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Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

If the line $y = mx$ bisects the angle between the line $ax^2 + 2h\ xy + by^2 = 0$ then $m$ is a root of the quadratic equation :

  1. $hx^2 + (a - b)x - h = 0$
  2. $x^2 +h(a - b)x - 1 = 0$
  3. $(a - b)x^2 + hx - (a - b) = 0$
  4. $(a - b)x^2 - hx - (a - b) = 0$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Equation of bisectors of the pair of straight lines $ax^2+2hxy+by^2=0$ is

$h(x^2-y^2)-(a-b)xy=0$......(1).

Since $y=mx $ is given to be the bisector of the pair of straight lines, then the line will satisfy the equation (1).

Then we get,
$h(1-m^2)-(a-b)m=0$

$hm^2+(a-b)m-h=0$.

So $m$ satisfies the equation $hx^2+(a-b)x-h=0$.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

Joint equation of perpendicular lines passing through $(0,0)$ one of which is parallel to $6x-4y+3=0$ is

  1. $6x^{2}-5xy-6y^{2}=0$
  2. $6x^{2}+5xy-6y^{2}=0$
  3. $5x^{2}+5xy-6y^{2}=0$
  4. $6x^{2}-5xy-5y^{2}=0$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The line parallel to 6x - 4y + 3 = 0 can be written as 3x - 2y = 0. Its perpendicular line passing through the origin has the slope negative reciprocal, giving 2x + 3y = 0. The joint equation of the line and its perpendicular is found by multiplying their linear equations, resulting in (3x - 2y)(2x + 3y) = 6x^2 + 5xy - 6y^2 = 0.

Multiple choice maths pair of straight lines bisection of angle pair of bisectors of angles bisector of angle between lines

The equation of the bisector of the obtuse angle between the lines 3x-4y+7=0 and 12x+5y-2=0 is: 

  1. 21 x+77y-101=0

  2. 21 x+77 y+101=0

  3. 21x-77y-101=0

  4. 21x-77y+101=0
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

To find the bisector of the obtuse angle, calculate the bisectors using the formula (3x - 4y + 7)/5 = +/- (12x + 5y - 2)/13. Test which one corresponds to the obtuse angle by checking the sign of the expression.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If $A+B=\dfrac{\pi}{3}$ and $\cos{A}+\cos{B}=1$, then which of the following is true

  1. $\cos{\left(A-B\right)}=\dfrac{1}{3}$
  2. $\left|\cos{A}-\cos{B}\right|=\sqrt{\dfrac{2}{3}}$
  3. $\cos{\left(A-B\right)}=-\dfrac{1}{3}$
  4. $\left|\cos{A}-\cos{B}\right|=\dfrac{1}{2\sqrt{3}}$
Reveal answer Fill a bubble to check yourself
B,C Correct answer
Explanation

$\cos{A}+\cos{B}=1$


$\Rightarrow 2\cos{\left(\dfrac{A+B}{2}\right)}\cos{\left(\dfrac{A-B}{2}\right)}=1$

Since $A+B=\dfrac{\pi}{3}\Rightarrow \dfrac{A+B}{2}=\dfrac{\pi}{6}$
Hence $\cos{\left(\dfrac{A+B}{3}\right)}=\cos{\left(\dfrac{\pi}{6}\right)}=\dfrac{\sqrt{3}}{2}$

$\Rightarrow 2\cos{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{\dfrac{\sqrt{3}}{2}}$

$\Rightarrow \cos{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{\sqrt{3}}$

Squaring both sides, we get

${\cos}^{2}{\left(\dfrac{A-B}{2}\right)}=\dfrac{1}{3}$

$\Rightarrow 2{\cos}^{2}{\left(\dfrac{A-B}{2}\right)}=\dfrac{2}{3}$

$\Rightarrow 2{\cos}^{2}{\left(\dfrac{A-B}{2}\right)}-1=\dfrac{2}{3}-1=\cos {(A-B)}=\dfrac{-1}{3}$

$\left|\cos{A}-\cos{B}\right|=2\sin{\left(\dfrac{A+B}{2}\right)}\sin{\left(\dfrac{B-A}{2}\right)}$

                    $=2\times\dfrac{1}{2}\sqrt{1-\dfrac{1}{3}}$

                    $=\sqrt{\dfrac{2}{3}}$ (on simplification)

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

If $R$ is the radius of circumscribing circle of a regular polygon of $n$ sides, then $R =?$

  1. $\dfrac{a}{2} sin (\dfrac{\pi}{n})$
  2. $\dfrac{a}{2} cos (\dfrac{\pi}{n})$
  3. $\dfrac{a}{2} cosec (\dfrac{\pi}{n})$
  4. $\dfrac{a}{2} cosec (\dfrac{\pi}{2n})$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
since, it is a regular polygon so its interior angle will be equal  

Hence, $nA=\pi\Rightarrow A=\dfrac{\pi}{n}$

and we know that 
$\dfrac{a}{sinA}=2R\Rightarrow R=\dfrac{a}{2}cosec(\dfrac{\pi}{n})$

therefore,Answer is $C$
Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

Two consecutive vertices of a regular hexagon $A _1A _2A _3A _4A _5A _6$ are $A _1\equiv (1, 0), A _2\equiv (3, 0)$. If the centre of hexagon lies above the x-axis, then equation of the circumcircle of the hexagon is?

  1. $x^2+y^2-4x-2\sqrt{3}y+\dfrac{17}{3}=0$
  2. $x^2+y^2-4x-2\sqrt{3}y+\dfrac{25}{3}=0$
  3. $x^2+y^2-4x-2\sqrt{3}y+3=0$

  4. None of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The side length is 2. The center of the hexagon is found by rotating the vector (A2-A1) by 60 degrees. With A1=(1,0) and A2=(3,0), the center is (2, sqrt(3)). The radius is 2. The circle equation is (x-2)^2 + (y-sqrt(3))^2 = 4, which simplifies to x^2 + y^2 - 4x - 2sqrt(3)y + 17/3 = 0.

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons
A polygon has $n$ sides. If all the sides and all the angles are same then this polygon is called a regular polygon. Let ${A} _{1},{A} _{2},{A} _{3},...{A} _{n}$ be a regular polygon of $n$ sides. Let $R$ be the radius of the circumscribed circle of a regular polygon and $r$ be the radius of the inscribed circle of a regular polygon.
If ${A} _{1}{A} _{2}={A} _{2}{A} _{3}={A} _{3}{A} _{4}=...={A} _{n}{A} _{1}=a$

Based on the above information, answer the question:

The area of a regular polygon of $n$ sides is

  1. $\dfrac{n{R}^{2}}{2}\sin{\left(\dfrac{2\pi}{n}\right)}$
  2. $n{R}^{2}\tan{\left(\dfrac{\pi}{n}\right)}$
  3. $\dfrac{n{r}^{2}}{2}\sin{\left(\dfrac{2\pi}{n}\right)}$
  4. $n{r}^{2}\tan{\left(\dfrac{\pi}{n}\right)}$
Reveal answer Fill a bubble to check yourself
A,D Correct answer
Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

Let ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ be a regular hexagon inscribed in a circle of unit radius.Then the product of the length of  ${A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}$ is

  1. $\dfrac{3}{4}$
  2. $3\sqrt{3}$
  3. $3$
  4. $\dfrac{3\sqrt{3}}{2}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given ${A} _{0}{A} _{1}{A} _{2}{A} _{3}{A} _{4}{A} _{5}$ is  a regular hexagon inscribed in a circle of unit radius
$\Rightarrow {A} _{0}{A} _{1}=1$
$\Rightarrow {A} _{0}{A} _{2}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\Rightarrow {A} _{0}{A} _{4}=2\sin{{60}^{0}}=2\times\dfrac{\sqrt{3}}{2}=\sqrt{3}$
$\therefore {A} _{0}{A} _{1}.{A} _{0}{A} _{2}.{A} _{0}{A} _{4}=1\times \sqrt{3}\times\sqrt{3}=3$

Multiple choice maths understanding 3d and 2d shapes defining regular polygons sum of exterior angles of a polygon regular polygons

The area of a regular polygon of n sides is (where r is inradius, R is circumradius, and a is side of the triangle)

  1. $\displaystyle \frac{nR^{2}}{2}\sin \left ( \frac{2\pi }{n} \right )$
  2. $\displaystyle nr^{2}\tan \left( \frac{\pi }{n} \right )$
  3. $\displaystyle \frac{na^{2}}{4}\cot \frac{\pi }{n} $
  4. $\displaystyle nR^{2}\tan(\frac {\pi}{n})$
Reveal answer Fill a bubble to check yourself
A,B,C Correct answer
Explanation

Area of the regular polygon will be 
$=\dfrac{nR^{2}}{2}sin(\dfrac{2\pi}{n})$.
Now 
$R=\dfrac{s}{2sin(\dfrac{\pi}{n})}$
Hence
$A=\dfrac{ns^{2}}{8sin^{2}\dfrac{\pi}{n}}.2sin(\dfrac{\pi}{n}).cos(\dfrac{\pi}{n})$

$=\dfrac{ns^{2}}{4}.cot(\dfrac{\pi}{n})$. where s is the side of the polygon.

$=nr^{2}.tan(\dfrac{\pi}{n})$ where r is the incentre.