Maths Select The Correct Option (MCQ)

$\text { Let } A\left(x_1, y_1\right)=A(-5,3) \text { and } B\left(x_2, y_2\right)=B(3,-5) \text {, }$

$a: b=1: 3$

$\therefore x_1=-5, y_1=3, x_2=3, y_2=-5, a=1, b=3 .$

$\therefore$ By section formula,

$\begin{array}{l|l}

\therefore \mathrm{x}=\frac{a x_2+b x_1}{a+b} & \therefore \mathrm{y}=\frac{a y_2+b y_1}{a+b} \\

\therefore \mathrm{x}=\frac{1(3)+3(-5)}{1+3} & \therefore \mathrm{y}=\frac{1(-5)+3(3)}{1+3} \\

\therefore \mathrm{x}=\frac{3-15}{4} & \therefore \mathrm{y}=\frac{-5+9}{4} \\

\therefore \mathrm{x}=\frac{-12}{4} & \therefore \mathrm{y}=\frac{4}{4} \\

\therefore \mathrm{x}=-3 & \therefore \mathrm{y}=1

\end{array}$

$\therefore$ Co-ordinates of $P$ are $(-3,1)$.

$\sin \theta=\frac{1}{2}$

$\therefore \theta=30^{\circ} \quad \ldots\left[\sin 30^{\circ}=\frac{1}{2}\right]$

$A\left(x_1, y_1\right)=A(-4,2), B\left(x_2, y_2\right)=B(6,2)$

Here, $x_1=-4, y_1=2, x_2=6, y_2=2$

$\therefore$ Co-ordinates of the midpoint of seg $A B$

$=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$=\left(\frac{-4+6}{2}, \frac{2+2}{2}\right)$

$=\left(\frac{2}{2}, \frac{4}{2}\right)$

$=(1,2)$

In $\triangle A B C$ and $\triangle L M N$,

$\triangle A B C \sim \triangle L M N$

$\therefore \angle A \cong \angle L \quad \ldots$ (Corresponding angles of similar triangles)

But $\angle A=60^{\circ} \quad \ldots$ (Given)

$\therefore \angle L=60^{\circ}$.

Since, seg $A B \| Y$-axis.

$\therefore \mathrm{x}$ co-ordinate of all points on seg $\mathrm{AB}$ will be the same. $x$ co-ordinate of $A(1,3)=1$

$x$ co-ordinate of $B(1,-3)=1$

$\therefore(1,-3)$ is option correct.

In $\triangle X Y Z$ and $\triangle P Q R$,

$\triangle X Y Z \sim \triangle P Q R \quad \ldots \text { (Given) }$

$\frac{X Y}{P Q}=\frac{Y Z}{Q R}=\frac{X Z}{P R} \quad$...[Corresponding sides of similar triangles. $]$

$\sin ^2 \theta+\sin ^2(90-\theta)=\sin ^2 \theta+\cos ^2 \theta=1$.

Let $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)=B(-1,-1)$

$P(x, y)=P(1,1)$ divides $A B$ in ratio $5: 2$.

$\therefore \mathrm{x}_1=1, \mathrm{y}_1=1, \mathrm{x}_2=-1, \mathrm{y}_2=-1, \mathrm{a}=5, \mathrm{~b}=2 \text {. }$

$\therefore$ By section formula,

$\therefore \mathrm{x}=\frac{a x_2+b x_1}{a+b}$

$\therefore 1=\frac{5(-1)+2 x_1}{5+2}$

$\therefore 7=-5+2 x_1$

$\therefore 7=-5+2 x_1$

$\therefore 2 \mathrm{x}_1=7+5$

$\therefore 2 \mathrm{x}_1=12$

$\therefore \mathrm{x}_1=\frac{12}{2}$

$\therefore \mathrm{x}_1=6$

$\therefore$ Co-ordinates of A are $(6,6)$.

$1+\tan ^2 \theta=\sec ^2 \theta$

$\because \sec ^2 \theta-\tan ^2 \theta=1 .$

Let $A\left(x_1, y_1\right)=A(-3,4)$ and $O\left(x_2, y_2\right)=O(0,0)$

Here, $x_1=-3, y_1=4, x_2=0, y_2=0$

By distance formula,

$d(A, O)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\therefore d(A, O)=\sqrt{[0-(-3)]^2+(0-4)^2}$

$\therefore d(A, O)=\sqrt{9+16}$

$\therefore d(A, O)=\sqrt{25}$

$\therefore d(A, O)=5 \mathrm{~cm}$

In $\triangle \mathrm{ABC}$ and $\triangle \mathrm{DEF}$,

$\triangle \mathrm{ABC} \sim \triangle \mathrm{DEF}$...(Given)

$\therefore \angle B \cong \angle \mathrm{E}$ ...(Corresponding angles of similar triangles)

But $\angle \mathrm{E}=35^{\circ}$...(Given)

$\therefore \angle B=35^{\circ} \text {. }$

$\cot \theta \cdot \tan \theta=\frac{1}{\tan \theta} \cdot \tan \theta=1$

Here, $x_1=x_1 y_1=7, x_2=1, y_2=15$

By distance formula,

$d(L, M)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\therefore d(L, M)=\sqrt{(1-x)^2+(15-7)^2}$

$\therefore 10=\sqrt{(1-x)^2+8^2}$

$\therefore 100=(1-x)^2+64$ $\quad$ $\ldots$[ Squaring both sides ]

$\therefore(1-x)^2=100-64$

$\therefore(1-x)^2=36$

$\therefore 1-x= \pm \sqrt{36}$ $\quad$ $\ldots$[ Taking square root of both sides ]

$\therefore 1-x= \pm 6$

$\therefore 1-x=6 \text { or } 1-x=-6$

$\therefore x=-5 \text { or } x=7$ $\quad$ $\ldots$[ Squaring both sides ]

$\therefore$ The value of $x$ is -5 or 7 .

Let $\triangle A B C$ and $\triangle P Q R$ be two similar triangles.

According to the given condition,

$\frac{\mathrm{A}(\triangle \mathrm{ABC})}{\mathrm{A}(\triangle \mathrm{PQR})}=\frac{9}{25}$

But $\frac{A(\triangle A B C)}{A(\triangle P Q R)}=\frac{A B^2}{P Q^2} \ldots(B y$ the theorem of areas of similar triangles)

$\therefore \frac{A B^2}{P Q^2}=\frac{9}{25}$

$\therefore \frac{A B}{P Q}=\frac{3}{5}$

$\therefore 3: 5$ is the ratio of their corresponding sides.

Let $P\left(x_1, y_1\right)=P(-1,1)$ and $Q\left(x_2, y_2\right)=Q(5,-7)$

Here, $x_1=-1, y_1=1, x_2=5, y_2=-7$

By distance formula,

$d(P, Q)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\therefore d(P, Q)=\sqrt{[5-(-1)]^2+(-7-1)^2}$

$\therefore d(P, Q)=\sqrt{36+64}$

$\therefore d(P, Q)=\sqrt{100}$

$\therefore d(P, Q)=10 \mathrm{~cm}$

Here, $15^2=225$

$6^2+14^2=36+196=232$

$\therefore 15^2 \neq 6^2+14^2$

The square of the largest number is not equal to the sum of the squares of the other two numbers.

$\therefore(6,14,15)$ is not a Pythagoras triplet.

Let $P\left(x_1, y_1\right)=P(2,2)$ and $Q\left(x_2, y_2\right)=Q(5, x)$

Here, $x_1=2, y_1=2, x_2=5, y_2=x$

By distance formula,

$d(P, Q)=\sqrt{\left(x_2-x_1\right)^2+\left(y_2-y_1\right)^2}$

$\therefore 5=\sqrt{(5-2)^2+(x-2)^2}$

$\therefore 5=\sqrt{9+x^2-4 x+4}$

$\therefore 5^2=x^2-4 x+13 \quad \ldots[\text { Squaring both sides] }$

$\therefore 25=x^2-4 x+13$

$\therefore x^2-4 x+13-25=0$

$\therefore x^2-4 x-12=0$

$\therefore(x-6)(x+2)=0$

$\therefore x-6=0 \text { or } x+2=0$

$\therefore x=6 \text { or } x=-2$ $\quad$[..Squaring both sides]

Here, $15^2=225$

$7^2+8^2=49+64=113$

$\therefore 15^2 \neq 7^2+8^2$

The square of the largest number is not equal to the sum of the squares of the other two numbers.

$\therefore(7,8,15)$ is not a Pythagoras triplet.

$\cos \theta \cdot \sec \theta=\cos \theta \cdot \frac{1}{\cos \theta}=1$.

$A\left(x_1, y_1\right)=A(-4,2), B\left(x_2, y_2\right)=B(6,2)$

Here, $x_1=-4, y_1=2, x_2=6, y_2=2$

$\therefore$ Co-ordinates of the midpoint of seg $A B$

$=\left(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}\right)$

$=\left(\frac{-4+6}{2}, \frac{2+2}{2}\right)$

$=\left(\frac{2}{2}, \frac{4}{2}\right)$

$=(1,2)$