09.2020
Co-Ordinate system
88260
The point on the line \(4 x-y-2=0\) which is equidistant from the points \((-5,6)\) and \((3,2)\) is
1 \((2,6)\)
2 \((4,14)\)
3 \((1,2)\)
4 \((3,10)\)
Explanation:
(B): Let, the point on line, \(4 x-y-2=0 \text { be } P(x, y)\) Let, \(\quad \mathrm{A}=(-5,6)\) and \(\mathrm{B}=(3,2)\) \(4 x-y-2=0 \tag{i}\) Point, \(\mathrm{P}\) is equivalent from point \(\mathrm{A}\) and \(\mathrm{B}\) \(\therefore \mathrm{AP}=\mathrm{PB}\) By distance formula \((x+5)^{2}+(y-6)^{2}=(x-3)^{2}+(y-2)^{2}\) \(x^{2}+25+10 x+y^{2}+36-12 y=x^{2}+9-6 x+y^{2}+4-4 y\) \(16 x-8 y+48=0\) \(4 x-2 y+12=0 \tag{ii}\) \(\text { subtracting eq }{ }^{n} .(\text { ii })-\text { eq }^{n} .(\text { (i) we get -........ (ii) }\) \(y=14\) \(x=4\) So, point on the line is \((4,14)\)
TS EAMCET-2021-04.08.2021
Co-Ordinate system
88261
The distance of the point \((3,5)\) from \(2 x+3 y\) \(-14=0\) measured parallel to \(x-2 y=1\) is
1 \(\frac{7}{\sqrt{5}}\)
2 \(\frac{7}{\sqrt{13}}\)
3 \(\sqrt{5}\)
4 \(\sqrt{13}\)
Explanation:
(C) : Let the equation of the line parallel to \(x-2 y=\) 1 is \(x-2 y+\lambda=0\) Since, it passes through \((3,5)\) \(3-10+\lambda=0 \Rightarrow \lambda=7\) \(\therefore\) The line is \(\mathrm{x}-2 \mathrm{y}+7=0\) The point of intersection of \(x-2 y+7=0\) and \(2 \mathrm{x}+3 \mathrm{y}-14=0\) is \((1,4)\) \(\therefore\) The distance between \((3,5)\) and \((1,4)\) \(=\sqrt{(3-1)^{2}+(5-4)^{2}}=\sqrt{4+1}=\sqrt{5}\)
