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.
AnswerThe 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$.
View full question & answer→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
AnswerPoint 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
$x = \frac{{ - 5}}{{22}}$.
View full question & answer→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.
AnswerLet 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$$ \Rightarrow 3y - 12 = - 20 \Rightarrow y = \frac{{ - 8}}{3}$
Thus required points are $\left( {0,\frac{{32}}{3}} \right)$ and $\left( {0,\frac{{ - 8}}{3}} \right)$.
View full question & answer→Question 42 Marks
Find the values of $\theta $ and p, if the equation $x\cos \theta + y\sin \theta = p$ is the normal form of the line $\sqrt 3 x + y + 2 = 0$
AnswerHere $\sqrt 3 x + y + 2 = 0$
$ \Rightarrow \sqrt 3 x + y = - 2 \Rightarrow - \sqrt 3 x - y = 2$
Dividing both sides by $\sqrt {{{( - \sqrt 3 )}^2} + {{( - 1)}^2}} = 2$, we have
$\frac{{ - \sqrt 3 }}{2}x - \frac{1}{2}y = 1$
Put $\cos \alpha = \frac{{ - \sqrt 3 }}{2}$ and $\sin \alpha= \frac{{ - 1}}{2}$
$ \Rightarrow \alpha $ lies in 3rd quadrant
$\therefore \cos \alpha = \frac{{ - \sqrt 3 }}{2} = - \cos 30^\circ $ = cos (180° + 30° )
$ \Rightarrow \alpha = 210^\circ $
$\therefore $ Equation of line in normal form is
$x\cos \frac{{7\pi }}{6} + y\sin \frac{{7\pi }}{6} = 1$
Comparing it with x cos $\alpha + y\ sin \alpha = p$, we have
$\alpha = \frac{{7\pi }}{6}$ and p = 1
View full question & answer→Question 52 Marks
Find the angles between the lines $\sqrt 3 x + y = 1$ and $x + \sqrt 3 y = 1$
AnswerWe have $\sqrt 3 x + y = 1$
$\Rightarrow y = - \sqrt 3 x + 1$
$\therefore {m_1} = - \sqrt 3$
Also $x + \sqrt 3 y = 1$
$\Rightarrow \sqrt 3 y = - x + 1$
$\Rightarrow y = \frac{{ - 1}}{{\sqrt 3 }}x + \frac{1}{{\sqrt 3 }}$
$\therefore {m_2} = \frac{{ - 1}}{{\sqrt 3 }}$
Let $\theta$ be the angle between the lines. Then,
$\tan \theta = \left| {\frac{{ - \sqrt 3 + \frac{1}{{\sqrt 3 }}}}{{1 + ( - \sqrt 3 )\left( {\frac{{ - 1}}{{\sqrt 3 }}} \right)}}} \right| = \left| {\frac{{\frac{{ - 3 + 1}}{{\sqrt 3 }}}}{{1 + 1}}} \right|$$= \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°
View full question & answer→Question 62 Marks
Find the equation of the line perpendicular to the line x - 7y + 5 = 0 and having x intercept 3.
AnswerEquation 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
View full question & answer→Question 72 Marks
Find the distance between parallel lines. $l(x + y) + p = 0$ and $l(x + y) - r = 0$
AnswerWe have the equation,
$lx + ly + p = 0$
and $lx + ly - r = 0$
where $a = 1, b = 1 , c_1 = p$ and $c_2 = -r$
$\therefore$ 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
View full question & answer→Question 82 Marks
Find the distance between parallel lines. $15x + 8y - 34 = 0$ and $15 x + 8y + 31 = 0$
AnswerThe 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
View full question & answer→Question 92 Marks
Find the distance of the point (-1, 1) from the line 12 (x + 6) = 5(y - 2).
AnswerHere 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.
View full question & answer→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.
AnswerHere 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 $p = 2\sqrt 2$
View full question & answer→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.
AnswerHere 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
View full question & answer→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.
AnswerAny 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.,
$\frac{a}{\sqrt{a^{2}+b^{2}}} x+\frac{b}{\sqrt{a^{2}+b^{2}}} y=\frac{c}{\sqrt{a^{2}+b^{2}}}$
$$Given equation of line is
$x-\sqrt{3} y+8=0$
$\Rightarrow \quad x-\sqrt{3} y=-8$
$\Rightarrow \quad-x+\sqrt{3} y=8 ...(ii)$
Now, divide by $\sqrt{(\text { coefficient of } x)^{2}+(\text { coefficient of } y)^{2}} = \sqrt{(-1)^{2}+(\sqrt{3})^{2}}=\sqrt{1+3}=2,$ we get
$-\frac{1}{2} x+\frac{\sqrt{3}}{2} y=\frac{8}{2}$
$\Rightarrow - \cos 60^o x + \sin 60^o y = 4 [\because$ convert in form of $x \cos \alpha + y \sin \alpha = p]$
$[\because \cos x$ is negative and $\sin x $is positive, it is possible in second quadrant$]$
$\Rightarrow x \cos(180^o - 60^o) + y \sin (180^o - 60^o) = 4$
$\Rightarrow x \cos 120^o + y \sin 120^o = 4$
$\left[\because \cos \left(180^{\circ}-\theta\right)=-\sin \theta \text { and } \sin \left(180^{\circ}-\theta\right)=\cos \theta\right]$
On comparing with $x \cos \alpha + y \sin\alpha = p,$ we get
$\alpha = 120^o, p = 4$
View full question & answer→Question 132 Marks
Reduce the given equation into the intercept form and find the intercept on the axis. 3 y + 2 = 0
AnswerHere 3 y + 2 = 0
$\Rightarrow 3y = - 2$
$ \Rightarrow \frac{{3y}}{{ - 2}} = 1 \Rightarrow \frac{{0x}}{{ - 2}} + \frac{{3y}}{{ - 2}} = 1 \Rightarrow \frac{{0x}}{{ - 2}} + \frac{y}{{\frac{{ - 2}}{3}}} = 1$
which is required intercept form.
Comparing it with $\frac{x}{a} + \frac{y}{b} = 1$ we have
a = 0 and b = -2/3.
View full question & answer→Question 142 Marks
Reduce the given equation into the intercept form and find the intercept on the axis. 4x - 3y = 6
AnswerHere 4x - 3y = 6
$\Rightarrow \frac{{4x}}{6} - \frac{{3y}}{6} = 1 \Rightarrow \frac{{2x}}{3} - \frac{y}{2} = 1]$$\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
View full question & answer→Question 152 Marks
Reduce the given equation into the intercept form and find the intercept on the axis.
3x + 2y - 12 = 0
AnswerHere 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
View full question & answer→Question 162 Marks
Find the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2).
AnswerMid 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.
View full question & answer→Question 172 Marks
Prove that the line through the point $(x_1, y_1)$ and parallel to the line $Ax + By + C = 0 is A(x - x_1) + B(y - y_1) = 0.$
AnswerEquation of the line parallel to line $Ax + By + C = 0 is Ax + By + K = 0 . . . (i)$
Since line (i) passes through $(x_1, y_1)$
$Ax_1 + By_1 + K = 0...(ii)$
Subtracting (ii) from (i), we have
$A(x - x_1) + B(y - y_1) = 0$
View full question & answer→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.
AnswerSlope 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 $\Rightarrow h = \frac{{22}}{9}$
View full question & answer→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 $30^\circ.$
AnswerHere, $p = 5$ and $\alpha = 30^\circ$
So, equation of line in normal form is
$x \cos \alpha + y \sin \alpha = p$
$\therefore x \cos 30^\circ + y \sin 30^\circ = 5$
$\Rightarrow \quad \frac{\sqrt{3}}{2} x+\frac{1}{2} y=5$
$\Rightarrow \quad \sqrt{3} x+y=10$ View full question & answer→Question 202 Marks
Find the equation of the line which satisfy the given condition:
The line passing through the points $(-1, 1)$ and $(2, - 4).$
AnswerGiven 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$
$\Rightarrow 5x + 3y + 2 = 0$
View full question & answer→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.
AnswerHere m = tan 30° = $\frac{1}{{\sqrt 3 }}$ and c = 2
Putting these values in y = mx + c, we have
$y = \frac{1}{{\sqrt 3 }}x + 2 \Rightarrow x - \sqrt 3 y + 2\sqrt 3 = 0$
View full question & answer→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 $-2.$
AnswerGiven, the line intersecting the $X-$axis to the left of the origin. It means it intersect the negative $X-$axis. Clearly, line $AB$ passes through the point $(-3, 0)$ and $m = -2.$
Equation of line in point slope from is
$y - y_1 = m(x - x_1) $
$\Rightarrow y - 0 = -2(x + 3)$
$\Rightarrow y = -2x - 6$
$\Rightarrow 2x + y + 6 = 0$
View full question & answer→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 $75^\circ.$
AnswerIt is given that the point $= (2, 2\sqrt3)$ and angle $\theta = 75^\circ $
Equation of line: $(y - y_1) = m (x - x_1)$
where, $m = $slope of line $= \tan \theta $ and $(x_1, y_1)$ are the points through which line passes
$\therefore 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 obtain$ \tan 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 $y – y_1 = m (x – x_1)$
$\therefore y-2 \sqrt{3}=(2+\sqrt{3})(x-2)$
$\Rightarrow y-2 \sqrt{3}=2 x-4+\sqrt{3} x-2 \sqrt{3}$
$\Rightarrow y=2 x-4+\sqrt{3} x$
$\Rightarrow(2+\sqrt{3}) x-y-4=0$
Therefore, the equation of the line is $(2 + \sqrt{3}) x – y - 4 = 0.$
View full question & answer→Question 242 Marks
Find the equation of the line passing through $(-4, 3)$ and having slope $\frac{1}{2}.$
AnswerEquation of the line passing through the point $(x_{0,} y_0)$ having slope m is
$y - y_1 = m (x - x_1) ...(i)$
Given, $m =$ slope of the line $= \frac{1}{2}$ and $x_1 = - 4, y_1 = 3$
From Eq. (i). required equation of the line is
$y - 3=\frac{1}{2}(x+4)$
$\Rightarrow 2y - 6 = x + 4$
$\Rightarrow x - 2y + 10 = 0$
View full question & answer→Question 252 Marks
The perpendicular from the origin to a line meets it at the point (-2, 9), find the equation of the line.
AnswerSlope 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
View full question & answer→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).
AnswerLet 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
View full question & answer→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).
AnswerLet 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
View full question & answer→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.
AnswerLet 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
View full question & answer→Question 292 Marks
Find the value of x for which the points (x, -1)(2, 1) and (4, 5) are collinear.
AnswerLet 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
View full question & answer→Question 302 Marks
Find the slope of the line, which makes an angle of $30^\circ$ with the positive direction of y-axis measured anticlockwise.
AnswerThe 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 $= \tan 120^\circ = \tan (90 + 30)$
$= -\cot 30^o = - \sqrt 3$
View full question & answer→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.
AnswerLet 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.
View full question & answer→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 $B(8, 0).$
AnswerIf 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 $R$
$R=\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right)=\left(\frac{0+8}{2}, \frac{-4+0}{2}\right)=(4,-2)$
$\therefore \text { Slope of } O R=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}=\frac{-2-0}{4-0}=\frac{-2}{4}=-\frac{1}{2}$
$\left[{\because x_{1}=0, y_{1}=0,} \ {x_{2}=4, y_{2}=-2}\right]$ View full question & answer→Question 332 Marks
Find a point on the x-axis, which is equidistant from the points $(7, 6)$ and $(3, 4).$
AnswerLet $P(x, 0)$ 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 $\left( {\frac{{15}}{2},0} \right)$.
View full question & answer→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?
AnswerThe 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.
View full question & answer→Question 352 Marks
If three points (h, 0), (a, b) and (0, k) lies on a line, show that $\frac{a}{h} + \frac{b}{k} = 1$
AnswerLet 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
$\frac{{ak}}{{hk}} + \frac{{bh}}{{hk}} = 1 \Rightarrow \frac{a}{h} + \frac{b}{k} = 1$.
View full question & answer→Question 362 Marks
A line passes through $(x_1, y_1)$ and $(h, k).$ If slope of the line is m, show that $k - {y_1} = m(h - {x_1})$.
AnswerLet $A(x_1, y_1)$ and $B(h, k)$ be two points.
$\therefore$ Slope of $AB = \frac{{k - {y_1}}}{{h - {x_1}}}$
Slope of $AB = m ($given$)$
$\therefore m = \frac{{k - {y_1}}}{{h - {x_1}}} \Rightarrow k - {y_1} = m(h - {x_1})$
View full question & answer→Question 372 Marks
The slope of a line is double of the slope of another line. If tangent of the angle between them is $\frac{1}{3}$, then find the slope of the lines.
AnswerIf slope of one line is $m$. Then, the slope of the other line is $2m.$
Let angle between these two lines be $\theta.$
Then, $\tan \theta=\frac{1}{3}$ [given)
$\Rightarrow \quad\left|\frac{2 m-m}{1+2 m \cdot m}\right|=\frac{1}{3}\left[\because \tan \theta=\left|\frac{m_{2}-m_{1}}{1+m_{1} \cdot m_{2}}\right|\right]$
$\Rightarrow \quad \frac{m}{1+2 m^{2}}=\frac{1}{3}$
$\Rightarrow 2m^2 - 3m + 1 = 0$
$\Rightarrow (2m - 1) (m - 1) = 0 \Rightarrow m=\frac{1}{2}, m=1$
Thus, the slope of these lines are $\frac{1}{2}$ and $1.$
View full question & answer→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.
AnswerHere, 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.
View full question & answer→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 $-\frac{3}{2}$
AnswerHere, 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.
View full question & answer→Question 402 Marks
Find the equation of the line through $(– 2, 3)$ with slope $– 4.$
AnswerHere $m = – 4$ 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.
View full question & answer→Question 412 Marks
Find the equations of the lines parallel to y axis and passing through (– 2, 3).
AnswerPosition 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.
View full question & answer→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 $x.$
AnswerSlope 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,
$\therefore m_1m_2 = -1 \Rightarrow \frac{1}{3} \times \frac{12}{x-8}=-1$
$\Rightarrow 4 = -(x - 8) \Rightarrow x = 4$
View full question & answer→Question 432 Marks
If the lines 2x + y - 3 = 0, 5x + ky - 3 = 0, 3x - y - 2 = 0 are concurrent, find the value of k.
AnswerGiven 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
View full question & answer→Question 442 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.
AnswerWe 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 $m =-\frac {1}{3}$
View full question & answer→Question 452 Marks
Find the distance between the parallel lines $3x – 4y + 7 = 0$ and $3x – 4y + 5 = 0$
AnswerHere, $A = 3, B = –4, C_1 = 7$ and $C_2 = 5.$
Hence, the required distance is $d=\frac{|7-5|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{2}{5}$
View full question & answer→Question 462 Marks
Find the distance of the point $(3, -5)$ from the line $3x - 4y - 26 = 0$
AnswerAccording 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
$d=\frac{\left|\mathrm{A} x_{1}+\mathrm{B} y_{1}+\mathrm{C}\right|}{\sqrt{\mathrm{A}^{2}+\mathrm{B}^{2}}}=\frac{|3.3+(-4)(-5)-26|}{\sqrt{3^{2}+(-4)^{2}}}=\frac{3}{5}$
View full question & answer→Question 472 Marks
Find the equation of a line perpendicular to the line x - 2y + 3 = 0 and passing through the point (1, – 2).
AnswerGiven 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
View full question & answer→Question 482 Marks
Find the angle between the lines $y-\sqrt{3} x-5=0$ and $\sqrt{3} y-x+6=0$.
AnswerGiven 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 $\theta = 30^\circ .$
Therefore,the angle between two lines is either $30^\circ$ or 1$80^\circ – 30^\circ = 150^\circ .$
View full question & answer→Question 492 Marks
Reduce the equation $\sqrt{3} x+y-8=0$ into normal form. Find the values of p and ω
AnswerWe 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°.
View full question & answer→Question 502 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
AnswerSuppose 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
View full question & answer→Question 512 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 $15^\circ .$
AnswerWe are given that, $p = 4$ and $\omega = 15^\circ$
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 $x \cos \omega + y \sin \omega = p$
$x \cos 15^{\circ}+y \sin 15^{\circ}$
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.
View full question & answer→Question 522 Marks
Find the equation of the line, which makes intercepts –3 and 2 on the x- and y-axes respectively.
AnswerHere 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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