Bisectors of Angle between Two Lines

Q. The vertices of a triangle are $A(-1,-7),B(5,1),$and $C(1,4).$ The equation of the bisector of $\angle ABC$ is ___

1. x-7y+2=0
2. 7x+y-2=0
3. 2y-7x-5=0
4. 2x+7y+5=0

Q. Lines $L_1:y-x=0$ and $L_2:2x+y=0$ intersect the line $L_3:y+2=0$ at $P$ and $Q$, respectively. The bisector of the acute angle between $L_1$ and $L_2$ intersects $L_3$ at R.

STATEMENT-1 : The ratio $PR:RQ$ equals $2\sqrt{2}:\sqrt{5}.$
because

STATEMENT-2: In any triangle, bisector of an angle divides the triangle into two similar triangles.

1. Statement-1 is True, Statement-2 is True; Statement-2 is NOT a correct explanation for Statement-1
2. Statement- 1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement-1
3. Statement-1 is True, Statement -2 is False
   
4. Statement-1 is False, Statement-2 is True.
   

Q. 4x-2y+6z=8;x+y-3z=-1;15x-3y+9z=21 solve by row echelon method

Q. $A(1,3)$ and $C(-2/5,-2/5)$ are the vertices of a triangle $ABC$ and the equation of the internal angle bisector of $\angle ABC$ is $x+y=2$, then

1. $BC:7x-3y+4=0$
2. $BC:7x+3y+4=0$
3. $B\equiv (5/2,9/2)$
4. $B\equiv (-5/2,9/2)$

Q. The sides of a rhombus $ABCD$ are parallel to the lines, $x-y+2=0$ and $7x-y+3=0$. If the diagonals of the rhombus intersect at $P(1,2)$ and the vertex $A$ (different from the origin) is on the $y-$axis, then the ordinate of $A$ is

1. $\frac{7}{4}$
2. $\frac{7}{2}$
3. $2$
4. $\frac{5}{2}$

Q. The equation of the bisector of the angle between the lines $2x+y-6=0$ and $2x-4y+7=0$ which contains the point $(1,2)$, is

1. $2x+6y-19=0$
2. $6x-2y-5=0$
3. $6x+2y-5=0$
4. $2x+6y+19=0$

Q.

If xyz=1 , x+1/z=5 , y+1/x=29 , then find z+1/y=

Q. Let $ABCD$ be a square of side length $1.$ Let $P,Q,R,S$ be points in the interiors of the sides $AD,BC,AB,CD,$ respectively, such that $PQ$ and $RS$ intersect at right angles. If $PQ=\frac{3\sqrt{3}}{4}$ then $RS$ equals

1. $\frac{2}{\sqrt{3}}$
2. $\frac{3\sqrt{3}}{4}$
3. $\frac{\sqrt{2}+1}{2}$
4. $4-2\sqrt{2}$

Q.

Find the position vector of a point R which divides the line joining two points P and Q whose position vectors are $\hat{i}+2\hat{j}-\hat{k}$ and $-\hat{i}+\hat{j}+\hat{k}$ respectively, in the ratio 2:1
externally.

Q. The acute angle bisector between the intersecting lines $\frac{x-1}{1}=\frac{y}{-1}=\frac{z-1}{3}$ and $\frac{x-1}{-3}=\frac{y}{-1}=\frac{z-1}{1}$ is

1. $\frac{x-1}{-1}=\frac{y}{1}=\frac{z-1}{2}$
2. $\frac{x-1}{-1}=\frac{y}{-1}=\frac{z-1}{2}$
3. $\frac{x-1}{2}=\frac{z-1}{1},y=0$
4. $\frac{x-1}{-2}=\frac{y}{-1}=\frac{z-1}{1}$

Q. Given $L_1$ = x-2y+11 = 0 and $L_2$ = 3x+6y+5 = 0

1. Equation of acute angle bisector of is y = .

2. Equation of acute angle bisector of is y = - .

3. Equation of obtuse angle bisector of is x = .

4. Equation of obtuse angle bisector of is x = - .

Q. The equation of the bisector of obtuse angle formed by the lines 3x+4y-11=0 and 12x-5y-2=0 will be .

1. 11x-3y-17=0
2. 11x+3y-17=0
3. 3x+11y-19=0
4. 3x-11y+19=0

Q.

Find X and Y, if

(i) and

(ii) and

Q. The image of a point $(3t+1,1-4t),\forall t\in R-\left\{0\right\}$ in a line, lies on $3x-4y+1=0.$ Then the slope of line(s) is (are)

1. $7$
2. $-7$
3. $\frac{1}{7}$
4. $-\frac{1}{7}$

Q. The equation of the bisector of the acute angle between $4x+3y-6=0$ and $5x+12y+9=0$ is

1. $9x+7y-3=0$
2. $9x-7y+41=0$
3. $9x-7y-41=0$
4. $7x+9y-3=0$

Q. The equations of three lines are given by :
15x-8y+1=0 , 12x+5y-3=0 and 21x-y-2=0 .
Show that third line bisects angle between other two lines.

Q. Find the class interval for ${10}^{th}$ class.

1. $3$
2. $4$
3. $5$
4. $7$

Q. The equation of straight line belongs to the family of lines $(4x-3y)+\lambda (x+2y-11)=0$ and parallel to the acute angle bisector between the lines $x-\sqrt{3}y+11=0$ and $2\sqrt{3}x-2y+7=0$ is

1. $x-y+7=0$
2. $x-y+1=0$
3. $3x+4y-25=0$
4. $3x+4y+7=0$

Q.

For the straight lines 4x + 3y – 6 = 0 and 5x + 12y + 9 = 0, the equation of the bisector of the obtuse angle between them is

1. 9x – 7y – 41 = 0

2. 7x + 9y – 3 = 0

3. 7x + 9y – 3 = 0

4. None of these

Q. Solve the following system of equations by matrix method:
(i) 5x + 7y + 2 = 0
4x + 6y + 3 = 0

(ii) 5x + 2y = 3
3x + 2y = 5

(iii) 3x + 4y − 5 = 0
x − y + 3 = 0

(iv) 3x + y = 19
3x − y = 23

(v) 3x + 7y = 4
x + 2y = −1

(vi) 3x + y = 23
5x + 3y = 12

Q. A ray of light coming from the point $P\equiv (2,3)$ is reflected at a point $Q$ on the $x$- axis and then passes through the point $R\equiv (4,5)$. The equation of $PQ$ is

1. $x+y=5$
   
2. $3x-y=3$
3. $4x+y=11$
4. $2x+y=7$

Q. If $A(1,3)$ and $C(-\frac{2}{5},-\frac{2}{5})$ are the vertices of a $△ABC$ and the equation of the internal angle bisector of $\angle ABC$ is $x+y=2$, then

1. the equation of line $BC$ is $7x-3y+4=0$
2. the equation of line $BC$ is $7x+3y+4=0$
3. $B\equiv (\frac{5}{2},\frac{9}{2})$
4. $B\equiv (-\frac{5}{2},\frac{9}{2})$

Q. If $A=\left[\begin{array}{ccc}2& -3& 5\\ 3& 2& -4\\ 1& 1& -2\end{array}\right]$, find A⁻¹ and hence solve the system of linear equations
2x − 3y + 5z = 11, 3x + 2y − 4z = −5, x + y + 2z = −3

Q. Two sides of a rhombus are along the lines $x-y+1=0$ and $7x-y-5=0$.
If its diagonals at $(-1,-2)$, then which one of the following is a vertex of this rhombus?

1. $(-3,-9)$
2. $(-3,-8)$
3. $(\frac{1}{3},-\frac{8}{3})$
4. $(-\frac{10}{3},-\frac{7}{3})$

Q. The equation(s) of the angle bisectors of the lines $3x-4y+7=0$ and $12x-5y-8=0$ is/are

1. $21x+27y+131=0$
2. $99x-77x-51=0$
3. $99x-77x+51=0$
4. $21x+27y-131=0$

Q. Find the equations to the straight lines which pass through the origin and are inclined at ${75}^o$ to the straight line
$(x+y)\sqrt{3}+y-x=a$.

Q. The equation of a straight line passing through the point $(1,2)$ and inclined at ${45}^{\circ }$ to the line $y=2x+1$ is

1. $5x+y=7$
2. $3x+y=5$
3. $x+y=3$
4. $x-y+1=0$

Q. The vertices of a triangle are $A(-1,-7),B(5,1),$and $C(1,4).$ The equation of the bisector of $\angle ABC$ is ___

1. x-7y+2=0
2. 7x+y-2=0
3. 2y-7x-5=0
4. 2x+7y+5=0

Q. Solve the following system of equations by matrix method:
(i) x + y − z = 3
2x + 3y + z = 10
3x − y − 7z = 1

(ii) x + y + z = 3
2x − y + z = − 1
2x + y − 3z = − 9

(iii) 6x − 12y + 25z = 4
4x + 15y − 20z = 3
2x + 18y + 15z = 10

(iv) 3x + 4y + 7z = 14
2x − y + 3z = 4
x + 2y − 3z = 0

(v)$$\begin{gathered} \frac{2}{x}-\frac{3}{y}+\frac{3}{z}=10 \\ \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=10 \\ \frac{3}{x}-\frac{1}{y}+\frac{2}{z}=13 \end{gathered}$$

(vi) 5x + 3y + z = 16
2x + y + 3z = 19
x + 2y + 4z = 25

(vii) 3x + 4y + 2z = 8
2y − 3z = 3
x − 2y + 6z = −2

(viii) 2x + y + z = 2
x + 3y − z = 5
3x + y − 2z = 6

(ix) 2x + 6y = 2
3x − z = −8
2x − y + z = −3

(x) x − y + z = 2
2x − y = 0
2y − z = 1

(xi) 8x + 4y + 3z = 18
2x + y +z = 5
x + 2y + z = 5

(xii) x + y + z = 6
x + 2z = 7
3x + y + z = 12

(xiii) $\frac{2}{x}+\frac{3}{y}+\frac{10}{z}=4,\frac{4}{x}-\frac{6}{y}+\frac{5}{z}=1,\frac{6}{x}+\frac{9}{y}-\frac{20}{z}=2;x,y,z\ne 0$

Q. The equations of the bisectors of the lines 3x-4y+7=0 and 12x-5y-8=0 are:

1. 21x+27y-131=0
2. 21x-27y+131=0
3. 99x-77y+51=0
4. 99x+77y-51=0