88785
If \(2 x+3 y+4=0\) is the perpendicular bisector of the line segment joining the points \(A(1,2)\) and \(B(\alpha, \beta)\), then the value of \(13 \alpha+13 \beta\) equals
1 -81
2 -99
3 99
4 81
Explanation:
(A):
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \tag{i}\)
\(\mathrm{~m}_{1}=\frac{\beta-2}{\alpha-1} \tag{ii}\)
\(2 x+3 y+4=0 \Rightarrow y=\frac{2}{3} x-\frac{4}{3}\)
\(\mathrm{m}_{2}=-\frac{2}{3} \tag{iii}\)
putting value of (ii), (iii) in (i), we get:-
\(\left(\frac{\beta-2}{\alpha-1}\right) \times\left(-\frac{2}{3}\right)=-1\)
\(2 \beta-4=3 \alpha-3\)
\(3 \alpha-2 \beta=-1 \tag{iv}\)
Mid point \(\left(\frac{\alpha+1}{2}, \frac{\beta+2}{2}\right)\) and it passes through the line
\((2 x+3 y+4)=0\)
\(2\left(\frac{\alpha+1}{2}\right)+3\left(\frac{\beta+2}{2}\right)+4=0\)
\(2 \alpha+2+3 \beta+6+8=0\)
\(2 \alpha+3 \beta=-16 \tag{v}\)
On solving eq \({ }^{\mathrm{n}}\) (iv) and (v), we get
\(\alpha=-\frac{35}{13}, \beta=\frac{-46}{13} \Rightarrow \alpha+\beta=-\frac{81}{13}\)
Hence \(13 \alpha+13 \beta=13(\alpha+\beta)\)
\(=13 \times\left(-\frac{81}{13}\right) \Rightarrow-81\)
AP EAMCET-2021-19.08.2021
Straight Line
88786
For three consecutive odd integers \(a, b\) and \(c\), if the variable line \(a x+b y+c=0\) always passes through the point \((\alpha, \beta)\), the value of \(\alpha^{2}+\beta^{2}\) equals
1 9
2 4
3 5
4 3
Explanation:
(C):
a, b, c are odd integer
For three consecutive odd integer are,
\(\mathrm{a}=\mathrm{a}, \quad \mathrm{b}=\mathrm{a}+2, \quad \mathrm{c}=\mathrm{a}+4\)
So,
\(a x+b y+c=0\)
\(a x+(a+2) y+(a+4)=0\)
\(a x+a y+a+2 y+4=0\)
\(2(y+2)+a(x+y+1)=0\)
\((y+2)+\frac{a}{2}(x+y+1)=0\)
\(L_{1}+\lambda L_{2}=0\)
So,
\(y+2=0 \Rightarrow y=-2\)
\(x+y+1=0 \Rightarrow x-2+1=0 \Rightarrow x=1\)
So, line passes through \((\alpha, \beta)=(1,-2)\)
\(a^{2}+\beta^{2}=(1)^{2}+(-2)^{2}\)
\(\alpha^{2}+\beta^{2}=5\)
AP EAMCET-2021-19.08.2021
Straight Line
88787
If the distance of a line from the origin is \(\sqrt{5}\) and having intercepts in the ratio of \(1: 2\) on axes, then the equations of line are .......... .
1 \(2 x-y \pm 5=0\)
2 \(2 x+y \pm 5=0\)
3 \(x-2 y \pm 5=0\)
4 \(x+2 y \pm 5=0\)
Explanation:
(B):
Given, Distance \(=\sqrt{5}\)
Intercept on axes \(=1: 2\)
\(\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \Rightarrow \mathrm{b}=2 \mathrm{a}\)
As we know, equation of line
i.e. \(\frac{x}{a}+\frac{y}{b}=1\)
i.e. \(\frac{x}{b / 2}+\frac{y}{b}=1 \Rightarrow 2 x+y=b\)
\(\Rightarrow 2 x+y-b=0 \tag{1}\)
Origin \((0,0)\) the distance between origin and line
\(\left|\frac{2(0)+0-b}{\sqrt{4+1}}\right|=\sqrt{5}\) (distance) \(\Rightarrow b= \pm 5\)
Put in (i), we get
\(2 x+y \pm 5=0\)
GUJCET-2011
Straight Line
88788
If \((a, b)\) is the centre of the circle passing through the vertices of the triangle formed by \(x\) \(+y=6,2 x+y=4\) and \(x+2 y=5\), then \((a, b)\) is
1 \((-17,-16)\)
2 \(\left(\frac{17}{2}, \frac{19}{2}\right)\)
3 \((17,18)\)
4 \(\left(\frac{-17}{2}, \frac{-19}{2}\right)\)
Explanation:
(B):
Let \(\mathrm{L}_{1}: \mathrm{x}+\mathrm{y}=6\)
\(\mathrm{L}_{2}: 2 \mathrm{x}+\mathrm{y}=4\)
...... (1)
\(\mathrm{L}_{3}: \mathrm{x}+2 \mathrm{y}=5\)
(1) Point of intersection of \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\)
On subtracting \(\mathrm{eq}^{\mathrm{n}}\). (2) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get \(2 \mathrm{x}-2=4-6\)
\(\therefore \mathrm{x}=-2\)
Put the value of \(x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(y=8\)
(2) Point of intersection \(\mathrm{L}_{2}\) and \(\mathrm{L}_{3}\)
Multiplying eq \(\mathrm{eq}^{\mathrm{n}}\). (3) by 2 we have -
\(\therefore \quad 2 \mathrm{x}+4 \mathrm{y}=10\)
\(\mathrm{Eq}^{\mathrm{n}}\). (4) - (2), we get \(\mathrm{y}=2\)
Put this value in (3) we get -
\(\mathrm{x}+2 \times 2=5\)
\(\therefore \mathrm{x}=1\)
Thus, co-ordinate of point B \((1,2)\)
Point of interaction of \(L_{1}\) and \(L_{3}\)
On subtracting eq \(\mathrm{e}^{\mathrm{n}}\). (3) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(2 y-y=5-6 \Rightarrow y=-1\)
put the value of \(y\) in eq \({ }^{n}\). (3)
\(\mathrm{x}=7\)
The co-ordinate of point \(C\) are \((7,-1)\)
Thus, circles passes through the point A, B, C,
Thus it must satisfies \(\mathrm{eq}^{\mathrm{n}}\). of circle
\(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \ldots\) (5)
Point A lies eq \({ }^{n}\). (5) we get -
\(4+64-4 \mathrm{~g}+16 \mathrm{f}+\mathrm{c}=0\)
\(\Rightarrow \quad \mathrm{c}=-68+4 g-16 \mathrm{f}\)
Point \(B\) lies eq \({ }^{\mathrm{n}}\). (5) we get-
\(1^{2}+2^{2}+2 \mathrm{~g}(1)+2 \mathrm{f}(2)+\mathrm{c}=0\)
from (6)
\(2 \mathrm{~g}-4 \mathrm{f}=21\)
Point \(C\) also lies on \(\mathrm{eq}^{\mathrm{n}}\). (5)
\(7^{2}+(-1)^{2}+2 \mathrm{~g} 7+2 \mathrm{f}(-1)+\mathrm{c}=0\)
we get-
\(50+14 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}=0\)
from eqn. (6),
\(50+14 \mathrm{~g}-2 \mathrm{f}-68+4 \mathrm{~g}-16 \mathrm{f}=0\)
\(18 \mathrm{~g}-18 \mathrm{f}-18=0\)
\(\therefore \mathrm{g}=\mathrm{f}+1\)
put this value in eq \({ }^{\mathrm{n}}\). (7)
\(2(\mathrm{f}+1)-4 \mathrm{f}=21\)
\(\therefore 2 \mathrm{f}+2-4 \mathrm{f}=21\)
\(\mathrm{f}=\frac{-19}{2} \left\lvert\, \begin{gathered}\text { put the value in eq } \mathrm{n} \\ \mathrm{g}=\frac{-17}{2}\end{gathered}\right.\)
Hence centre of circle \(=(-\mathrm{g},-\mathrm{f})=\left(\frac{17}{2}, \frac{19}{2}\right)\)
NEET Test Series from KOTA - 10 Papers In MS WORD
WhatsApp Here
Straight Line
88785
If \(2 x+3 y+4=0\) is the perpendicular bisector of the line segment joining the points \(A(1,2)\) and \(B(\alpha, \beta)\), then the value of \(13 \alpha+13 \beta\) equals
1 -81
2 -99
3 99
4 81
Explanation:
(A):
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \tag{i}\)
\(\mathrm{~m}_{1}=\frac{\beta-2}{\alpha-1} \tag{ii}\)
\(2 x+3 y+4=0 \Rightarrow y=\frac{2}{3} x-\frac{4}{3}\)
\(\mathrm{m}_{2}=-\frac{2}{3} \tag{iii}\)
putting value of (ii), (iii) in (i), we get:-
\(\left(\frac{\beta-2}{\alpha-1}\right) \times\left(-\frac{2}{3}\right)=-1\)
\(2 \beta-4=3 \alpha-3\)
\(3 \alpha-2 \beta=-1 \tag{iv}\)
Mid point \(\left(\frac{\alpha+1}{2}, \frac{\beta+2}{2}\right)\) and it passes through the line
\((2 x+3 y+4)=0\)
\(2\left(\frac{\alpha+1}{2}\right)+3\left(\frac{\beta+2}{2}\right)+4=0\)
\(2 \alpha+2+3 \beta+6+8=0\)
\(2 \alpha+3 \beta=-16 \tag{v}\)
On solving eq \({ }^{\mathrm{n}}\) (iv) and (v), we get
\(\alpha=-\frac{35}{13}, \beta=\frac{-46}{13} \Rightarrow \alpha+\beta=-\frac{81}{13}\)
Hence \(13 \alpha+13 \beta=13(\alpha+\beta)\)
\(=13 \times\left(-\frac{81}{13}\right) \Rightarrow-81\)
AP EAMCET-2021-19.08.2021
Straight Line
88786
For three consecutive odd integers \(a, b\) and \(c\), if the variable line \(a x+b y+c=0\) always passes through the point \((\alpha, \beta)\), the value of \(\alpha^{2}+\beta^{2}\) equals
1 9
2 4
3 5
4 3
Explanation:
(C):
a, b, c are odd integer
For three consecutive odd integer are,
\(\mathrm{a}=\mathrm{a}, \quad \mathrm{b}=\mathrm{a}+2, \quad \mathrm{c}=\mathrm{a}+4\)
So,
\(a x+b y+c=0\)
\(a x+(a+2) y+(a+4)=0\)
\(a x+a y+a+2 y+4=0\)
\(2(y+2)+a(x+y+1)=0\)
\((y+2)+\frac{a}{2}(x+y+1)=0\)
\(L_{1}+\lambda L_{2}=0\)
So,
\(y+2=0 \Rightarrow y=-2\)
\(x+y+1=0 \Rightarrow x-2+1=0 \Rightarrow x=1\)
So, line passes through \((\alpha, \beta)=(1,-2)\)
\(a^{2}+\beta^{2}=(1)^{2}+(-2)^{2}\)
\(\alpha^{2}+\beta^{2}=5\)
AP EAMCET-2021-19.08.2021
Straight Line
88787
If the distance of a line from the origin is \(\sqrt{5}\) and having intercepts in the ratio of \(1: 2\) on axes, then the equations of line are .......... .
1 \(2 x-y \pm 5=0\)
2 \(2 x+y \pm 5=0\)
3 \(x-2 y \pm 5=0\)
4 \(x+2 y \pm 5=0\)
Explanation:
(B):
Given, Distance \(=\sqrt{5}\)
Intercept on axes \(=1: 2\)
\(\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \Rightarrow \mathrm{b}=2 \mathrm{a}\)
As we know, equation of line
i.e. \(\frac{x}{a}+\frac{y}{b}=1\)
i.e. \(\frac{x}{b / 2}+\frac{y}{b}=1 \Rightarrow 2 x+y=b\)
\(\Rightarrow 2 x+y-b=0 \tag{1}\)
Origin \((0,0)\) the distance between origin and line
\(\left|\frac{2(0)+0-b}{\sqrt{4+1}}\right|=\sqrt{5}\) (distance) \(\Rightarrow b= \pm 5\)
Put in (i), we get
\(2 x+y \pm 5=0\)
GUJCET-2011
Straight Line
88788
If \((a, b)\) is the centre of the circle passing through the vertices of the triangle formed by \(x\) \(+y=6,2 x+y=4\) and \(x+2 y=5\), then \((a, b)\) is
1 \((-17,-16)\)
2 \(\left(\frac{17}{2}, \frac{19}{2}\right)\)
3 \((17,18)\)
4 \(\left(\frac{-17}{2}, \frac{-19}{2}\right)\)
Explanation:
(B):
Let \(\mathrm{L}_{1}: \mathrm{x}+\mathrm{y}=6\)
\(\mathrm{L}_{2}: 2 \mathrm{x}+\mathrm{y}=4\)
...... (1)
\(\mathrm{L}_{3}: \mathrm{x}+2 \mathrm{y}=5\)
(1) Point of intersection of \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\)
On subtracting \(\mathrm{eq}^{\mathrm{n}}\). (2) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get \(2 \mathrm{x}-2=4-6\)
\(\therefore \mathrm{x}=-2\)
Put the value of \(x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(y=8\)
(2) Point of intersection \(\mathrm{L}_{2}\) and \(\mathrm{L}_{3}\)
Multiplying eq \(\mathrm{eq}^{\mathrm{n}}\). (3) by 2 we have -
\(\therefore \quad 2 \mathrm{x}+4 \mathrm{y}=10\)
\(\mathrm{Eq}^{\mathrm{n}}\). (4) - (2), we get \(\mathrm{y}=2\)
Put this value in (3) we get -
\(\mathrm{x}+2 \times 2=5\)
\(\therefore \mathrm{x}=1\)
Thus, co-ordinate of point B \((1,2)\)
Point of interaction of \(L_{1}\) and \(L_{3}\)
On subtracting eq \(\mathrm{e}^{\mathrm{n}}\). (3) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(2 y-y=5-6 \Rightarrow y=-1\)
put the value of \(y\) in eq \({ }^{n}\). (3)
\(\mathrm{x}=7\)
The co-ordinate of point \(C\) are \((7,-1)\)
Thus, circles passes through the point A, B, C,
Thus it must satisfies \(\mathrm{eq}^{\mathrm{n}}\). of circle
\(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \ldots\) (5)
Point A lies eq \({ }^{n}\). (5) we get -
\(4+64-4 \mathrm{~g}+16 \mathrm{f}+\mathrm{c}=0\)
\(\Rightarrow \quad \mathrm{c}=-68+4 g-16 \mathrm{f}\)
Point \(B\) lies eq \({ }^{\mathrm{n}}\). (5) we get-
\(1^{2}+2^{2}+2 \mathrm{~g}(1)+2 \mathrm{f}(2)+\mathrm{c}=0\)
from (6)
\(2 \mathrm{~g}-4 \mathrm{f}=21\)
Point \(C\) also lies on \(\mathrm{eq}^{\mathrm{n}}\). (5)
\(7^{2}+(-1)^{2}+2 \mathrm{~g} 7+2 \mathrm{f}(-1)+\mathrm{c}=0\)
we get-
\(50+14 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}=0\)
from eqn. (6),
\(50+14 \mathrm{~g}-2 \mathrm{f}-68+4 \mathrm{~g}-16 \mathrm{f}=0\)
\(18 \mathrm{~g}-18 \mathrm{f}-18=0\)
\(\therefore \mathrm{g}=\mathrm{f}+1\)
put this value in eq \({ }^{\mathrm{n}}\). (7)
\(2(\mathrm{f}+1)-4 \mathrm{f}=21\)
\(\therefore 2 \mathrm{f}+2-4 \mathrm{f}=21\)
\(\mathrm{f}=\frac{-19}{2} \left\lvert\, \begin{gathered}\text { put the value in eq } \mathrm{n} \\ \mathrm{g}=\frac{-17}{2}\end{gathered}\right.\)
Hence centre of circle \(=(-\mathrm{g},-\mathrm{f})=\left(\frac{17}{2}, \frac{19}{2}\right)\)
88785
If \(2 x+3 y+4=0\) is the perpendicular bisector of the line segment joining the points \(A(1,2)\) and \(B(\alpha, \beta)\), then the value of \(13 \alpha+13 \beta\) equals
1 -81
2 -99
3 99
4 81
Explanation:
(A):
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \tag{i}\)
\(\mathrm{~m}_{1}=\frac{\beta-2}{\alpha-1} \tag{ii}\)
\(2 x+3 y+4=0 \Rightarrow y=\frac{2}{3} x-\frac{4}{3}\)
\(\mathrm{m}_{2}=-\frac{2}{3} \tag{iii}\)
putting value of (ii), (iii) in (i), we get:-
\(\left(\frac{\beta-2}{\alpha-1}\right) \times\left(-\frac{2}{3}\right)=-1\)
\(2 \beta-4=3 \alpha-3\)
\(3 \alpha-2 \beta=-1 \tag{iv}\)
Mid point \(\left(\frac{\alpha+1}{2}, \frac{\beta+2}{2}\right)\) and it passes through the line
\((2 x+3 y+4)=0\)
\(2\left(\frac{\alpha+1}{2}\right)+3\left(\frac{\beta+2}{2}\right)+4=0\)
\(2 \alpha+2+3 \beta+6+8=0\)
\(2 \alpha+3 \beta=-16 \tag{v}\)
On solving eq \({ }^{\mathrm{n}}\) (iv) and (v), we get
\(\alpha=-\frac{35}{13}, \beta=\frac{-46}{13} \Rightarrow \alpha+\beta=-\frac{81}{13}\)
Hence \(13 \alpha+13 \beta=13(\alpha+\beta)\)
\(=13 \times\left(-\frac{81}{13}\right) \Rightarrow-81\)
AP EAMCET-2021-19.08.2021
Straight Line
88786
For three consecutive odd integers \(a, b\) and \(c\), if the variable line \(a x+b y+c=0\) always passes through the point \((\alpha, \beta)\), the value of \(\alpha^{2}+\beta^{2}\) equals
1 9
2 4
3 5
4 3
Explanation:
(C):
a, b, c are odd integer
For three consecutive odd integer are,
\(\mathrm{a}=\mathrm{a}, \quad \mathrm{b}=\mathrm{a}+2, \quad \mathrm{c}=\mathrm{a}+4\)
So,
\(a x+b y+c=0\)
\(a x+(a+2) y+(a+4)=0\)
\(a x+a y+a+2 y+4=0\)
\(2(y+2)+a(x+y+1)=0\)
\((y+2)+\frac{a}{2}(x+y+1)=0\)
\(L_{1}+\lambda L_{2}=0\)
So,
\(y+2=0 \Rightarrow y=-2\)
\(x+y+1=0 \Rightarrow x-2+1=0 \Rightarrow x=1\)
So, line passes through \((\alpha, \beta)=(1,-2)\)
\(a^{2}+\beta^{2}=(1)^{2}+(-2)^{2}\)
\(\alpha^{2}+\beta^{2}=5\)
AP EAMCET-2021-19.08.2021
Straight Line
88787
If the distance of a line from the origin is \(\sqrt{5}\) and having intercepts in the ratio of \(1: 2\) on axes, then the equations of line are .......... .
1 \(2 x-y \pm 5=0\)
2 \(2 x+y \pm 5=0\)
3 \(x-2 y \pm 5=0\)
4 \(x+2 y \pm 5=0\)
Explanation:
(B):
Given, Distance \(=\sqrt{5}\)
Intercept on axes \(=1: 2\)
\(\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \Rightarrow \mathrm{b}=2 \mathrm{a}\)
As we know, equation of line
i.e. \(\frac{x}{a}+\frac{y}{b}=1\)
i.e. \(\frac{x}{b / 2}+\frac{y}{b}=1 \Rightarrow 2 x+y=b\)
\(\Rightarrow 2 x+y-b=0 \tag{1}\)
Origin \((0,0)\) the distance between origin and line
\(\left|\frac{2(0)+0-b}{\sqrt{4+1}}\right|=\sqrt{5}\) (distance) \(\Rightarrow b= \pm 5\)
Put in (i), we get
\(2 x+y \pm 5=0\)
GUJCET-2011
Straight Line
88788
If \((a, b)\) is the centre of the circle passing through the vertices of the triangle formed by \(x\) \(+y=6,2 x+y=4\) and \(x+2 y=5\), then \((a, b)\) is
1 \((-17,-16)\)
2 \(\left(\frac{17}{2}, \frac{19}{2}\right)\)
3 \((17,18)\)
4 \(\left(\frac{-17}{2}, \frac{-19}{2}\right)\)
Explanation:
(B):
Let \(\mathrm{L}_{1}: \mathrm{x}+\mathrm{y}=6\)
\(\mathrm{L}_{2}: 2 \mathrm{x}+\mathrm{y}=4\)
...... (1)
\(\mathrm{L}_{3}: \mathrm{x}+2 \mathrm{y}=5\)
(1) Point of intersection of \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\)
On subtracting \(\mathrm{eq}^{\mathrm{n}}\). (2) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get \(2 \mathrm{x}-2=4-6\)
\(\therefore \mathrm{x}=-2\)
Put the value of \(x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(y=8\)
(2) Point of intersection \(\mathrm{L}_{2}\) and \(\mathrm{L}_{3}\)
Multiplying eq \(\mathrm{eq}^{\mathrm{n}}\). (3) by 2 we have -
\(\therefore \quad 2 \mathrm{x}+4 \mathrm{y}=10\)
\(\mathrm{Eq}^{\mathrm{n}}\). (4) - (2), we get \(\mathrm{y}=2\)
Put this value in (3) we get -
\(\mathrm{x}+2 \times 2=5\)
\(\therefore \mathrm{x}=1\)
Thus, co-ordinate of point B \((1,2)\)
Point of interaction of \(L_{1}\) and \(L_{3}\)
On subtracting eq \(\mathrm{e}^{\mathrm{n}}\). (3) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(2 y-y=5-6 \Rightarrow y=-1\)
put the value of \(y\) in eq \({ }^{n}\). (3)
\(\mathrm{x}=7\)
The co-ordinate of point \(C\) are \((7,-1)\)
Thus, circles passes through the point A, B, C,
Thus it must satisfies \(\mathrm{eq}^{\mathrm{n}}\). of circle
\(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \ldots\) (5)
Point A lies eq \({ }^{n}\). (5) we get -
\(4+64-4 \mathrm{~g}+16 \mathrm{f}+\mathrm{c}=0\)
\(\Rightarrow \quad \mathrm{c}=-68+4 g-16 \mathrm{f}\)
Point \(B\) lies eq \({ }^{\mathrm{n}}\). (5) we get-
\(1^{2}+2^{2}+2 \mathrm{~g}(1)+2 \mathrm{f}(2)+\mathrm{c}=0\)
from (6)
\(2 \mathrm{~g}-4 \mathrm{f}=21\)
Point \(C\) also lies on \(\mathrm{eq}^{\mathrm{n}}\). (5)
\(7^{2}+(-1)^{2}+2 \mathrm{~g} 7+2 \mathrm{f}(-1)+\mathrm{c}=0\)
we get-
\(50+14 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}=0\)
from eqn. (6),
\(50+14 \mathrm{~g}-2 \mathrm{f}-68+4 \mathrm{~g}-16 \mathrm{f}=0\)
\(18 \mathrm{~g}-18 \mathrm{f}-18=0\)
\(\therefore \mathrm{g}=\mathrm{f}+1\)
put this value in eq \({ }^{\mathrm{n}}\). (7)
\(2(\mathrm{f}+1)-4 \mathrm{f}=21\)
\(\therefore 2 \mathrm{f}+2-4 \mathrm{f}=21\)
\(\mathrm{f}=\frac{-19}{2} \left\lvert\, \begin{gathered}\text { put the value in eq } \mathrm{n} \\ \mathrm{g}=\frac{-17}{2}\end{gathered}\right.\)
Hence centre of circle \(=(-\mathrm{g},-\mathrm{f})=\left(\frac{17}{2}, \frac{19}{2}\right)\)
88785
If \(2 x+3 y+4=0\) is the perpendicular bisector of the line segment joining the points \(A(1,2)\) and \(B(\alpha, \beta)\), then the value of \(13 \alpha+13 \beta\) equals
1 -81
2 -99
3 99
4 81
Explanation:
(A):
\(\mathrm{m}_{1} \cdot \mathrm{m}_{2}=-1 \tag{i}\)
\(\mathrm{~m}_{1}=\frac{\beta-2}{\alpha-1} \tag{ii}\)
\(2 x+3 y+4=0 \Rightarrow y=\frac{2}{3} x-\frac{4}{3}\)
\(\mathrm{m}_{2}=-\frac{2}{3} \tag{iii}\)
putting value of (ii), (iii) in (i), we get:-
\(\left(\frac{\beta-2}{\alpha-1}\right) \times\left(-\frac{2}{3}\right)=-1\)
\(2 \beta-4=3 \alpha-3\)
\(3 \alpha-2 \beta=-1 \tag{iv}\)
Mid point \(\left(\frac{\alpha+1}{2}, \frac{\beta+2}{2}\right)\) and it passes through the line
\((2 x+3 y+4)=0\)
\(2\left(\frac{\alpha+1}{2}\right)+3\left(\frac{\beta+2}{2}\right)+4=0\)
\(2 \alpha+2+3 \beta+6+8=0\)
\(2 \alpha+3 \beta=-16 \tag{v}\)
On solving eq \({ }^{\mathrm{n}}\) (iv) and (v), we get
\(\alpha=-\frac{35}{13}, \beta=\frac{-46}{13} \Rightarrow \alpha+\beta=-\frac{81}{13}\)
Hence \(13 \alpha+13 \beta=13(\alpha+\beta)\)
\(=13 \times\left(-\frac{81}{13}\right) \Rightarrow-81\)
AP EAMCET-2021-19.08.2021
Straight Line
88786
For three consecutive odd integers \(a, b\) and \(c\), if the variable line \(a x+b y+c=0\) always passes through the point \((\alpha, \beta)\), the value of \(\alpha^{2}+\beta^{2}\) equals
1 9
2 4
3 5
4 3
Explanation:
(C):
a, b, c are odd integer
For three consecutive odd integer are,
\(\mathrm{a}=\mathrm{a}, \quad \mathrm{b}=\mathrm{a}+2, \quad \mathrm{c}=\mathrm{a}+4\)
So,
\(a x+b y+c=0\)
\(a x+(a+2) y+(a+4)=0\)
\(a x+a y+a+2 y+4=0\)
\(2(y+2)+a(x+y+1)=0\)
\((y+2)+\frac{a}{2}(x+y+1)=0\)
\(L_{1}+\lambda L_{2}=0\)
So,
\(y+2=0 \Rightarrow y=-2\)
\(x+y+1=0 \Rightarrow x-2+1=0 \Rightarrow x=1\)
So, line passes through \((\alpha, \beta)=(1,-2)\)
\(a^{2}+\beta^{2}=(1)^{2}+(-2)^{2}\)
\(\alpha^{2}+\beta^{2}=5\)
AP EAMCET-2021-19.08.2021
Straight Line
88787
If the distance of a line from the origin is \(\sqrt{5}\) and having intercepts in the ratio of \(1: 2\) on axes, then the equations of line are .......... .
1 \(2 x-y \pm 5=0\)
2 \(2 x+y \pm 5=0\)
3 \(x-2 y \pm 5=0\)
4 \(x+2 y \pm 5=0\)
Explanation:
(B):
Given, Distance \(=\sqrt{5}\)
Intercept on axes \(=1: 2\)
\(\Rightarrow \frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{2} \Rightarrow \mathrm{b}=2 \mathrm{a}\)
As we know, equation of line
i.e. \(\frac{x}{a}+\frac{y}{b}=1\)
i.e. \(\frac{x}{b / 2}+\frac{y}{b}=1 \Rightarrow 2 x+y=b\)
\(\Rightarrow 2 x+y-b=0 \tag{1}\)
Origin \((0,0)\) the distance between origin and line
\(\left|\frac{2(0)+0-b}{\sqrt{4+1}}\right|=\sqrt{5}\) (distance) \(\Rightarrow b= \pm 5\)
Put in (i), we get
\(2 x+y \pm 5=0\)
GUJCET-2011
Straight Line
88788
If \((a, b)\) is the centre of the circle passing through the vertices of the triangle formed by \(x\) \(+y=6,2 x+y=4\) and \(x+2 y=5\), then \((a, b)\) is
1 \((-17,-16)\)
2 \(\left(\frac{17}{2}, \frac{19}{2}\right)\)
3 \((17,18)\)
4 \(\left(\frac{-17}{2}, \frac{-19}{2}\right)\)
Explanation:
(B):
Let \(\mathrm{L}_{1}: \mathrm{x}+\mathrm{y}=6\)
\(\mathrm{L}_{2}: 2 \mathrm{x}+\mathrm{y}=4\)
...... (1)
\(\mathrm{L}_{3}: \mathrm{x}+2 \mathrm{y}=5\)
(1) Point of intersection of \(\mathrm{L}_{1}\) and \(\mathrm{L}_{2}\)
On subtracting \(\mathrm{eq}^{\mathrm{n}}\). (2) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get \(2 \mathrm{x}-2=4-6\)
\(\therefore \mathrm{x}=-2\)
Put the value of \(x\) in \(\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(y=8\)
(2) Point of intersection \(\mathrm{L}_{2}\) and \(\mathrm{L}_{3}\)
Multiplying eq \(\mathrm{eq}^{\mathrm{n}}\). (3) by 2 we have -
\(\therefore \quad 2 \mathrm{x}+4 \mathrm{y}=10\)
\(\mathrm{Eq}^{\mathrm{n}}\). (4) - (2), we get \(\mathrm{y}=2\)
Put this value in (3) we get -
\(\mathrm{x}+2 \times 2=5\)
\(\therefore \mathrm{x}=1\)
Thus, co-ordinate of point B \((1,2)\)
Point of interaction of \(L_{1}\) and \(L_{3}\)
On subtracting eq \(\mathrm{e}^{\mathrm{n}}\). (3) \(-\mathrm{eq}^{\mathrm{n}}\). (1) we get -
\(2 y-y=5-6 \Rightarrow y=-1\)
put the value of \(y\) in eq \({ }^{n}\). (3)
\(\mathrm{x}=7\)
The co-ordinate of point \(C\) are \((7,-1)\)
Thus, circles passes through the point A, B, C,
Thus it must satisfies \(\mathrm{eq}^{\mathrm{n}}\). of circle
\(\mathrm{x}^{2}+\mathrm{y}^{2}+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \ldots\) (5)
Point A lies eq \({ }^{n}\). (5) we get -
\(4+64-4 \mathrm{~g}+16 \mathrm{f}+\mathrm{c}=0\)
\(\Rightarrow \quad \mathrm{c}=-68+4 g-16 \mathrm{f}\)
Point \(B\) lies eq \({ }^{\mathrm{n}}\). (5) we get-
\(1^{2}+2^{2}+2 \mathrm{~g}(1)+2 \mathrm{f}(2)+\mathrm{c}=0\)
from (6)
\(2 \mathrm{~g}-4 \mathrm{f}=21\)
Point \(C\) also lies on \(\mathrm{eq}^{\mathrm{n}}\). (5)
\(7^{2}+(-1)^{2}+2 \mathrm{~g} 7+2 \mathrm{f}(-1)+\mathrm{c}=0\)
we get-
\(50+14 \mathrm{~g}-2 \mathrm{f}+\mathrm{c}=0\)
from eqn. (6),
\(50+14 \mathrm{~g}-2 \mathrm{f}-68+4 \mathrm{~g}-16 \mathrm{f}=0\)
\(18 \mathrm{~g}-18 \mathrm{f}-18=0\)
\(\therefore \mathrm{g}=\mathrm{f}+1\)
put this value in eq \({ }^{\mathrm{n}}\). (7)
\(2(\mathrm{f}+1)-4 \mathrm{f}=21\)
\(\therefore 2 \mathrm{f}+2-4 \mathrm{f}=21\)
\(\mathrm{f}=\frac{-19}{2} \left\lvert\, \begin{gathered}\text { put the value in eq } \mathrm{n} \\ \mathrm{g}=\frac{-17}{2}\end{gathered}\right.\)
Hence centre of circle \(=(-\mathrm{g},-\mathrm{f})=\left(\frac{17}{2}, \frac{19}{2}\right)\)
