(Each question 2 marks) - MATHS STD 11 Science Questions - Vidyadip

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55 questions · self-marked practice — reveal the answer and mark yourself.

Question 12 Marks

Find the value of p so that three lines 3x + y - 2 = 0, px + 2y - 3 = 0 and 2x - y - 3 = 0 may intersect at one point.

Answer

The equation of lines are
3x + y - 2 = 0, px + 2y - 3 = 0 and 2x - y - 3 = 0.
We know that three lines are concurrent if
${a_3}({b_1}{c_2} - {b_2}{c_1}) + {b_3}({c_1}{a_2} - {c_2}{a_1})$$ + {c_3}({a_1}{b_2} - {a_2}{b_1}) = 0$
$\therefore 2[1 \times ( - 3) - 2 \times ( - 2)] + ( - 1)[ - 2$$ \times p - ( - 3) \times 3] + ( - 3)[3 \times 2 - p \times 1] = 0$
$ \Rightarrow 2[ - 3 + 4] - 1[ - 2p + 9] - 3[6 - p] = 0$
$ \Rightarrow 2 + 2p - 9 - 18 + 3p = 0$
$ \Rightarrow 5p - 25 = 0 \Rightarrow p = 5$.

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Question 22 Marks

Find the equation of the line parallel to y-axis and drawn through the point of intersection of the lines x - 7y + 5 = 0 and 3x + y = 0

Answer

Point of intersection of lines x - 7y + 5 = 0 and 3x + y = 0 obtained by solving these equations has coordinates.
$\therefore x = \frac{{ - 5}}{{22}}$ and $y = \frac{{15}}{{22}}$.
Since the required line is parallel to y-axis, so the equation of required line is
$x = \frac{{ - 5}}{{22}}$.

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Question 32 Marks

What are the points on the y-axis whose distance from the line $\frac{x}{3} + \frac{y}{4} = 1$ is 4 units.

Answer

Let point on y-axis be (0, y).
The given equation of line is $\frac{x}{3} + \frac{y}{4} = 1$.
$ \Rightarrow 4x + 3y = 12$ $ \Rightarrow 4x + 3y - 12 = 0$
Now perpendicular distance from point (0, y) to line 4x + 3y - 12 = 0 is
$\left| {\frac{{4 \times 0 + 3y - 12}}{{\sqrt {{{(4)}^2} + {{(3)}^2}} }}} \right| = \left| {\frac{{3y - 12}}{{\sqrt {25} }}} \right| = \left| {\frac{{3y - 12}}{5}} \right|$
It is given that
$\left| {\frac{{3y - 12}}{5}} \right| = 4 \Rightarrow \left| {\frac{{3y - 12}}{5}} \right| = \pm 4$
When $\frac{{3y - 12}}{5} = 4$ $ \Rightarrow 3y - 12 = 20 \Rightarrow y = \frac{{32}}{3}$
When $\frac{{3y - 12}}{5} = - 4$$ \Rightarrow 3y - 12 = - 20 \Rightarrow y = \frac{{ - 8}}{3}$
Thus required points are $\left( {0,\frac{{32}}{3}} \right)$ and $\left( {0,\frac{{ - 8}}{3}} \right)$.

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Question 42 Marks

Find the values of $\theta $ and p, if the equation $x\cos \theta + y\sin \theta = p$ is the normal form of the line $\sqrt 3 x + y + 2 = 0$

Answer

Here $\sqrt 3 x + y + 2 = 0$
$ \Rightarrow \sqrt 3 x + y = - 2 \Rightarrow - \sqrt 3 x - y = 2$
Dividing both sides by $\sqrt {{{( - \sqrt 3 )}^2} + {{( - 1)}^2}} = 2$, we have
$\frac{{ - \sqrt 3 }}{2}x - \frac{1}{2}y = 1$
Put $\cos \alpha = \frac{{ - \sqrt 3 }}{2}$ and $\sin \alpha= \frac{{ - 1}}{2}$
$ \Rightarrow \alpha $ lies in 3rd quadrant
$\therefore \cos \alpha = \frac{{ - \sqrt 3 }}{2} = - \cos 30^\circ $ = cos (180° + 30° )
$ \Rightarrow \alpha = 210^\circ $
$\therefore $ Equation of line in normal form is
$x\cos \frac{{7\pi }}{6} + y\sin \frac{{7\pi }}{6} = 1$
Comparing it with x cos $\alpha + y\ sin \alpha = p$, we have
$\alpha = \frac{{7\pi }}{6}$ and p = 1

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Question 52 Marks

Find the angles between the lines $\sqrt 3 x + y = 1$ and $x + \sqrt 3 y = 1$

Answer

We have $\sqrt 3 x + y = 1$
$\Rightarrow y = - \sqrt 3 x + 1$
$\therefore {m_1} = - \sqrt 3$
Also $x + \sqrt 3 y = 1$
$\Rightarrow \sqrt 3 y = - x + 1$
$\Rightarrow y = \frac{{ - 1}}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}$
$\therefore {m_2} = \frac{{ - 1}}{{\sqrt 3 }}$
Let $\theta$ be the angle between the lines. Then,
$\tan \theta = \left| {\frac{{ - \sqrt 3 + \frac{1}{{\sqrt 3 }}}}{{1 + ( - \sqrt 3 )\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)}}} \right| = \left| {\frac{{\frac{{ - 3 + 1}}{{\sqrt 3 }}}}{{1 + 1}}} \right|$$= \left| {\frac{{ - 2}}{{\sqrt 3 }} \times \frac{1}{2}} \right| = \left| {\frac{{ - 1}}{{\sqrt 3 }}} \right| = \frac{1}{{\sqrt 3 }}$
$\tan \theta = \tan 30^\circ$ and $\tan (180^\circ - 30^\circ )$
$\Rightarrow \theta = 30^\circ$ and 150°

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Question 62 Marks

Find the equation of the line perpendicular to the line x - 7y + 5 = 0 and having x intercept 3.

Answer

Equation of any line which is perpendicular to the line
x - 7y + 5 = 0 is 7x + y + k = 0
Since this line passes through point (3, 0)
$\therefore 7 \times 3 + 0 + k = 0 \Rightarrow $ k = -21
Thus equation of required line is 7x + y - 21 = 0

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Question 72 Marks

Find the distance between parallel lines. l(x + y) + p = 0 and l(x + y) - r = 0

Answer

We have the equation,
$\mathrm{l} x+\mathrm{l} y+\mathrm{p}=0$
$\text { and } \mathrm{l} x+\mathrm{l} y-\mathrm{r}=0$
$\text { where } \mathrm{a}=1, \mathrm{~b}=1, \mathrm{c}_1=\mathrm{p} \text { and } \mathrm{c}_2=-\mathrm{r}$
$\therefore$ The distance between two parallel lines
d = $\frac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
$\Rightarrow \frac{{\left| {p + r} \right|}}{{\sqrt {{1^2} + {1^2}} }}$
$\Rightarrow \frac{1}{{\sqrt 2 }}\left| {\frac{{p + r}}{1}} \right|$ units

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Question 82 Marks

Find the distance between parallel lines. 15x + 8y - 34 = 0 and 15 x + 8y + 31 = 0

Answer

The given equations are
15x + 8y - 34 = 0
and 15 x + 8y + 31 = 0
where $a = 15, b = 8, c_1 = -34$ and $c_2 = 31$
$\therefore$ The distance between two parallel lines
$d = \frac{{\left| {{c_1} - {c_2}} \right|}}{{\sqrt {{a^2} + {b^2}} }}$
$\Rightarrow \frac{{\left| { - 34 - 31} \right|}}{{\sqrt {{{(15)}^2} + {{(8)}^2}} }}$
$\Rightarrow$ $\frac{{65}}{{17}}$ units

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Question 92 Marks

Find the distance of the point (-1, 1) from the line 12 (x + 6) = 5(y - 2).

Answer

Here given line is 12(x + 6) = 5(y - 2)
$\Rightarrow$ 12x + 72 = 5y - 10 $\Rightarrow$ 12x - 5y + 82 = 0
$\therefore$ Perpendicular distance of the point (-1, 1) from the line 12x - 5y + 82 = 0 is
$\left| {\frac{{12( - 1) - 5(1) + 82}}{{\sqrt {{{(12)}^2} + {{( - 5)}^2}} }}} \right| = \left| {\frac{{ - 12 - 5 + 82}}{{\sqrt {144 + 25} }}} \right| = \left| {\frac{{65}}{{13}}} \right|$ = 5 units.

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Question 102 Marks

Reduce the equation x - y = 4 into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive X-axis.

Answer

Here x - y = 4
Dividing both sides by $\sqrt {{{(1)}^2} + {{( - 1)}^2}} = \sqrt 2$ we have
$\frac{x}{{\sqrt 2 }} - \frac{y}{{\sqrt 2 }} = \frac{4}{{\sqrt 2 }} \Rightarrow \frac{1}{{\sqrt 2 }}x - \frac{1}{{\sqrt 2 }}y = 2\sqrt 2$
Put $\cos \alpha = \frac{1}{{\sqrt 2 }}$ and $\sin \alpha = \frac{{ - 1}}{{\sqrt 2 }}$
$\Rightarrow \alpha$ lies in IVth quadrant.
$\therefore \cos \alpha = \frac{1}{{\sqrt 2 }} = \cos \left( {2\pi - \frac{\pi }{4}} \right) \Rightarrow \alpha = \frac{{7\pi }}{4}$
$\therefore$ Equation of line in normal form is
$x\cos \frac{{7\pi }}{4} + y\sin \frac{{7\pi }}{4} = 2\sqrt 2$
Comparing it with $x\cos \alpha + y\sin \alpha = p$, we have
$\alpha = \frac{{7\pi }}{4}$ and $p = 2\sqrt 2$

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Question 112 Marks

Reduce the equation y - 2 = 0 into normal form. Find their perpendicular distance from the origin and angle between perpendicular and the positive X-axis.

Answer

Here y - 2 = 0
$\Rightarrow$ y = 2 $\Rightarrow$ 0x + y = 2
Dividing both sides by $\sqrt {{{(0)}^2} + {{(1)}^2}} = 1$ we have
0x + y = 2
Put $\cos \alpha = 0$ and $\sin \alpha = 1$
$\Rightarrow \alpha = \frac{\pi }{2}$
$\therefore$ Equation of line in normal form is
$x\cos \frac{\pi }{2} + y\sin \frac{\pi }{2} = 2$
Comparing it with $x\cos \alpha + y\sin a = p,$ we have
$\alpha = \frac{\pi }{2}$ and p = 2

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Question 122 Marks

Reduce the equation $x-\sqrt{3} y+8=0$ into normal form. Find the perpendicular distance from the origin and angle between the perpendicular and the positive X-axis.

Answer

Any equation of the form ax + by + c = 0 can be converted into normal form (x cos$\alpha$ + y sin$\alpha$ = p) dividing each term by $\sqrt{a^{2}+b^{2}}$ i.e.,
$\frac{a}{\sqrt{a^{2}+b^{2}}} x+\frac{b}{\sqrt{a^{2}+b^{2}}} y=\frac{c}{\sqrt{a^{2}+b^{2}}}$
Given equation of line is
$x-\sqrt{3} y+8=0$
$\Rightarrow \quad x-\sqrt{3} y=-8$
$\Rightarrow \quad-x+\sqrt{3} y=8$ ...(ii)
Now, divide by $\sqrt{(\text { coefficient of } x)^{2}+(\text { coefficient of } y)^{2}}$ = $\sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2,$ we get
$-\frac{1}{2} x+\frac{\sqrt{3}}{2} y=\frac{8}{2}$
$\Rightarrow-\cos 60^{\circ} x+\sin 60^{\circ} y=4[\because$ convert in form of $x \cos \alpha+y \sin \alpha=p]$
[ $\because \cos x$ is negative and $\sin x$ is positive, it is possible in second quadrant]
$\Rightarrow x \cos \left(180^{\circ}-60^{\circ}\right)+y \sin \left(180^{\circ}-60^{\circ}\right)=4$
$\Rightarrow x \cos 120^{\circ}+y \sin 120^{\circ}=4$
${\left[\because \cos \left(180^{\circ}-\theta\right)=-\sin \theta \text { and } \sin \left(180^{\circ}-\theta\right)=\cos \theta\right]}$
On comparing with $x \cos \alpha+y \sin \alpha=p$, we get
$\alpha=120^{\circ}, p=4$

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Question 132 Marks

Reduce the given equation into the intercept form and find the intercept on the axis. 3 y + 2 = 0

Answer

Here 3 y + 2 = 0
$\Rightarrow 3y = - 2$
$ \Rightarrow \frac{{3y}}{{ - 2}} = 1 \Rightarrow \frac{{0x}}{{ - 2}} + \frac{{3y}}{{ - 2}} = 1 \Rightarrow \frac{{0x}}{{ - 2}} + \frac{y}{{\frac{{ - 2}}{3}}} = 1$
which is required intercept form.
Comparing it with $\frac{x}{a} + \frac{y}{b} = 1$ we have
a = 0 and b = -2/3.

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Question 142 Marks

Reduce the given equation into the intercept form and find the intercept on the axis. 4x - 3y = 6

Answer

Here 4x - 3y = 6
$\Rightarrow \frac{{4x}}{6} - \frac{{3y}}{6} = 1 \Rightarrow \frac{{2x}}{3} - \frac{y}{2} = 1]$$\Rightarrow \frac{x}{{\frac{3}{2}}} + \frac{y}{{ - 2}} = 1$
which is required intercept form,
Comparing it with $\frac{x}{a} + \frac{y}{b} = 1$, we have
$a = \frac{3}{2}$ and b = -2

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Question 152 Marks

Reduce the given equation into the intercept form and find the intercept on the axis.
3x + 2y - 12 = 0

Answer

Here 3x + 2y - 12 = 0
$\Rightarrow$ 3x + 2y = 12
$\Rightarrow \frac{{3x}}{{12}} + \frac{{2y}}{{12}} = 1 \Rightarrow \frac{x}{4} + \frac{y}{6} = 1$
which is required intercept form.
Comparing it with $\frac{x}{a} + \frac{y}{b} = 1$, we have
a = 4 and b = 6

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Question 162 Marks

Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).

Answer

Mid point of the line segment joining the points (3, 4) and (-1, 2) is $\left( {\frac{{3 - 1}}{2},\frac{{4 + 2}}{2}} \right)$ i.e. (1, 3).
Slope of the line joining points (3, 4) and (-1, 2)
$= \frac{{2 - 4}}{{ - 1 - 3}} = \frac{{ - 2}}{{ - 4}} = \frac{1}{2}$
$\therefore$ Slope of the required line is -2.
Thus the required line passes through point (1, 3) having slope -2.
$\therefore$ Equation of required line is y - 3 = -2(x - 1) $\Rightarrow$ y - 3 = -2x + 2 $\Rightarrow$ 2x + y - 5 = 0.

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Question 172 Marks

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

Answer

Equation of the line parallel to line $A x+B y+C=0$ is $A x+B y+K=0 \ldots$ (i)
Since line (i) passes through ( $\mathrm{x}_1, \mathrm{y}_1$ )
$A x_1+B y_1+K=0 \ldots \text { (ii) }$
Subtracting (ii) from (i), we have
$A\left(x-x_1\right)+B\left(y-y_1\right)=0$

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Question 182 Marks

The line through the points (h, 3) and (4, 1) intersects the line 7x - 9y - 19 = 0 at right angle. Find the value of h.

Answer

Slope of the line passing through the points (h, 3) and (4, 1) is
$= \frac{{1 - 3}}{{4 - h}} = \frac{{ - 2}}{{4 - h}}$
Also slope of the line 7x - 9y - 19 = 0 is $\frac{7}{9}$
Since two lines are perpendicular to each other
$\therefore \frac{{ - 2}}{{4 - h}} \times \frac{7}{9} = - 1 \Rightarrow \frac{{ - 14}}{{36 - 9h}} = - 1$ $\Rightarrow$ -14 = -36 + 9h
$\Rightarrow$ 9h = 36 - 14 $\Rightarrow h = \frac{{22}}{9}$

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Question 192 Marks

Find the equation of the line which satisfy the given condition:
Perpendicular distance from the origin is 5 units and the angle made by the perpendicular with the positive X-axis is $30^{\circ}$.

Answer

Here, p = 5 and $\alpha$ = $30^o$

So, equation of line in normal form is
x cos$\alpha$ + y sin$\alpha$ = p
$\therefore x \cos 30^{\circ}+y \sin 30^{\circ}=5$
$\Rightarrow \quad \frac{\sqrt{3}}{2} x+\frac{1}{2} y=5$
$\Rightarrow \quad \sqrt{3} x+y=10$

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Question 202 Marks

Find the equation of the line which satisfy the given condition:
The line passing through the points (-1, 1) and (2, - 4).

Answer

Given points are $A(x_1, y_1) = (-1, 1)$ and $B(x_2, y_2) = (2, -4)$, then equation of line AB is
$y-y_{1}=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}\left(x-x_{1}\right)$
$\Rightarrow \quad y-1=\frac{-4-1}{2+1}(x+1)$$\left[\because x_{1}=-1, y_{1}=1, x_{2}=2, y_{2}=-4\right]$
$\Rightarrow \quad y-1=\frac{-5}{3}(x+1) \Rightarrow 3 y-3=-5 x-5$
$\Rightarrow$ 5x + 3y + 2 = 0

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Question 212 Marks

Find the equation of the line which satisfy the given condition:
The line intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30° with positive direction of the x-axis.

Answer

Here m = tan 30° = $\frac{1}{{\sqrt 3 }}$ and c = 2
Putting these values in y = mx + c, we have
$y = \frac{1}{{\sqrt 3 }}x + 2 \Rightarrow x - \sqrt 3 y + 2\sqrt 3 = 0$

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Question 222 Marks

Find the equation of the line which satisfy the given condition:
The line intersecting the X-axis at a distance of 3 units to the left of origin with slope -2.

Answer

Given, the line intersecting the $X$-axis to the left of the origin. It means it intersect the negative $X$-axis. Clearly, line $A B$ passes through the point $(-3,0)$ and $m=-2$.
Equation of line in point slope from is
$y-y_1=m\left(x-x_1\right) \Rightarrow y-0=-2(x+3)$
$\Rightarrow y=-2 x-6$
$\Rightarrow 2 x+y+6=0$

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Question 232 Marks

Find the equation of the line which satisfy the given conditions:
Passing through $(2,2 \sqrt{3})$ and inclined with the x-axis at an angle of $75^o$.

Answer

It is given that the point $=(2,2 \sqrt{ } 3)$ and angle $\theta=75^{\circ}$
Equation of line: $\left(y-y_1\right)=m\left(x-x_1\right)$
where, $\mathrm{m}=$ slope of line $=\tan \theta$ and $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ are the points through which line passes
$\therefore \mathrm{m}=\tan 75^{\circ}$
$75^{\circ}=45^{\circ}+30^{\circ}$ Applying the formula: $\tan (A+B)=\frac{\tan A+\tan B}{1-\tan A \cdot \tan B}$
$\tan \left(45^{\circ}+30^{\circ}\right)=\frac{\tan 45^{\circ}+\tan 30^{\circ}}{1-\tan 45^{\circ} \cdot \tan 30^{\circ}}=\frac{1+\frac{1}{\sqrt{3}}}{1-\frac{1}{\sqrt{3}}}$
$\tan 75^{\circ}=\frac{\sqrt{3}+1}{\sqrt{3}-1}$
Rationalizing we obtaintan $75^{\circ}=\frac{3+1+2 \sqrt{3}}{3-1}=2+\sqrt{3}$
We know that the point ( $x, y$ ) lies on the line with slope $m$ through the fixed point ( $x_1, y_1$ ), if and only if, its coordinates satisfy the equation $\mathrm{y}-\mathrm{y}_1=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$
$\therefore y-2 \sqrt{3}=(2+\sqrt{3})(x-2)$
$\Rightarrow y-2 \sqrt{3}=2 x-4+\sqrt{3} x-2 \sqrt{3}$
$\Rightarrow y=2 x-4+\sqrt{3} x$
$\Rightarrow(2+\sqrt{3}) x-y-4=0$
Therefore, the equation of the line is $(2+\sqrt{3}) x-y-4=0$.

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Question 242 Marks

Find the equation of the line which satisfy the given condition:
Passing through (0, 0) with slope m.

Answer

Given point $=(0,0)$ and slope $=\mathrm{m}$
We know that the point ( $x, y$ ) lies on the line with slope $m$ through the fixed point ( $x_0, y_0$ ), if and only if, its coordinates satisfy the equation $\mathrm{y}-\mathrm{y}_0=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_0\right)$
$\therefore y-0=m(x-0)$
$\Rightarrow y=m x$
$\Rightarrow y-m x=0$
Therefore, the required equation of the line is $y-m x=0$.

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Question 252 Marks

The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.

Answer

Slope of line $OP = \frac{{9 - 0}}{{ - 2 - 0}} = \frac{{ - 9}}{2}$
Since the required line is perpendicular to OP,

$\therefore$ Slope of required line $ = \frac{2}{9}$
The required line passing through point (-2, 9) having slope $\frac{2}{9}$.
So equation of required line is
$y - 9 = \frac{2}{9}(x + 2) \Rightarrow$ 9y - 81 = 2 x + 4
$\Rightarrow$ 2x - 9y + 85 = 0

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Question 262 Marks

Find the equation of a line that cuts off equal intercepts on the coordinate axis and passes through the point (2, 3).

Answer

Let equal intercepts on the coordinate axis be a and the line passes through point (2, 3).
$\therefore \frac{2}{a} + \frac{3}{a} = 1 \Rightarrow $ a = 5
Thus equation of required line is
$\frac{x}{5} + \frac{y}{5} = 1 \Rightarrow$ x + y = 5

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Question 272 Marks

Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).

Answer

Let A (2, 5) and B(-3, 6) be any two points.
$\therefore$ Slope of AB $= \frac{{6 - 5}}{{ - 3 - 2}} = - \frac{1}{5}$
Since the required line is perpendicular to AB.
$\therefore$ Slope of required line m = 5.
Now the required line passing through point (-3, 5) having slope 5
$\therefore$ y - 5 = 5(x + 3) $\Rightarrow$ y - 5 = 5x + 15
$\Rightarrow$ 5x - y + 20 = 0

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Question 282 Marks

Without using distance formula, show that the points (-2, -1), (4, 0), (3, 3) and (-3, 2) are the vertices of a parallelogram.

Answer

Let A(-2, -1), B (4, 0), C (3, 3) and D(-3, 2) be vertices of a quadrilateral ABCD.
$\therefore$ Slope of AB $ = \frac{{0 - ( - 1)}}{{4 - ( - 2)}} = \frac{1}{6}$
Slope of BC $= \frac{{3 - 0}}{{3 - 4}} = \frac{3}{{ - 1}} = - 3$
Slope of DC $= \frac{{3 - 2}}{{3 - ( - 3)}} = \frac{1}{6}$
Slope of AD $= \frac{{2 - ( - 1)}}{{ - 3 - ( - 2)}} = \frac{3}{{ - 1}} = - 3$
$\therefore$ Slope of AB = Slope of DC $\Rightarrow$ AB|| DC
And Slope of BC = Slope of AD $\Rightarrow$ BC || AD
Thus ABCD is a parallelogram

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Question 292 Marks

Find the value of x for which the points (x, -1)(2, 1) and (4, 5) are collinear.

Answer

Let A(x, -1), B (2, 1) and C (4, 5) be three collinear points.
$\therefore$ Slope of AB $= \frac{{1 - ( - 1)}}{{2 - x}} = \frac{{1 + 1}}{{2 - x}} = \frac{2}{{2 - x}}$
Slope of BC = $\frac{{5 - 1}}{{4 - 2}} = \frac{4}{2} = 2$
Since points A,Band C are collinear therefore slope of AB=slope of BC
2/2-x=2
Hence x=2

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Question 302 Marks

Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

Answer

The line makes an angle of $30^{\circ}$ with the positive direction of $y$-axis.
Now the line makes an angle of $(90+30)^{\circ}=120^{\circ}$ with the positive direction $x$-axis.
$\therefore$ Slope of the line $=\tan 120^{\circ}=\tan (90+30)$
$=-\cot 30^{\circ}=-\sqrt{3}$

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Question 312 Marks

Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and C(-1, -1) are the vertices of a right angled triangle.

Answer

Let A (4, 4), B (3, 5) and C(-1, -1) be three vertices of a $\Delta {\rm A}{\rm B}C$.
$\therefore$ Slope of AB $ = \frac{{5 - 4}}{{3 - 4}} = \frac{1}{{ - 1}} = - 1$
$\therefore$ Slope of BC $= \frac{{ - 1 - 5}}{{ - 1 - 3}} = \frac{{ - 6}}{{ - 4}} = \frac{3}{2}$
$\therefore$ Slope of AC $ = \frac{{ - 1 - 4}}{{ - 1 - 4}} = \frac{{ - 5}}{{ - 5}} = 1$
Now slope of AB $\times$ slope of AC = -1$\times$1 = -1
This shows thatAB$\bot$AC. Thus $\Delta {\rm A}{\rm B}C$ is right angled at point A.

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Question 322 Marks

Find the slope of a line, which passes through the origin and mid-point of the line segment joining the points P(0, - 4) and B(8, 0).

Answer

If two points are given, then slope m = $\frac{y_{2}-y_{1}}{x_{2}-x_{1}}$
Given points are P (0, -4) and Q (8, 0).
$\therefore x_1 = 0, y_1 = -4, x_2 = 8, y_2 = 0$
These points plotted in XY - plane are given below.
Mid-point of PQ is R

$R=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)=(4,-2)$
$\therefore \text { Slope of } O R=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}$
$\left[{\because x_{1}=0, y_{1}=0,} \ {x_{2}=4, y_{2}=-2}\right]$

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Question 332 Marks

Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4).

Answer

Let P(x, 0) be any point on the x-axis which is equidistant from Q(7, 6) and R (3, 4).
Then $PQ = \sqrt {{{(x - 7)}^2} + {{(0 - 6)}^2}}$ $= \sqrt {{x^2} - 14x + 49 + 36}$
$ = \sqrt {{x^2} - 14x + 85}$
$PR = \sqrt {{{(x - 3)}^2} + {{(0 - 4)}^2}} = \sqrt {{x^2} - 6x + 9 + 16}$
$= \sqrt {{x^2} - 6x + 25}$
Since PQ = PR
$\therefore \sqrt {{x^2} - 14x + 85} = \sqrt {{x^2} - 6x + 25}$
Squaring both sides, we have
$x^2 - 14x + 85 = x^2 - 6x + 25$
$\Rightarrow$ -14x + 6x = 25 - 85 $\Rightarrow$ 8x = -60
$\Rightarrow x = \frac{{15}}{2}$
Thus coordinates of point on the x-axis is $\left( {\frac{{15}}{2},0} \right)$.

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Question 342 Marks

Consider the following population and year graph, find the slope of the line AB and using it, find what will be the population in the year 2010?

Answer

The points on the line are A(1985, 92) and B(1995, 97)
$\therefore$ Slope of AB $= \frac{{97 - 92}}{{1995 - 1985}} = \frac{5}{{10}} = \frac{1}{2}$
Let the population in year 2010 be x crores.
Then C (2010, x) lies on the line
$\therefore$ Slope of BC $= \frac{{y - 97}}{{2010 - 1995}} = \frac{{y - 97}}{{15}}$
Since points A, B and C lie on a line.
$\therefore$ Slopes of AB = Slope of BC
$\therefore \frac{1}{2} = \frac{{y - 97}}{{15}} \Rightarrow$ 15 = 2y - 194
$\Rightarrow$ 2y = 209 $\Rightarrow y = \frac{{209}}{2}$= 104.5
Thus population in 2010 will be 104.5 crores.

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Question 352 Marks

If three points (h, 0), (a, b) and (0, k) lies on a line, show that $\frac{a}{h} + \frac{b}{k} = 1$

Answer

Let A(h, 0), B (a, b) and C (0, k) be three points lie on a line.
$\therefore$ Slope of AB $= \frac{{b - 0}}{{a - h}} = \frac{b}{{a - h}}$
Slope of BC $= \frac{{k - b}}{{0 - a}} = \frac{{b - k}}{a}$
Slope of AB = Slope of BC (given)
$\therefore \frac{b}{{a - h}} = \frac{{b - k}}{a} \Rightarrow$ ab = ab - ak - bh + hk
$\Rightarrow$ ak + bh = hk
Dividing both sides by hk
$\frac{{ak}}{{hk}} + \frac{{bh}}{{hk}} = 1 \Rightarrow \frac{a}{h} + \frac{b}{k} = 1$.

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Question 362 Marks

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

Answer

Let $A(x_1, y_1)$ and $B(h, k)$ be two points.
$\therefore$ Slope of AB $= \frac{{k - {y_1}}}{{h - {x_1}}}$
Slope of AB = m (given)
$\therefore m = \frac{{k - {y_1}}}{{h - {x_1}}} \Rightarrow k - {y_1} = m(h - {x_1})$

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Question 372 Marks

Find the angle between the x-axis and the line joining the points (3, -1) and (4, -2).

Answer

Let A(3, -1) and B(4, -2) be two points. Let Q be the angle which AB makes with positive direction of x-axis.
$\therefore$ Slope of AB = $\tan \theta$
Also Slope of AB $= \frac{{ - 2( - 1)}}{{4 - 3}} = \frac{{ - 1}}{1} = - 1$
Now $\tan \theta = - 1 = - \tan 45^\circ = \tan (180 - 45)^\circ$
$\Rightarrow \theta = 135^\circ$

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Question 382 Marks

Write the equation of the lines for which tan $\theta=\frac{1}{2}$, where θ is the inclination of the line and x-intercept is 4.

Answer

Here, we have m = tan θ = $\frac{1}{2}$ and d = 4.
the equation of the line is y = m(x – d)
$y=\frac{1}{2}(x-4) \text { or } 2 y-x+4=0$
which is the required equation.

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Question 392 Marks

Write the equation of the lines for which tan $\theta=\frac{1}{2}$, where θ is the inclination of the line and y-intercept is $-\frac{3}{2}$

Answer

Here, slope of the line is m = tan θ = $\frac{1}{2}$ and y - intercept c =$-\frac{3}{2}$
the equation of the line is y = mx +c
$y=\frac{1}{2} x-\frac{3}{2} \text { or } 2 y-x+3=0$
which is the required equation.

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Question 402 Marks

Find the equation of the line through (– 2, 3) with slope – 4.

Answer

Here $m = – 4$ and given point $(x_0 , y_0)$ is $(– 2, 3)$.
By slope-intercept form formula, then we have
$m=\frac{y-y_{0}}{x-x_{0}}, \text { i.e., } y-y_{0}=m\left(x-x_{0}\right)$
$y – 3 = – 4 (x + 2) or 4x + y + 5 = 0.$
which is the required equation.

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Question 412 Marks

Find the equations of the lines parallel to y axis and passing through (– 2, 3).

Answer

Position of the lines is shown in Fig 10.12.
They-coordinate of every point on the line parallel to the x-axis is 3, thus, the equation of the line parallel to the x-axis and passing through (– 2, 3) is y = 3.
Similarly, also the equation of the line parallel to the y-axis and passing through (– 2, 3) is x = – 2.

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Question 422 Marks

Line through the points (-2, 6) and (4, 8) is perpendicular to the line through the points (8, 12) and (x, 24). Find the value of x.

Answer

Slope of the line through the points (-2, 6) and (4, 8) is
$m_{1}=\frac{8-6}{4-(-2)}=\frac{2}{6}=\frac{1}{3}$
Slope of the line through the points (8, 12) and (x, 24) is
$m_{2}=\frac{24-12}{x-8}=\frac{12}{x-8}$
Since, two lines are perpendicular,
$\therefore m_1m_2$ = -1 $\Rightarrow \frac{1}{3} \times \frac{12}{x-8}=-1$
$\Rightarrow$ 4 = -(x - 8) $\Rightarrow$ x = 4

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Question 432 Marks

Show that the area of the triangle formed by the lines $y = m_1 x + c_1 , y = m_2x + c_2$ and $x = 0$ is $\frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$

Answer

Given lines are
$y=m_1 x+c_1 \ldots \text { (1) }$
$y=m_2 x+c_2 \ldots \text { (2) }$
$x=0 \ldots \text { (3) }$
We know that line $y=m x+c$ meets the line $x=0$ ( $y$-axis) at the point $(0, c)$. Thus, two vertices of the triangle formed by lines (1) to (3) are $P\left(0, c_1\right)$ and $Q\left(0, c_2\right)$. Third vertex can be obtained by solving equations (1) and (2). Solving (1) and (2), we obtain
$x=\frac{\left(c_{2}-c_{1}\right)}{\left(m_{1}-m_{2}\right)} \text { and } y=\frac{\left(m_{1} c_{2}-m_{2} c_{1}\right)}{\left(m_{1}-m_{2}\right)}$
Thus, third vertex of the triangle is R $\left(\frac{\left(c_{2}-c_{1}\right)}{\left(m_{1}-m_{2}\right)}, \frac{\left(m_{1} c_{2}-m_{2} c_{1}\right)}{\left(m_{1}-m_{2}\right)}\right)$
Now, the area of the triangle is given
$=\frac{1}{2} | 0\left(\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}-c_{2}\right)+\frac{c_{2}-c_{1}}{m_{1}-m_{2}}\left(c_{2}-c_{1}\right)+0\left(c_{1}-\frac{m_{1} c_{2}-m_{2} c_{1}}{m_{1}-m_{2}}\right)|=\frac{\left(c_{2}-c_{1}\right)^{2}}{2\left|m_{1}-m_{2}\right|}$

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Question 442 Marks

Assuming that straight lines work as the plane mirror for a point, find the image of the point (1, 2) in the line x – 3y + 4 = 0

Answer

Suppose Q (h, k) is the image of the point P (1, 2) in the line
x – 3y + 4 = 0 ... (1)

Thus,the line (1) is the perpendicular bisector of line segment PQ
Therefore,Slope of line PQ=$\frac{-1}{\text { Slope of line } x-3 y+4=0}$
so that $\frac{k-2}{h-1}=\frac{-1}{\frac{1}{3}}$ or 3h+k=5
and the mid-point of PQ, i.e., point $\left(\frac{h+1}{2}, \frac{k+2}{2}\right)$ will satisfy the equation (1) so that
$\frac{h+1}{2}-3\left(\frac{k+2}{2}\right)+4=0 \text { or } h-3 k=-3$ .........(3)
Solving (2) and (3), we obtain h=$\frac{6}{5}$ and k=$\frac{7}{5}$
Therefore, the image of the point (1, 2) in the line (1) is $\left(\frac{6}{5}, \frac{7}{5}\right)$

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Question 452 Marks

Find the distance of the line 4x – y = 0 from the point P (4, 1) measured along the line making an angle of 135° with the positive x-axis.

Answer

Given that the line is 4x – y = 0 ... (1)
In order to find the distance of the line (1) from the point P (4, 1) along another line, we have to find the point of intersection of both the lines.
For this purpose, we will first find the equation of the second line.
Slope of second line is tan 135° = –1.
Equation of the line with slope -1 through the point P (4, 1) is

y – 1 = – 1 (x – 4) or x + y – 5 = 0 ... (2)
Solving (1) and (2), we get x = 1 and y = 4 so that point of intersection of the two lines Q (1, 4).
Now, distance of line (1) from the point P (4, 1) along the line (2)
= the distance between the points P (4, 1) and Q (1, 4).
$=\sqrt{(1-4)^{2}+(4-1)^{2}}=3 \sqrt{2} \text { units. }$
Therefore, the required distance is $3 \sqrt{2}$ units.

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Question 462 Marks

If the lines 2x + y - 3 = 0, 5x + ky - 3 = 0, 3x - y - 2 = 0 are concurrent, find the value of k.

Answer

Given that three lines are concurrent, if they pass through a common point, i.e., point of intersection of any two lines lies on the third line. Here given lines are
2x + y – 3 = 0 ....(1)
5x + ky – 3 = 0 ....(2)
3x – y – 2 = 0 ... (3)
Solving (1) and (3) by cross-multiplication method, we get $\frac{x}{-2-3}=\frac{y}{-9+4}=\frac{1}{-2-3}$ or x = 1, y = 1
Therefore, the point of intersection of two lines is (1, 1).
Since the above three lines are concurrent, the point (1, 1) will satisfy equation (2) so that
5.1 + k.1 – 3 = 0
$\Rightarrow$ k = – 2

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Question 472 Marks

If the angle between two lines is $\frac{\pi}{4}$ and slope of one of the lines is $\frac{1}{2}$, then find the slope of the other line.

Answer

We know that, the acute angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by
$\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$ ...(i)
Let $m_{1}=\frac{1}{2}, m_2 = m$ and $ \theta=\frac{\pi}{4}$
Now, putting these values in Eq. (i). we get
$\frac{m-\frac{1}{2}}{1+\frac{1}{2} m}=1 $ or $ \frac{m-\frac{1}{2}}{1+\frac{1}{2} m}=-1$
$\Rightarrow m-\frac{1}{2}=1+\frac{1}{2} m $ or $m-\frac{1}{2}=-1-\frac{1}{2} m$
$\Rightarrow \left(1-\frac{1}{2}\right) m=1+\frac{1}{2} \text { or } m\left(1+\frac{1}{2}\right)=-1+\frac{1}{2}$
$\Rightarrow$ m = 3 or m = $-\frac {1}{3}$

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Question 482 Marks

Find the distance between the parallel lines 3x – 4y + 7 = 0 and 3x – 4y + 5 = 0

Answer

Here, $A=3, B=-4, C_1=7$ and $C_2=5$.
Hence, the required distance is $d=\frac{|7-5|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{2}{5}$

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Question 492 Marks

Find the distance of the point (3, -5) from the line 3x - 4y - 26 = 0

Answer

According to the question,
Given line is 3x – 4y –26 = 0 ...(1)
Comparing (1) with general equation of line Ax + By + C = 0, we get
A = 3, B = – 4 and C = – 26.
Given point is $(x_{1,} y_1) = (3, –5)$. The distance of the given point from the given line is
$d=\frac{\left|\mathrm{A} x_{1}+\mathrm{B} y_{1}+\mathrm{C}\right|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|3.3+(-4)(-5)-26|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{3}{5}$

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Question 502 Marks

Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1, – 2).

Answer

Given line x - 2y + 3 = 0 can be written as
$y=\frac{1}{2} x+\frac{3}{2}$ ........(1)
Slope of the line (1) is $m_{1}=\frac{1}{2}$. Thus,slope of the line perpendicular to line (1) is
$m_{2}=-\frac{1}{m_{1}}=-2$
Equation of the line with slope – 2 and passing through the point (1, – 2) is
$y-(-2)=-2(x-1) \text { or } y=-2 x$
This is the required equition of line

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Question 512 Marks

Find the angle between the lines $y-\sqrt{3} x-5=0$ and $\sqrt{3} y-x+6=0$.

Answer

Given lines are
$y-\sqrt{3} x-5=0 \text { or } y=\sqrt{3} x+5$ .........(1)
and $\sqrt{3} y-x+6=0 \text { or } y=\frac{1}{\sqrt{3}} x-2 \sqrt{3}$......(2)
Slope of line (1) is $m_{1}=\sqrt{3}$ and slope of line (2) is $m_{2}=\frac{1}{\sqrt{3}}$
The acute angle (say) $\theta$ between two lines is given by
$\tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} m_{2}}\right|$......(3)
Substituting,the values of $m_1$ and $m_2$ in (3), we obtain
$\tan \theta=\left|\frac{\frac{1}{\sqrt{3}}-\sqrt{3}}{1+\sqrt{3} \times \frac{1}{\sqrt{3}}}\right|=\left|\frac{1-3}{2 \sqrt{3}}\right|=\frac{1}{\sqrt{3}}$
which gives $\theta$ = 30°.
Therefore,the angle between two lines is either 30°or 180° – 30° = 150°.

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Question 522 Marks

Reduce the equation $\sqrt{3} x+y-8=0$ into normal form. Find the values of p and ω

Answer

We have,
$\sqrt{3} x+y-8=0$ ... (1)
Dividing (1) by $\sqrt{(\sqrt{3})^{2}+(1)^{2}}=2$, we obtain
$\frac{\sqrt{3}}{2} x+\frac{1}{2} y=4$ or $\cos 30^{\circ} x+\sin 30^{\circ} y=4$ ....(2)
Comparing (2) with x cos $\omega$ + y sin $\omega$ = p, then we get p = 4 and $\omega$ = 30°.

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Question 532 Marks

The Fahrenheit temperature F and absolute temperature K satisfy a linear equation. Given that K = 273 when F = 32 and that K = 373 when F = 212. Express K in terms of F and find the value of F, when K = 0

Answer

Suppose that F along x-axis and K along y-axis, we have two points (32, 273) and (212, 373) in XY-plane. By two-point form, the point (F, K) satisfies the equation,then we get
$\mathrm{K}-273=\frac{373-273}{212-32}(\mathrm{F}-32)$ or $\mathrm{K}-273=\frac{100}{180}(\mathrm{F}-32)$
or $K=\frac{5}{9}(F-32)+273$ .......(1)
which is the required relation.
When K = 0, Equation (1) gives
$0=\frac{5}{9}(\mathrm{F}-32)+273$ or $\mathrm{F}-32=-\frac{273 \times 9}{5}=-491.4$ or F= -459.4

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Question 542 Marks

Find the equation of the line whose perpendicular distance from the origin is 4 units and the angle which the normal makes with positive direction of x-axis is 15°.

Answer

We are given that, p = 4 and ω = $15^{\circ}$
Now, $\cos 15^{\circ}=\frac{\sqrt{3}+1}{2 \sqrt{2}}$
and $\sin 15^{\circ}=\frac{\sqrt{3}-1}{2 \sqrt{2}}$
The equation of the line is x cos ω + y sin ω = p
$x \cos 15^{\circ}+y \sin 15^{\circ}$
or $\frac{\sqrt{3}+1}{2 \sqrt{2}} x+\frac{\sqrt{3}-1}{2 \sqrt{2}} y=4$
or $(\sqrt{3}+1) x+(\sqrt{3}-1) y=8 \sqrt{2}$
This is the required equation.

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Question 552 Marks

Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively.

Answer

Here a = –3 and b = 2.
We know that equation of the line is $\frac{x}{a}+\frac{y}{b}=1$
$\frac{x}{-3}+\frac{y}{2}=1$ or 2x - 3y + 6 = 0

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