5 Marks Questions

Question 15 Marks

The equation of the line through the intersection of the the lines $2x - 3y = 0$ and $4x - 5y = 2$ and

	column $I$		column $II$
$(a)$	Throught the point $(2, 1)$ is	$(a)$	$2x - y = 4$
$(b)$	perpendicular to the line $x + 2y + 1 = 0$ is	$(b)$	$x + y - 5 = 0$
$(c)$	parpallel to the line $3x + 4y + 5 = 0$	$(c)$	$x - y - 1$
$(d)$	Equally inlined to the axis is	$(d)$	$3x - 4y - 1 = 0$

Answer

Given equcations are $2 - 3y = 0 ...(i)$

and $4x - 5y = 2 ...(ii)$ equation of line passing through $eq. (i)$ and $(ii)$ we get $(2x - 3y) + k(4x - 5y - 2) = 0 ...(iii)$ If $eq.(iii)$ passing through$(2, 1)$ we get $(2 \times 2 - 3 \times 1) + k(4 \times 2 - 5 \times -2) = 0$
$\Rightarrow (4 - 3) + k(8 - 5 - 2) = 0$
$ \Rightarrow 1 + k(8 - 7)= 0$
$ \Rightarrow k = -1$
So, the required equation is $(2x - 3y) -1 (4x - 5y - 2) = 0$
$ \Rightarrow 2x - 3y - 4x + 5y + 2 = 0$
$ \Rightarrow - 2x + 2y + 2 = 0$
$ \Rightarrow x - y - 1 = 0$
Hence$, (a) \leftrightarrow (iii)$

Equation of any line passing through the ponit of intersection of the line $2x - 3y = 0$ and $4x - 5y = 2$ is

$(2x - 3y) + k(4x - 5y - 2) = 0 ...(i)$
$ \Rightarrow (2 + 4k) x + (-3 - 5k) y - 2k = 0$
Slope $=\frac{-(2+4\text{k})}{-3-5\text{k}}=\frac{2+4\text{k}}{3+5\text{k}}$
Slope of the given line $x + 2y + 1 = 0$ is $-\frac{1}{2}$
If they are perpendicular to each other then
$\Rightarrow\frac{1}{2}\bigg(\frac{2+4\text{k}}{3+5\text{k}}\bigg)=-1$
$\Rightarrow\frac{2+2\text{k}}{3+5\text{k}}=1$
$\Rightarrow1+2\text{k}=3+5\text{k}$
$ \Rightarrow3\text{k}=-2$
$\Rightarrow \text{k}\frac{-2}{3}$
$(2\text{x}-3\text{y})-\frac{2}{3}(4\text{x}-5\text{y}-2)=0$
$\Rightarrow 6x - 3y - 8x + 10y + 4 = 0$
$ \Rightarrow -2x + y + 4$
$ \Rightarrow 2x - y = 4$
Hence $, (b) \leftrightarrow (i)$

Given equation are

$2x - 3y = 0 ...(i) 4x - 5y = 2 ...(ii)$
Equation of line pssing through $eq. (i)$ and $(ii)$
we get $(2x - 3y) + k(4x - 5y - 2) = 0$
$\Rightarrow (2 + 4k) x + (-3 - 5k) y - 2k = 0$
Slope $=\frac{-(2+4\text{k})}{-3-5\text{k}}=\frac{2+4\text{k}}{3+5\text{k}}$
slope of the given line $3x - 4y + 5 = 0$ is $\frac{3}{4}$ . If the two Equation are parallel , then
$\Rightarrow\frac{2+4\text{k}}{3+5\text{k}}=\frac{3}{4}$
$ \Rightarrow 8 + 16k = 9 + 15k$
$ \Rightarrow 16k - 15k = 9 - 8$
$ \Rightarrow k = 1$
So, the equation of the requied line is $(2x - 3y) + 1(4x - 5y - 2) = 0 2x - 3y + 4x - 5y - 2 = 0$
$\Rightarrow 6x - 8y - 2 = 0$
$ \Rightarrow 3x - 4y - 1 = 0$
Hence$, (c) \leftrightarrow (iv)$

Given equation are

$2x - 3y = 0 ...(i) 4x - 5y - 2 = 0...(ii)$
Equation of the passing through the intersection of eq. $(i)$ and $(ii)$ we get $(2x - 3y) + k(4x - 5y - 2) = 0$
$ \Rightarrow (2 + 4k)x + (-3 - 5k) y - 2k = 0$
Slope $=\frac{2+4\text{k}}{3+5\text{k}}$
since the equation is equally inclined with axes
$\therefore\ =\tan135^\circ=\tan{(180^\circ-45)}=-\tan45^\circ=-1$
So $\frac{2+4\text{k}}{3+5\text{k}}=-1$
$\Rightarrow2+4\text{k}=-3-5\text{k}$
$\Rightarrow 9k = -5$
$\Rightarrow\text{k}=\frac{-5}9{}$ Requied equation is $(2\text{x}-3\text{y})-\frac{5}{9}(4\text{x}-5\text{y}-2)=0$
$\Rightarrow 18x - 27y - 20x + 25y + 10 = 0$
$\Rightarrow -2x - 2y + 10 = 0$
$ \Rightarrow x + y - 5 = 0$
Hence$, (d) \leftrightarrow (ii).$

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Question 25 Marks

Match the questions given under $\text{Column}\ C_1$ with their appropriate answers given under the $\text{Column}\ C_2$: The value of the $\lambda ,$ if the lines $(2x + 3y + 4) + \lambda (6x - y + 12) = 0$ are:

	$\text{Column}\ C_1$		$\text{Column}\ C_2$
$(a)$	Parallel to $y-$axis is	$(i)$	$\lambda=-\frac{3}{4}$
$(b)$	Perpendicular to $7x + y - 4 = 0$ is	$(ii)$	$\lambda=-\frac{1}{3}$
$(c)$	Passes through $(1, 2)$ is	$(iii)$	$\lambda=-\frac{17}{41}$
$(d)$	Parallel to $x$ axis is	$(iv)$	$\lambda=3$

Answer

	$\text{Column}\ C_1$		$\text{Column}\ C_2$
$(a)$	Parallel to $y-$axis is	$(i)$	$\lambda=-\frac{3}{4}$
$(b)$	Perpendicular to $7x + y - 4 = 0$ is	$(ii)$	$\lambda=-\frac{1}{3}$
$(c)$	Passes through $(1, 2)$ is	$(iii)$	$\lambda=-\frac{17}{41}$
$(d)$	Parallel to $x$ axis is	$(iv)$	$\lambda=3$

Given equcation is
$(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0$
$\Rightarrow(2+6\lambda)\text{x}+(3-\lambda)\text{y}+4+12\lambda=0\ \ ...\text{(i)}$ If $eq.(i)$ is parallel to $y-$axis,
then $3- \lambda=0$
$\Rightarrow\lambda=3$
Hence$, (a)$ ⇔ $(iv)$
Given lines are
$(2\text{x}+3\text{y}+4)+\lambda(6\text{x}-\text{y}+12)=0 ..(i)$
$\Rightarrow(2+6\lambda)\text{x}+3(3-\lambda)\text{y}+12\lambda=0$ Slope $=-\bigg(\frac{2+6\lambda}{3-\lambda}\bigg)$ Second equation is $7\text{x}+\text{y}-4=0\ \ ...\text{(ii)}$ Slope $= -7$ If $eq. (i)\ eq. (ii)$ are perpendilcular to each other
$\therefore(-7)\bigg[-\bigg(\frac{2+6\lambda}{3-\lambda}\bigg)\bigg]=1$
$\Rightarrow\frac{14+42\lambda}{3-\lambda} = -1$
$41\lambda=-17$
$\lambda=-\frac{17}{41}$
Hence$, (b)$ ⇔ $(iii)$
Given equation is $(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0\ \ ...\text{(i)}$
If $eq.(i)$ passes throgh the given ponit $(1, 2)$ then $(2\times1+3\times2+4) +\lambda(6\times1-0+12)=0$
$\Rightarrow(2+6+4) +\lambda(6-2+12)=0$
$\Rightarrow 12+16\lambda=0$
$\Rightarrow=\frac{-12}{16}=\frac{-3}{4}$
Hence$, (c)$ ⇔ $(i)$
The equation is $(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0$
$\Rightarrow(2\text{x}+3\text{y}+4)+\lambda(6\text{x}- \text{y}+12)=0$ If $eq. (i)$ is parallel to $x-$axis$,$ then $2+6\lambda$
$\Rightarrow\lambda=\frac{-1}{3}$
Hence$, (d)$ ⇔ $(ii)$

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Question 35 Marks

Find the equation of the line which passes through the point (-4, 3) and the portion of the line intercepted between the axes is divided internally in the ratio 5 : 3 by this point.

Answer

Let the line through the point P(-A, 3) meets axis at A(h, 0) and 0(0, k)
Now according to the question AP : BP = 5 : 3
$\therefore (-4, 3)\equiv\Big(\frac{3\times\text{h}+5\times}{5+3},\frac{3\times0+5\times\text{k}}{5+3}\Big)\equiv\Big(\frac{3\text{h}}{8},\frac{5\text{h}}{8}\Big)$
$\Rightarrow -4=\frac{3\text{h}}{8}$ and $3=\frac{5\text{k}}{8}$
$\Rightarrow \text{h}=-\frac{32}{3}$ and $\text{k}=\frac{24}{5}$
So, equation of the required line in intercept form is:
$\frac{\text{x}}{-\frac{32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$
$\Rightarrow 9\text{x}-20\text{y}+96=0$

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Question 45 Marks

Find the angle between the lines $\text{y} = (2 - \sqrt{3})(\text{x} + 5)$ and $\text{y} = (2 + \sqrt{3})(\text{x} - 7).$

Answer

The given equation are $\text{y}\ =\ (2-\sqrt{3})(\text{x}+5)\ .....(\text{i})$
and $\text{y}=(2+\sqrt{3})(\text{x}-7)$
Slope of $eq. (i)\ m_1\ ($say$)\ =\ (2-\sqrt{3})$
and slope of $eq. (ii)\ m_2\ ($say$)\ =\ (2+\sqrt{3})$
let $\theta$ be the angle between the two given lines
$\therefore \tan\theta=\Big|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\Big|=\Big|\frac{2-\sqrt{3}-2-\sqrt{3}}{1+(2-\sqrt{3})(2+\sqrt{3})}\Big|$
$=\Big|\frac{-2\sqrt{3}}{1+4-3}\Big|=\Big|\frac{-2\sqrt{3}}{2}\Big|=|-\sqrt{3}|$
$\Rightarrow \tan\theta=\sqrt{3}\text{ or }-\sqrt{3}$
$\therefore \theta = 60^\circ\text{ or }120^\circ$
Hence$,$ the required angle is $60^\circ$ or $120^\circ .$

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Question 55 Marks

If the equation of the base of an equilateral triangle is x + y = 2 and the vertex is (2, - 1), then find the length of the side of the triangle.
[Hint: Find length of perpendicular (p) from (2, - 1) to the line and use $\text{p}=\text{l}\sin60^\circ$ where l is the length of side of the triangle].

Answer

Equation of the base AB of a $\triangle\text{ABC}$ is x + y = 2
In $\triangle\text{ABD},$
$\sin 60^\circ = \frac{\text{AD}}{\text{AB}}$
$\Rightarrow \frac{\sqrt{3}}{2}=\frac{\text{AD}}{\text{AB}}\Rightarrow \text{AD}=\frac{\sqrt{3}}{2}\text{AB}$
Lenght of perpendicular from A(2, -1) to the line x + y = 2 is
$\text{AD}=\bigg|\frac{1\times2+1\times-1-2}{\sqrt{(1)^2+(1)^2}}\bigg|$

$\Rightarrow \frac{\sqrt{3}}{2}\text{AB}=\bigg|\frac{2-1-2}{\sqrt{2}}\bigg|=\bigg|\frac{-1}{\sqrt{2}}\bigg|$
$\Rightarrow \frac{\sqrt{3}}{2}\text{AB}=\frac{1}{\sqrt{2}}$
$\Rightarrow \text{AB}=\frac{\sqrt{2}}{\sqrt{3}}$
Hence, the required length of side $=\frac{\sqrt{2}}{\sqrt{3}}.$

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Question 65 Marks

Find the equation of a straight line on which length of perpendicular from the origin is four units and the line makes an angle of 120° with the positive direction of x-axis.
[Hint: Use normal form, here $\omega =30^\circ.$]

Answer

Given that:
OM = 4 units
$\angle \text{BAZ}=120^\circ$
$\therefore \angle \text{BAO}= 180^\circ - 120^\circ$ or $\angle\text{MAO}=60^\circ$
$\angle\text{MOA}+\angle\text{MAO}=90^\circ$
$\therefore \theta=30^\circ$

So, equation of AB in its normal form
$\Rightarrow \text{x}\cos\theta+\text{y}\sin\theta=\text{p}$
$\Rightarrow \text{x}\cos30^\circ+\text{y}\sin 30^\circ=4$
$\Rightarrow \text{x}\times\frac{\sqrt{3}}{2}+\text{y}\times\frac{1}{2}=4$
$\Rightarrow \sqrt{3}\text{x}+\text{y}=8$
Hence, the required equation is $\sqrt{3}\text{x}+\text{y}=8.$

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Question 75 Marks

Find the equations of the lines through the point of intersection of the lines $x - y + 1 = 0$ and $2x - 3y + 5 = 0$ and whose distance from the point $(3, 2)$ is $\frac{7}{5}$.

Answer

Given lines are: $x - y + 1 = 0 .....(i)$
and $2x - 3y + 5 = 0 .....(ii)$
Solving these lines$,$ we get point of intersection as $(2, 3).$
So$,$ equation of line is:
$y - 3 = m(x - 2)$
$\Rightarrow 3m - y + 3 - 2m = 0$
Distance of this line from $(3, 2) = \frac{7}{5}$
$\therefore \frac{7}{5}\begin{vmatrix} \frac{3\text{m}-2+3-2\text{m}}{\sqrt{1+\text{m}^2}} \end{vmatrix}\Rightarrow \frac{49}{25}=\frac{(\text{m}+1)^2}{1+\text{m}^2}$
$\Rightarrow 49 + 49m^2 = 25(m^2 + 2m + 1)$
$\Rightarrow 24m^2 - 5m + 24 = 0$
$\Rightarrow 12m^2 - 25m + 12 = 0$
$\Rightarrow (3m - 4)(4m - 3) = 0$
$\therefore \text{m}=\frac{4}{3}$ or $\frac{3}{4}$
So$,$ the equation of the line can be:
$\Rightarrow \text{y}-3=\frac{4}{3}(\text{x}-2)$ or $\text{y}-3=\frac{3}{4}(\text{x}-2)$
$\Rightarrow 4x - 3y + 1 = 0$ or $3x - 4y + 6 = 0$

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Question 85 Marks

$P_1, P_2$ are points on either of the two lines $\text{y} - \sqrt{3}|\text{x}| = 2$ at a distance of $5$ units from their point of intersection. Find the coordinates of the foot of perpendiculars drawn from$ P1 , P2$ on the bisector of the angle between the given lines.
$[$Hint: Lines are $\text{y} = \sqrt{3}\text{x} + 2$ and $\text{y} = -\sqrt{3}\text{x} + 2$ according as $\text{x} \geq 0$ or $x < 0. y-$ axis is the bisector of the angles between the lines. $P_1, P_2$ are the points on these lines at a distance of $5$ units from the point of intersection of these lines which have a point on $y-$axis as common foot of perpendiculars from these points. The y$-$coordinate of the foot of the perpendicular is given by $2 + 5 \cos30^\circ.]$

Answer

Given lines are: $\text{y}-\sqrt{3}\text{x}=2,$ for $\text{x}\geq0\ .....(\text{i})$
and $\text{y}+\sqrt{3}\text{x}=2,$ for $\text{x}\leq0\ .....(\text{ii})$
Clearly, lines intersect ar $A (, 2).$
Line $(i)$ is inclined at an angle of $60^\circ$ with $+$ ve direction of $x-$axis.
Line $(ii)$ is inclined at an angle of $120^\circ$ with $+$ ve direction of $x-$axis.

$P_1$ and $P_2$ are points at distance $5$ units from point $A$ on the lines.
Clearly$,$ angle bisector of lines is $y-$axis.
Foot of perpendicular from $P_1$ and $P_2$ on $y-$axis is $B.$
Now$, AP_1 = 5$
$\therefore$ In $\triangle \text{ABP}_1,\frac{\text{AB}}{\text{AP}_1}=\cos30^\circ$
$\therefore \text{AB}=\frac{5\sqrt{3}}{2}$
$\therefore \text{OB}=2+\frac{5\sqrt{3}}{2}$
So$,$ the coordinates of the perpendicular are $\Big(0,2+\frac{5\sqrt{3}}{2}\Big).$

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Question 95 Marks

Find the equation of one of the sides of an isosceles right angled triangle whose hypotenuse is given by 3x + 4y = 4 and the opposite vertex of the hypotenuse is (2, 2).

Answer

As shown in the figure, hypotenuse is along the line 3x + 4y + 4 = 0.
$\therefore$ Slope of $\text{AC}=\frac{-3}{4}.$
Since ABC is isosceles righta angled triangle,

Now, let the slope of the line making an angle 45° with AC be m.
$\therefore \tan 45^\circ=\begin{vmatrix}\frac{\text{m}-\Big(-\frac{3}{4}\Big)}{1+\text{m}\Big(-\frac{3}{4}\Big)} \end{vmatrix}$
$\Rightarrow \frac{4\text{m}+3}{4-3\text{m}}=\pm1$
$\Rightarrow 4\text{m}+3=4-3\text{m}$ or $4\text{m}+3=3\text{m}-4$
$\Rightarrow \text{m}=\frac{1}{7}$ or m = -7
So, if the slope of line BC is $\frac{1}{7}$ then the slope of line AB is -7.
So, equation of BC is: $\text{y}-2=\Big(\frac{1}{7}\Big)(\text{x}-2)$
⇒ x - 7y + 12 = 0.
Equation of AB is: y - 2 = -7(x - 2)
⇒ 7x + y - 16 = 0.

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Question 105 Marks

Find the equation of lines passing through $(1, 2)$ and making angle $30^\circ$ with $y-$axis.

Answer

Given that the makes angle $30^\circ$ with $y-$axis.
$\therefore$ Angle made by the line with $x-$axis is $60^\circ$
$\therefore$ Slope of the line
$\text{M}=\tan60^\circ$
$\Rightarrow \text{m}=\sqrt{3}$

So$,$ the equation of the line passing therought the point $(1, 2)$ and slop $\sqrt{3}$ is
$y - y_1 = m(x - x_1)$
$\Rightarrow \text{y}-2=\sqrt{3}(\text{x}-1)$
$\Rightarrow \text{y}-2=\sqrt{3}\text{x}-\sqrt{3}$
$\Rightarrow \text{y}-\sqrt{3}\text{x}+\sqrt{3}-2=0$
Hence$,$ the required equation of line is $\text{y}-\sqrt{3}\text{x}+\sqrt{3}-2=0.$

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Question 115 Marks

In what direction should a line be drawn through the point (1, 2) so that its point of intersection with the line x + y = 4 is at a distance $\frac{\sqrt{6}}{3}$ from the given point.

Answer

Let the given line x + y = 4 and required line 'l' intersect at B(a, b).
Slope of line 'l' is given by $\text{m}=\frac{\text{b}-2}{\text{a}-1}=\tan\theta\ .....(\text{i})$
Givne that $\text{AB}=\frac{\sqrt{6}}{3}$
So, by distance formula for point A(1, 2) and B(a, b), we get
$\sqrt{(\text{a}-1)^2+(\text{b}-2)^2}=\frac{\sqrt{6}}{3}$
On squaring both the side
$\text{a}^2+ -2\text{a}+\text{b}^2+4-4\text{b}=\frac{6}{9}$
$\text{a}^2\text{b}^2-2\text{a}-4\text{b}+5=\frac{2}{3}\ .....(\text{ii})$
Point B(a, b) also satisfies the eqn. x + y = 4
$\therefore \text{a}+\text{b}\ .....(\text{iii})$
On solving (ii) and (iii), we get $\text{a}=\frac{3\sqrt{3}+1}{2\sqrt{3}},\text{b}=\frac{5\sqrt{3}-1}{2\sqrt{3}}$
Putting values of a and b in eqn. (i), we have
$\tan\theta=\frac{\frac{5\sqrt{3}-1}{2\sqrt{3}}}{\frac{3\sqrt{3}+1}{2\sqrt{3}}}=\frac{5\sqrt{3}-1-4\sqrt{3}}{3\sqrt{3}+1-2\sqrt{3}}=\frac{\sqrt{3}-1}{\sqrt{3}+1}$
$\therefore \tan\theta=\tan15^\circ$
$\Rightarrow \theta = 15^\circ$

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