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Straight Line

88774 If $A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$ are vertices of a triangle, then the coordinates of the point in which the bisector of the angle $A$ meet the side $B C$ is

1

(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})

2

(\frac{17}{8}, \frac{22}{8}, \frac{45}{8})

3

(\frac{- 22}{8}, \frac{- 17}{8}, \frac{45}{8})

4

(\frac{- 17}{8}, \frac{22}{8}, \frac{45}{8})

Explanation:

(A) : Given points are
$A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$
$AB = \sqrt{(5 - 4)^{2} + (4 - 3)^{2} + (6 - 2)^{2}}$
$AB = \sqrt{(1)^{2} + (1)^{2} + (4)^{2}} = \sqrt{18}$
$AB = 3 \sqrt{2}$
$A C = \sqrt{(- 1 - 4)^{2} + (- 1 - 3)^{2} + (5 - 2)^{2}}$
$= \sqrt{(- 5)^{2} + (- 4)^{2} + (3)^{2}} = \sqrt{25 + 16 + 9} = \sqrt{50}$
$AC = 5 \sqrt{2}$
$AB : AC = 3 \sqrt{2} : 5 \sqrt{2} = 3 : 5$
Let say the bisector of angle $A$ meet the side $BC$ that point $D (x_{D}, y_{D}, z_{D})$ and $D$ divides $BC$ side in ratio $3 : 5$ internally
$x_{D} = \frac{{mx}_{C} + {nx}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times 5}{3 + 5} = \frac{22}{8}$
$y_{D} = \frac{{mx}_{C} + {ny}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times (4)}{3 + 5} = \frac{17}{8}$
$z_{D} = \frac{{mz}_{C} + {nz}_{B}}{m + n} = \frac{3 \times (5) + 5 \times (6)}{3 + 5} = \frac{45}{8}$
Therefore, coordinates of point $D$ is $(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})$ ,

TS EAMCET-2020-11.09.2020

Straight Line

88775 Let $P$ be the point of intersection of the lines $L_{1} \equiv x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0 . A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are points on the lines $l_{1} = 0$ and $l_{2}$ $= 0$ respectively such that $P A = 3 \sqrt{2}, P B =$ $\sqrt{2}, x_{1}, y_{1} \geq 0, x_{2} y_{2} \geq 0$ , then the angle made by the line segment $A B$ at the origin is

1

\frac{π}{4}

2

\frac{π}{2}

3

\cos^{- 1} (\frac{3}{4})

4

\cos^{- 1} (\frac{9}{\sqrt{85}})

Explanation:

(D) : The point of intersection of lines
$L_{1} = x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0$ is $(6, - 1)$
Now, for point $A (x_{1}, y_{1})$ such that $PA = 3 \sqrt{2}$ , take the equation of line $L_{1}$ in symmetric form
$L_{1} : \frac{x - 6}{\cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm 3 \sqrt{2}$
$\Rightarrow x_{1} = 6 \pm 3 and y_{1} = - 1 \pm 3$
$∵ x_{1}, y_{1} \geq 0, so A (x_{1}, y_{1}) is (9, 2)$
Similarly, for point $B (x_{2}, y_{2})$ such that $PB = \sqrt{2}$ take the equation of line $L_{2}$ in symmetric form
$L_{2} : \frac{x - 6}{- \cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm \sqrt{2}$
$\Rightarrow x_{2} = 6 \mp 1$ and $y_{2} = - 1 \pm 1$
$∵ x_{2}, y_{2} \geq 0$ , so $B (x_{2}, y_{2})$ is $(5, 0)$
$∴$ Slope of line joining origin and point $A$ is $m_{1} = \frac{2}{9}$
and slope of line joining origin and point $B$ is $m_{2} = 0$
$∴ ∠ BOA = θ = \tan^{- 1} (\frac{2 / 9 - 0}{1 + 0}) = \tan^{- 1} (\frac{2}{9})$
$= \cos^{- 1} (\frac{9}{\sqrt{85}})$

TS EAMCET-2020-11.09.2020

Straight Line

88776 A bisector of the angle between the normal's of the planes $4 x + 3 y = 5$ and $x + 2 y + 2 z = 4$ is along the vector

1

(17 \hat{i} + 9 \hat{j} - 12 \hat{k})

2

(17 \hat{i} - 9 \hat{j} + 12 \hat{k})

3

(17 \hat{i} - \hat{j} + 10 \hat{k})

4

(7 \hat{i} - \hat{j} - 10 \hat{k})

Explanation:

(D) : Bisector of the angle between the normal's to the planes $4 x + 3 y - 5 = 0$ and $x + 2 y + 2 z - 4 = 0$ is given by-
$\frac{4 x + 3 y - 5}{\sqrt{(4)^{2} + (3)^{2}}} = \pm \frac{x + 2 y + 2 z - 4}{\sqrt{(1)^{2} + (2)^{2} + (2)^{2}}}$
$\frac{4 x + 3 y - 5}{5} = \pm \frac{x + 2 y + 2 z - 4}{3}$
(i) Taking + ve sign, we get-
$3 (4 x + 3 y - 5) = 5 (x + 2 y + 2 z - 4)$
$7 x - y - 10 z + 5 = 0 \Rightarrow 7 \hat{i} - \hat{j} - 10 \hat{k}$
(ii) -ve sign will gives-
$3 (4 x + 3 y - 5) = 5 (- x - 2 y - 2 z + 4)$
$17 x + 19 y - 35 = 0$
Hence, from case (i) Bisector of the angle between normal of the given plane is represented by the vectors $7 \hat{i} - \hat{j} - 10 \hat{k}$ or $17 \hat{i} + 19 \hat{j}$

TS EAMCET-2022-19.07.2022

Straight Line

88777 If the vertices of a triangle $ABC$ are $A (1, 7), B$ $(- 5, - 1)$ and $C (7, 4)$ , then the equation of a bisector of $△ ABC$ is

1

7 x - 9 y + 26 = 0

2

9 x - 7 y + 38 = 0

3

7 x + 9 y + 44 = 0

4

9 x + 7 y + 52 = 0

Explanation:

(A):
$AB = \sqrt{(- 5 - 1)^{2} + (- 1 - 7)^{2}}$
$= \sqrt{36 + 64}$
$AB = 10$
$BC = \sqrt{(7 + 5)^{2} + (4 + 1)^{2}}$
$= \sqrt{144 + 25}$
$= \sqrt{169}$
$BC = 13$
$BD$ divide the line $AC$ in ratio $10 : 13$
$Point D = (\frac{10 \times 7 + 13 \times 1}{23}, \frac{10 \times 4 + 13 \times 7}{23})$
$D = (\frac{83}{23}, \frac{131}{23})$
Hence the equation of bisector
$y + 1 = \frac{- 1 - \frac{131}{23}}{- 5 - \frac{83}{23}} (x + 5)$
$\Rightarrow y + 1 = \frac{- 23 - 131}{- 115 - 83} (x + 5)$
$\Rightarrow y + 1 = \frac{- 154}{- 198} (x + 5)$
$\Rightarrow 9 (y + 1) = 7 (x + 5)$
$\Rightarrow 9 y + 9 = 7 x + 35$
$\Rightarrow 7 x - 9 y + 26 = 0$

TS EAMCET-2021-04.08.2021

Straight Line

88774 If $A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$ are vertices of a triangle, then the coordinates of the point in which the bisector of the angle $A$ meet the side $B C$ is

1

(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})

2

(\frac{17}{8}, \frac{22}{8}, \frac{45}{8})

3

(\frac{- 22}{8}, \frac{- 17}{8}, \frac{45}{8})

4

(\frac{- 17}{8}, \frac{22}{8}, \frac{45}{8})

Explanation:

(A) : Given points are
$A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$
$AB = \sqrt{(5 - 4)^{2} + (4 - 3)^{2} + (6 - 2)^{2}}$
$AB = \sqrt{(1)^{2} + (1)^{2} + (4)^{2}} = \sqrt{18}$
$AB = 3 \sqrt{2}$
$A C = \sqrt{(- 1 - 4)^{2} + (- 1 - 3)^{2} + (5 - 2)^{2}}$
$= \sqrt{(- 5)^{2} + (- 4)^{2} + (3)^{2}} = \sqrt{25 + 16 + 9} = \sqrt{50}$
$AC = 5 \sqrt{2}$
$AB : AC = 3 \sqrt{2} : 5 \sqrt{2} = 3 : 5$
Let say the bisector of angle $A$ meet the side $BC$ that point $D (x_{D}, y_{D}, z_{D})$ and $D$ divides $BC$ side in ratio $3 : 5$ internally
$x_{D} = \frac{{mx}_{C} + {nx}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times 5}{3 + 5} = \frac{22}{8}$
$y_{D} = \frac{{mx}_{C} + {ny}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times (4)}{3 + 5} = \frac{17}{8}$
$z_{D} = \frac{{mz}_{C} + {nz}_{B}}{m + n} = \frac{3 \times (5) + 5 \times (6)}{3 + 5} = \frac{45}{8}$
Therefore, coordinates of point $D$ is $(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})$ ,

TS EAMCET-2020-11.09.2020

Straight Line

88775 Let $P$ be the point of intersection of the lines $L_{1} \equiv x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0 . A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are points on the lines $l_{1} = 0$ and $l_{2}$ $= 0$ respectively such that $P A = 3 \sqrt{2}, P B =$ $\sqrt{2}, x_{1}, y_{1} \geq 0, x_{2} y_{2} \geq 0$ , then the angle made by the line segment $A B$ at the origin is

1

\frac{π}{4}

2

\frac{π}{2}

3

\cos^{- 1} (\frac{3}{4})

4

\cos^{- 1} (\frac{9}{\sqrt{85}})

Explanation:

(D) : The point of intersection of lines
$L_{1} = x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0$ is $(6, - 1)$
Now, for point $A (x_{1}, y_{1})$ such that $PA = 3 \sqrt{2}$ , take the equation of line $L_{1}$ in symmetric form
$L_{1} : \frac{x - 6}{\cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm 3 \sqrt{2}$
$\Rightarrow x_{1} = 6 \pm 3 and y_{1} = - 1 \pm 3$
$∵ x_{1}, y_{1} \geq 0, so A (x_{1}, y_{1}) is (9, 2)$
Similarly, for point $B (x_{2}, y_{2})$ such that $PB = \sqrt{2}$ take the equation of line $L_{2}$ in symmetric form
$L_{2} : \frac{x - 6}{- \cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm \sqrt{2}$
$\Rightarrow x_{2} = 6 \mp 1$ and $y_{2} = - 1 \pm 1$
$∵ x_{2}, y_{2} \geq 0$ , so $B (x_{2}, y_{2})$ is $(5, 0)$
$∴$ Slope of line joining origin and point $A$ is $m_{1} = \frac{2}{9}$
and slope of line joining origin and point $B$ is $m_{2} = 0$
$∴ ∠ BOA = θ = \tan^{- 1} (\frac{2 / 9 - 0}{1 + 0}) = \tan^{- 1} (\frac{2}{9})$
$= \cos^{- 1} (\frac{9}{\sqrt{85}})$

TS EAMCET-2020-11.09.2020

Straight Line

88776 A bisector of the angle between the normal's of the planes $4 x + 3 y = 5$ and $x + 2 y + 2 z = 4$ is along the vector

1

(17 \hat{i} + 9 \hat{j} - 12 \hat{k})

2

(17 \hat{i} - 9 \hat{j} + 12 \hat{k})

3

(17 \hat{i} - \hat{j} + 10 \hat{k})

4

(7 \hat{i} - \hat{j} - 10 \hat{k})

Explanation:

(D) : Bisector of the angle between the normal's to the planes $4 x + 3 y - 5 = 0$ and $x + 2 y + 2 z - 4 = 0$ is given by-
$\frac{4 x + 3 y - 5}{\sqrt{(4)^{2} + (3)^{2}}} = \pm \frac{x + 2 y + 2 z - 4}{\sqrt{(1)^{2} + (2)^{2} + (2)^{2}}}$
$\frac{4 x + 3 y - 5}{5} = \pm \frac{x + 2 y + 2 z - 4}{3}$
(i) Taking + ve sign, we get-
$3 (4 x + 3 y - 5) = 5 (x + 2 y + 2 z - 4)$
$7 x - y - 10 z + 5 = 0 \Rightarrow 7 \hat{i} - \hat{j} - 10 \hat{k}$
(ii) -ve sign will gives-
$3 (4 x + 3 y - 5) = 5 (- x - 2 y - 2 z + 4)$
$17 x + 19 y - 35 = 0$
Hence, from case (i) Bisector of the angle between normal of the given plane is represented by the vectors $7 \hat{i} - \hat{j} - 10 \hat{k}$ or $17 \hat{i} + 19 \hat{j}$

TS EAMCET-2022-19.07.2022

Straight Line

88777 If the vertices of a triangle $ABC$ are $A (1, 7), B$ $(- 5, - 1)$ and $C (7, 4)$ , then the equation of a bisector of $△ ABC$ is

1

7 x - 9 y + 26 = 0

2

9 x - 7 y + 38 = 0

3

7 x + 9 y + 44 = 0

4

9 x + 7 y + 52 = 0

Explanation:

(A):
$AB = \sqrt{(- 5 - 1)^{2} + (- 1 - 7)^{2}}$
$= \sqrt{36 + 64}$
$AB = 10$
$BC = \sqrt{(7 + 5)^{2} + (4 + 1)^{2}}$
$= \sqrt{144 + 25}$
$= \sqrt{169}$
$BC = 13$
$BD$ divide the line $AC$ in ratio $10 : 13$
$Point D = (\frac{10 \times 7 + 13 \times 1}{23}, \frac{10 \times 4 + 13 \times 7}{23})$
$D = (\frac{83}{23}, \frac{131}{23})$
Hence the equation of bisector
$y + 1 = \frac{- 1 - \frac{131}{23}}{- 5 - \frac{83}{23}} (x + 5)$
$\Rightarrow y + 1 = \frac{- 23 - 131}{- 115 - 83} (x + 5)$
$\Rightarrow y + 1 = \frac{- 154}{- 198} (x + 5)$
$\Rightarrow 9 (y + 1) = 7 (x + 5)$
$\Rightarrow 9 y + 9 = 7 x + 35$
$\Rightarrow 7 x - 9 y + 26 = 0$

TS EAMCET-2021-04.08.2021

Straight Line

88774 If $A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$ are vertices of a triangle, then the coordinates of the point in which the bisector of the angle $A$ meet the side $B C$ is

1

(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})

2

(\frac{17}{8}, \frac{22}{8}, \frac{45}{8})

3

(\frac{- 22}{8}, \frac{- 17}{8}, \frac{45}{8})

4

(\frac{- 17}{8}, \frac{22}{8}, \frac{45}{8})

Explanation:

(A) : Given points are
$A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$
$AB = \sqrt{(5 - 4)^{2} + (4 - 3)^{2} + (6 - 2)^{2}}$
$AB = \sqrt{(1)^{2} + (1)^{2} + (4)^{2}} = \sqrt{18}$
$AB = 3 \sqrt{2}$
$A C = \sqrt{(- 1 - 4)^{2} + (- 1 - 3)^{2} + (5 - 2)^{2}}$
$= \sqrt{(- 5)^{2} + (- 4)^{2} + (3)^{2}} = \sqrt{25 + 16 + 9} = \sqrt{50}$
$AC = 5 \sqrt{2}$
$AB : AC = 3 \sqrt{2} : 5 \sqrt{2} = 3 : 5$
Let say the bisector of angle $A$ meet the side $BC$ that point $D (x_{D}, y_{D}, z_{D})$ and $D$ divides $BC$ side in ratio $3 : 5$ internally
$x_{D} = \frac{{mx}_{C} + {nx}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times 5}{3 + 5} = \frac{22}{8}$
$y_{D} = \frac{{mx}_{C} + {ny}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times (4)}{3 + 5} = \frac{17}{8}$
$z_{D} = \frac{{mz}_{C} + {nz}_{B}}{m + n} = \frac{3 \times (5) + 5 \times (6)}{3 + 5} = \frac{45}{8}$
Therefore, coordinates of point $D$ is $(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})$ ,

TS EAMCET-2020-11.09.2020

Straight Line

88775 Let $P$ be the point of intersection of the lines $L_{1} \equiv x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0 . A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are points on the lines $l_{1} = 0$ and $l_{2}$ $= 0$ respectively such that $P A = 3 \sqrt{2}, P B =$ $\sqrt{2}, x_{1}, y_{1} \geq 0, x_{2} y_{2} \geq 0$ , then the angle made by the line segment $A B$ at the origin is

1

\frac{π}{4}

2

\frac{π}{2}

3

\cos^{- 1} (\frac{3}{4})

4

\cos^{- 1} (\frac{9}{\sqrt{85}})

Explanation:

(D) : The point of intersection of lines
$L_{1} = x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0$ is $(6, - 1)$
Now, for point $A (x_{1}, y_{1})$ such that $PA = 3 \sqrt{2}$ , take the equation of line $L_{1}$ in symmetric form
$L_{1} : \frac{x - 6}{\cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm 3 \sqrt{2}$
$\Rightarrow x_{1} = 6 \pm 3 and y_{1} = - 1 \pm 3$
$∵ x_{1}, y_{1} \geq 0, so A (x_{1}, y_{1}) is (9, 2)$
Similarly, for point $B (x_{2}, y_{2})$ such that $PB = \sqrt{2}$ take the equation of line $L_{2}$ in symmetric form
$L_{2} : \frac{x - 6}{- \cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm \sqrt{2}$
$\Rightarrow x_{2} = 6 \mp 1$ and $y_{2} = - 1 \pm 1$
$∵ x_{2}, y_{2} \geq 0$ , so $B (x_{2}, y_{2})$ is $(5, 0)$
$∴$ Slope of line joining origin and point $A$ is $m_{1} = \frac{2}{9}$
and slope of line joining origin and point $B$ is $m_{2} = 0$
$∴ ∠ BOA = θ = \tan^{- 1} (\frac{2 / 9 - 0}{1 + 0}) = \tan^{- 1} (\frac{2}{9})$
$= \cos^{- 1} (\frac{9}{\sqrt{85}})$

TS EAMCET-2020-11.09.2020

Straight Line

88776 A bisector of the angle between the normal's of the planes $4 x + 3 y = 5$ and $x + 2 y + 2 z = 4$ is along the vector

1

(17 \hat{i} + 9 \hat{j} - 12 \hat{k})

2

(17 \hat{i} - 9 \hat{j} + 12 \hat{k})

3

(17 \hat{i} - \hat{j} + 10 \hat{k})

4

(7 \hat{i} - \hat{j} - 10 \hat{k})

Explanation:

(D) : Bisector of the angle between the normal's to the planes $4 x + 3 y - 5 = 0$ and $x + 2 y + 2 z - 4 = 0$ is given by-
$\frac{4 x + 3 y - 5}{\sqrt{(4)^{2} + (3)^{2}}} = \pm \frac{x + 2 y + 2 z - 4}{\sqrt{(1)^{2} + (2)^{2} + (2)^{2}}}$
$\frac{4 x + 3 y - 5}{5} = \pm \frac{x + 2 y + 2 z - 4}{3}$
(i) Taking + ve sign, we get-
$3 (4 x + 3 y - 5) = 5 (x + 2 y + 2 z - 4)$
$7 x - y - 10 z + 5 = 0 \Rightarrow 7 \hat{i} - \hat{j} - 10 \hat{k}$
(ii) -ve sign will gives-
$3 (4 x + 3 y - 5) = 5 (- x - 2 y - 2 z + 4)$
$17 x + 19 y - 35 = 0$
Hence, from case (i) Bisector of the angle between normal of the given plane is represented by the vectors $7 \hat{i} - \hat{j} - 10 \hat{k}$ or $17 \hat{i} + 19 \hat{j}$

TS EAMCET-2022-19.07.2022

Straight Line

88777 If the vertices of a triangle $ABC$ are $A (1, 7), B$ $(- 5, - 1)$ and $C (7, 4)$ , then the equation of a bisector of $△ ABC$ is

1

7 x - 9 y + 26 = 0

2

9 x - 7 y + 38 = 0

3

7 x + 9 y + 44 = 0

4

9 x + 7 y + 52 = 0

Explanation:

(A):
$AB = \sqrt{(- 5 - 1)^{2} + (- 1 - 7)^{2}}$
$= \sqrt{36 + 64}$
$AB = 10$
$BC = \sqrt{(7 + 5)^{2} + (4 + 1)^{2}}$
$= \sqrt{144 + 25}$
$= \sqrt{169}$
$BC = 13$
$BD$ divide the line $AC$ in ratio $10 : 13$
$Point D = (\frac{10 \times 7 + 13 \times 1}{23}, \frac{10 \times 4 + 13 \times 7}{23})$
$D = (\frac{83}{23}, \frac{131}{23})$
Hence the equation of bisector
$y + 1 = \frac{- 1 - \frac{131}{23}}{- 5 - \frac{83}{23}} (x + 5)$
$\Rightarrow y + 1 = \frac{- 23 - 131}{- 115 - 83} (x + 5)$
$\Rightarrow y + 1 = \frac{- 154}{- 198} (x + 5)$
$\Rightarrow 9 (y + 1) = 7 (x + 5)$
$\Rightarrow 9 y + 9 = 7 x + 35$
$\Rightarrow 7 x - 9 y + 26 = 0$

TS EAMCET-2021-04.08.2021

Straight Line

88774 If $A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$ are vertices of a triangle, then the coordinates of the point in which the bisector of the angle $A$ meet the side $B C$ is

1

(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})

2

(\frac{17}{8}, \frac{22}{8}, \frac{45}{8})

3

(\frac{- 22}{8}, \frac{- 17}{8}, \frac{45}{8})

4

(\frac{- 17}{8}, \frac{22}{8}, \frac{45}{8})

Explanation:

(A) : Given points are
$A (4, 3, 2), B (5, 4, 6), C (- 1, - 1, 5)$
$AB = \sqrt{(5 - 4)^{2} + (4 - 3)^{2} + (6 - 2)^{2}}$
$AB = \sqrt{(1)^{2} + (1)^{2} + (4)^{2}} = \sqrt{18}$
$AB = 3 \sqrt{2}$
$A C = \sqrt{(- 1 - 4)^{2} + (- 1 - 3)^{2} + (5 - 2)^{2}}$
$= \sqrt{(- 5)^{2} + (- 4)^{2} + (3)^{2}} = \sqrt{25 + 16 + 9} = \sqrt{50}$
$AC = 5 \sqrt{2}$
$AB : AC = 3 \sqrt{2} : 5 \sqrt{2} = 3 : 5$
Let say the bisector of angle $A$ meet the side $BC$ that point $D (x_{D}, y_{D}, z_{D})$ and $D$ divides $BC$ side in ratio $3 : 5$ internally
$x_{D} = \frac{{mx}_{C} + {nx}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times 5}{3 + 5} = \frac{22}{8}$
$y_{D} = \frac{{mx}_{C} + {ny}_{B}}{m + n} = \frac{3 \times (- 1) + 5 \times (4)}{3 + 5} = \frac{17}{8}$
$z_{D} = \frac{{mz}_{C} + {nz}_{B}}{m + n} = \frac{3 \times (5) + 5 \times (6)}{3 + 5} = \frac{45}{8}$
Therefore, coordinates of point $D$ is $(\frac{22}{8}, \frac{17}{8}, \frac{45}{8})$ ,

TS EAMCET-2020-11.09.2020

Straight Line

88775 Let $P$ be the point of intersection of the lines $L_{1} \equiv x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0 . A (x_{1}, y_{1})$ and $B (x_{2}, y_{2})$ are points on the lines $l_{1} = 0$ and $l_{2}$ $= 0$ respectively such that $P A = 3 \sqrt{2}, P B =$ $\sqrt{2}, x_{1}, y_{1} \geq 0, x_{2} y_{2} \geq 0$ , then the angle made by the line segment $A B$ at the origin is

1

\frac{π}{4}

2

\frac{π}{2}

3

\cos^{- 1} (\frac{3}{4})

4

\cos^{- 1} (\frac{9}{\sqrt{85}})

Explanation:

(D) : The point of intersection of lines
$L_{1} = x - y - 7 = 0$ and $L_{2} \equiv x + y - 5 = 0$ is $(6, - 1)$
Now, for point $A (x_{1}, y_{1})$ such that $PA = 3 \sqrt{2}$ , take the equation of line $L_{1}$ in symmetric form
$L_{1} : \frac{x - 6}{\cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm 3 \sqrt{2}$
$\Rightarrow x_{1} = 6 \pm 3 and y_{1} = - 1 \pm 3$
$∵ x_{1}, y_{1} \geq 0, so A (x_{1}, y_{1}) is (9, 2)$
Similarly, for point $B (x_{2}, y_{2})$ such that $PB = \sqrt{2}$ take the equation of line $L_{2}$ in symmetric form
$L_{2} : \frac{x - 6}{- \cos \frac{π}{4}} = \frac{y + 1}{\sin \frac{π}{4}} = \pm \sqrt{2}$
$\Rightarrow x_{2} = 6 \mp 1$ and $y_{2} = - 1 \pm 1$
$∵ x_{2}, y_{2} \geq 0$ , so $B (x_{2}, y_{2})$ is $(5, 0)$
$∴$ Slope of line joining origin and point $A$ is $m_{1} = \frac{2}{9}$
and slope of line joining origin and point $B$ is $m_{2} = 0$
$∴ ∠ BOA = θ = \tan^{- 1} (\frac{2 / 9 - 0}{1 + 0}) = \tan^{- 1} (\frac{2}{9})$
$= \cos^{- 1} (\frac{9}{\sqrt{85}})$

TS EAMCET-2020-11.09.2020

Straight Line

88776 A bisector of the angle between the normal's of the planes $4 x + 3 y = 5$ and $x + 2 y + 2 z = 4$ is along the vector

1

(17 \hat{i} + 9 \hat{j} - 12 \hat{k})

2

(17 \hat{i} - 9 \hat{j} + 12 \hat{k})

3

(17 \hat{i} - \hat{j} + 10 \hat{k})

4

(7 \hat{i} - \hat{j} - 10 \hat{k})

Explanation:

(D) : Bisector of the angle between the normal's to the planes $4 x + 3 y - 5 = 0$ and $x + 2 y + 2 z - 4 = 0$ is given by-
$\frac{4 x + 3 y - 5}{\sqrt{(4)^{2} + (3)^{2}}} = \pm \frac{x + 2 y + 2 z - 4}{\sqrt{(1)^{2} + (2)^{2} + (2)^{2}}}$
$\frac{4 x + 3 y - 5}{5} = \pm \frac{x + 2 y + 2 z - 4}{3}$
(i) Taking + ve sign, we get-
$3 (4 x + 3 y - 5) = 5 (x + 2 y + 2 z - 4)$
$7 x - y - 10 z + 5 = 0 \Rightarrow 7 \hat{i} - \hat{j} - 10 \hat{k}$
(ii) -ve sign will gives-
$3 (4 x + 3 y - 5) = 5 (- x - 2 y - 2 z + 4)$
$17 x + 19 y - 35 = 0$
Hence, from case (i) Bisector of the angle between normal of the given plane is represented by the vectors $7 \hat{i} - \hat{j} - 10 \hat{k}$ or $17 \hat{i} + 19 \hat{j}$

TS EAMCET-2022-19.07.2022

Straight Line

88777 If the vertices of a triangle $ABC$ are $A (1, 7), B$ $(- 5, - 1)$ and $C (7, 4)$ , then the equation of a bisector of $△ ABC$ is

1

7 x - 9 y + 26 = 0

2

9 x - 7 y + 38 = 0

3

7 x + 9 y + 44 = 0

4

9 x + 7 y + 52 = 0

Explanation:

(A):
$AB = \sqrt{(- 5 - 1)^{2} + (- 1 - 7)^{2}}$
$= \sqrt{36 + 64}$
$AB = 10$
$BC = \sqrt{(7 + 5)^{2} + (4 + 1)^{2}}$
$= \sqrt{144 + 25}$
$= \sqrt{169}$
$BC = 13$
$BD$ divide the line $AC$ in ratio $10 : 13$
$Point D = (\frac{10 \times 7 + 13 \times 1}{23}, \frac{10 \times 4 + 13 \times 7}{23})$
$D = (\frac{83}{23}, \frac{131}{23})$
Hence the equation of bisector
$y + 1 = \frac{- 1 - \frac{131}{23}}{- 5 - \frac{83}{23}} (x + 5)$
$\Rightarrow y + 1 = \frac{- 23 - 131}{- 115 - 83} (x + 5)$
$\Rightarrow y + 1 = \frac{- 154}{- 198} (x + 5)$
$\Rightarrow 9 (y + 1) = 7 (x + 5)$
$\Rightarrow 9 y + 9 = 7 x + 35$
$\Rightarrow 7 x - 9 y + 26 = 0$

TS EAMCET-2021-04.08.2021

Question Bank Series

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