2 Marks Questions

Question 12 Marks

Prove that the line through the point $\left(x_1, y_1\right)$ and parallel to line $A x+B y+C=0$ is $A\left(x-x_1\right)+B(y-$ $\left.y_1\right)=0$.

Answer

The given line is
$A x+B y+C=0\ldots(i)$
⇒ $y=-\frac{A}{B} x-\frac{C}{A}$
$\therefore$ Slope, $m=-\frac{A}{B}$
Hence, equation of the line parallel to (i) and passing through $\left(x_1, y_1\right)$ is
$y-y_1=\frac{-A}{B}\left(x-x_1\right)$
⇒$A\left(x-x_1\right)+B\left(y-y_1\right)=0$.

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Question 22 Marks

Find the equation of the line passing through $(1,2)$ and parallel to the line $y=3 x-1$.

Answer

Given line : $y=3 x-1$
i.e., $3 x-y-1=0$
Any line parallel to it is written as
$3 x-y+\lambda=0$
Since, this line passes through $(1,2)$, we get
$3(1)-2+\lambda=0$
$\Rightarrow \quad \lambda=-1$
Hence, $3 x-y-1$ is the equation of required line, which is the equation of given line itself.

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Question 32 Marks

Reduce the equation $\sqrt{3} x+y=4$ into normal form and find the values of $p$ and $\alpha$.

Answer

The given equation is
$\sqrt{3} x+y=4$, where $A=\sqrt{3}, B=1, C=-4$
$\therefore \frac{\sqrt{3} x}{\sqrt{(\sqrt{3})^2+1^2}}+\frac{y}{\sqrt{(\sqrt{3})^2+1^2}}=\frac{4}{\sqrt{(\sqrt{3})^2+1^2}}$,
$\Rightarrow\frac{\sqrt{3}}{2} x+\frac{1}{2} y=\frac{4}{2}$
$\Rightarrow x \cos 30^{\circ}+y \sin 30^{\circ}=2$, is the required equation of line in normal form where $p=2, \alpha=30^{\circ}$.

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Question 42 Marks

Reduce the equation $3 x+4 y=5$ into slope intercept form and find its slope.

Answer

Given equation is
$3 x+4 y=5$
Here, slope, $m=-\frac{A}{B}=-\frac{3}{4}$ and the slope intercept form is
$y=-\frac{A}{B} x-\frac{C}{B}$
⇒$y=-\frac{3}{4} x-\frac{5}{4}$
Slop of line $=-\frac{3}{4}$.

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Question 52 Marks

Find the acute angle between the lines $x+y=0$ and $y=0$.

Answer

The given lines are
$x+y=0 \text { and } y=0$
Here, $A_1=1, B_1=1, A_2=0, B_2=1$.
Angle ' $\theta$ ' between the lines is given by
$\begin{aligned} \theta & =\tan ^{-1}\left|\frac{A_1 B_2-A_2 B_1}{A_1 A_2+B_1 B_2}\right| \\ & =\tan ^{-1}\left|\frac{1-0}{0+1}\right| \\ & =\tan ^{-1}(1) \\ & =45^{\circ} .\end{aligned}$

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Question 62 Marks

Find the normal form of the line $x+y=2$.

Answer

Given equation of the line is
$x+y=2$
$\Rightarrow$ $\frac{x}{\sqrt{2}}+\frac{y}{\sqrt{2}}=\sqrt{2}$
$\Rightarrow$ $x\left(\frac{1}{\sqrt{2}}\right)+y\left(\frac{1}{\sqrt{2}}\right)=\sqrt{2}$
$\Rightarrow$ $x \cos 45^{\circ}+y \sin 45^{\circ}=\sqrt{2}\quad$ $\left[\because \sin 45^{\circ}=\cos 45^{\circ}=\frac{1}{\sqrt{2}}\right]$
Hence, the required normal form of the given line is $x \cos 45^{\circ}+y \sin 45^{\circ}=\sqrt{2}.$

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Question 72 Marks

A line passes through $P(1,2)$ such that its intercept between the axes is bisected at $P$. Find the equation of line.

Answer

Equation of a line in intercept form is
$\frac{x}{a}+\frac{y}{b}=1$
Given, $ \frac{a+0}{2}=1$ and $\frac{0+b}{2}=2$
$\Rightarrow$ $a=2$ and $b=4$
$\therefore$ $\frac{x}{2}+\frac{y}{4}=1$
$\Rightarrow 2 x+y-4=0$, is the required equation of the line.

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Question 82 Marks

Find the equation of the line for which $p=2$, $\sin \alpha=\frac{4}{5}$.

Answer

Equation of a line in normal form is
$x \cos \alpha+y \sin \alpha=p$
Given, $ p=2, \sin \alpha=\frac{4}{5}$
$\Rightarrow$ $\cos \alpha=\sqrt{1-\left(\frac{4}{5}\right)^2}=\frac{3}{5}\quad$$\left[\because \sin ^2 \theta+\cos ^2 \theta=1\right]$
$\Rightarrow$ $x \cdot \frac{3}{5}+y \cdot \frac{4}{5}=2$
$\Rightarrow 3 x+4 y-10=0$, is the required equation of the line.

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Question 92 Marks

Find the equation of a line, which is equidistant from y = - 3 and y = 13.

Answer

The given lines are y = - 3 and y = 13.
So, the resulting line is also of the form y = a,
where $a=\frac{-3+13}{2}=5$
$\therefore$ The resulting line is y = 5.

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Question 102 Marks

By using slope method, show that the points P(4,8) Q(5, 12) and R(9, 28) are collinear.

Answer

The given points are $P\left(x_1, y_1\right)=(4,8), Q\left(x_2, y_2\right)=$ $(5,12)$ and $R\left(x_3, y_3\right)=(9,28)$.
$\therefore$ Slope of line $P Q, \quad m_1=\frac{12-8}{5-4}=4$
and Slope of line PR, $m_2=\frac{28-8}{9-4}=4$
Since, slope of PQ = Slope of PR.
$\therefore$ The points P, Q and R are collinear.

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Question 112 Marks

A line passes through $\left(x_1, y_1\right)$ and $(h, k)$. If slope of the line is $m$, show that $\left(k-y_1\right)=m\left(h-x_1\right)$.

Answer

Slope of the line joining $\left(x_1, y_1\right)$ and $(h, k)$ is
$m=\frac{k-y_1}{h-x_1}$
⇒ $m\left(h-x_1\right)=k-y_1$
⇒ $k-y_1=m\left(h-x_1\right)$. Hence Proved.

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Question 122 Marks

Show that the points $(a, 0),(0, b)$ and $(3 a,-2 b)$ are collinear.

Answer

Let the given points be $P\left(x_1, y_1\right)=(a, 0), Q\left(x_2, y_2\right)=$ $(0, b)$ and $R\left(x_3, y_3\right)=(3 a,-2 b)$.
$\therefore$ Area of $\triangle P Q R$
$\begin{array}{l}=\frac{1}{2}\left|x_1\left(y_2-y_3\right)+x_2\left(y_3-y_1\right)+x_3\left(y_1-y_2\right)\right| \\ =\frac{1}{2}|a(b+2 b)+0(-2 b-0)+3 a(0-b)| \\ =\frac{1}{2}|3 a b-3 a b| \\ =0\end{array}$
Thus, points (a, 0), (0, b) and (3a,2b) are collinear

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Question 132 Marks

Find the distance between the points $(3,-2)$ and $(-1,-1)$.

Answer

Let, the point be $A\left(x_1, y_1\right)=(3,-2)$ and $B\left(x_2, y_2\right)=$ $(-1,-1)$.
Then, the distance
$A B=\left|\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}\right|$
$=\left|\sqrt{(3+1)^2+(-2+1)^2}\right|$
$=\left|\sqrt{4^2+(-1)^2}\right|$
$=\sqrt{17}$ units.

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Applied Maths 2 Marks Questions

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