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Maths 3 Marks Question

Straight Lines · CBSE STD 11 Science (CBSE - English Medium). 9 questions.

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Question 13 Marks
Match the questions given under $\text{Column} \ C_1$ with their appropriate answers given under the $\text{Column} \ C_2:$
  $\text{Column} \ C_1$   $\text{Column}\  C_2$
$(a)$ The coordinates of the points $P$ and $Q$ on the line $x + 5y = 13$ which are at a distance of $2$ units from the line $12x - 5y + 26 = 0$ are (i) $(3, 1), (-7, 11) $
$(b)$ The coordinates of the point on the line $x + y = 4,$ which are at a unit distance from the line $4x + 3y - 10 = 0$ are (ii) $-\frac{1}{3},\frac{11}{3},\frac{4}{3},\frac{7}{3}$
$(c)$ The coordinates of the point on the line joining $A (-2, 5)$ and $B (3, 1)$ such that$ AP = PQ = QB$ are (iii) $1,\frac{12}{5},-3,\frac{16}{5}$
Answer
  $\text{Column} C_1$   $\text{Column}\  C_2$
$(a)$ The coordinates of the points $P$ and $Q$ on the line $x + 5y = 13$ which are at a distance of 2 units from the line $12x - 5y + 26 = 0$ are $(iii)$ $1,\frac{12}{5},-3,\frac{16}{5}$
$(b)$ The coordinates of the point on the line $x + y = 4,$ which are at a unit distance from the line $4x + 3y - 10 = 0$ are $(i)$ $(3, 1), (-7, 11) $
$(c)$ The coordinates of the point on the line joining $A (-2, 5)$ and $B (3, 1)$ such that $AP = PQ = QB$ are $(ii)$ $-\frac{1}{3},\frac{11}{3},\frac{4}{3},\frac{7}{3}$

Let $P(x_1, y_1)$ be any point of the given line
$x + 5y = 13 \therefore x_1 + 5y_1 = 13$ Distance of line $12x - 5y + 26 = 0$ from the point $P(x_1, y_1) 2=\bigg|\frac{12\text{x}_1-5\text{y}_1+26}{\sqrt{(12)^2+(-5)^2}}\bigg|$
$\Rightarrow 2=\bigg|\frac{12\text{x}_1-(13-\text{x}_1)+26}{13}\bigg|$
$\Rightarrow 2=\bigg|\frac{12\text{x}-1-13+\text{x}_1+26}{13}\bigg|$
$\Rightarrow 2=\Big|\frac{13\text{x}_1+13}{13}\Big|$
$\Rightarrow 2=\pm(\text{x}_1+1)$
$\Rightarrow 2=\text{x}_1+1$
$\Rightarrow x_1 = 1 ($Taking $(+)$ sign$)$ and $2 = -x_1 - 1$
$\Rightarrow x_1 = -3 ($Taking $(-)$ sing$)$ Putting the values of $x_1$ in eq. $x_1 + 5y = 13.$ We get $\text{y}_1=\frac{12}{5}$ and $\frac{16}{5}.$
So$,$ the required points are $\Big(1,\frac{12}{5}\Big)$ and $\Big(-3,\frac{16}{5}\Big)$
Hence$, (a)$ ⇔$ (iii).$
Let $P (x_1, y_1)$ be any point on the given line
$x + y = 4$
$\therefore x_1 + y_1 = 4 .....(i)$ Distance of the line $4x + 3y - 10 = 0$ from the point $P(x_1, y_1)$
$\Rightarrow 1 =\bigg|\frac{4\text{x}_1+3\text{y}_1-10}{\sqrt{(4)^2+(3)^2}}\bigg|$
$\Rightarrow 1=\Big|\frac{4\text{x}_1+3(4-\text{x}_1)-10}{5}\Big|$
$\Rightarrow 1 = \Big|\frac{4\text{x}_1+12-3\text{x}_1-10}{5}\Big|$
$\Rightarrow 1=\Big|\frac{\text{x}_1+2}{5}\Big|$
$\Rightarrow 1=\pm\Big(\frac{\text{x}_1+2}{5}\Big)$
$\Rightarrow \frac{\text{x}_1+2}{5}=1 ($Taking $(+)$ sign$)$
$\Rightarrow \text{x}_!+2=5$
$\Rightarrow \text{x}_1=3$ and $\frac{\text{x}_1+2}{5}=-1 ($Taking $(-)$ sign$)$
$\Rightarrow \text{x}_1+2=-5\Rightarrow \text{x}_1=-7$ Putting the values of $x_1$ in eq. $(i)$ we get $x_1 + y_1 = 4$ at$ x_1 = 3, y_1 = 1$ at $x_1 = -7, y_1 = 11$
So$,$ the required are $(3, 1)$ and $B(-7, 11)$
Hence$, (b)$ ⇔ $(i).$
Given that $AP = PQ = QB$
Equation of line joining $A(-2, 5)$ and $B(3, 1)$ is $\text{y}-5=\frac{1-5}{3+2}(\text{x}+2)$


$\Rightarrow \text{y}-5=\frac{-4}{5}(\text{x}+2)$
$\Rightarrow 5y - 25 = -4x - 8$
$\Rightarrow 4x + 5y - 17 = 0$ Let $P(x_1, y_1)$ and $Q(x_2, y_2)$ be any two points on the  divides the line the in the ratio $1 : 2$
$\therefore \text{x}_1=\frac{1.3+2(-2)}{1+2}=\frac{3-4}{3}=\frac{-1}{3}$
$\text{y}_1=\frac{1.1+2.5}{1+2}=\frac{1+10}{3}=\frac{11}{3}$ So, the coordinates of $\text{P}(\text{x}_!,\text{y}_1)=\Big(\frac{-1}{3},\frac{11}{3}\Big).$ Now point$ Q(x_2, y_2)$ is the mid point of $PB$ $\therefore \text{x}_2=\frac{3-\frac{1}{3}}{2}=\frac{4}{3}$
$\text{y}_2=\frac{1+\frac{11}{3}}{2}=\frac{7}{3}$
Hence$,$ the coordinates of $\text{Q}(\text{x}_2,\text{y}_2)=\Big(\frac{4}{3},\frac{7}{3}\Big)$
Hence$, (c)$ ⇔$ (ii).$

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Question 23 Marks
Find the equation of the lines which passes through the point (3, 4) and cuts off intercepts from the coordinate axes such that their sum is 14.
Answer
Equation of line in intercept form is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
Given that, a + b = 14 ⇒ b = 14 - a
So, equation of line is: $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{14-\text{a}}=1$
Since it passes through the point (3, 4), we have
$\Rightarrow\frac{3}{\text{a}}+\frac{4}{14-\text{a}}=1$
⇒ a2 - 13a + 42 = 0
⇒ (a - 7)(a - 6) = 0
$\therefore$ a = 7, then b = 7
When a = 7, then b = 7
When a = 6, then b = 8
Thus, equation of line is: $\frac{\text{x}}{7}+\frac{\text{y}}{7}=1,$
i.e., x + y = 7 or $\frac{\text{x}}{6}+\frac{\text{y}}{8}=1$
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Question 33 Marks
For what values of a and b the intercepts cut off on the coordinate axes by the line ax + by + 8 = 0 are equal in length but opposite in signs to those cut off by the line 2x - 3y + 6 = 0 on the axes.
Answer
The given equation are ax + by + 8 = 0 .....(i)
⇒ ax + by + 8 = 0
⇒ ax + by = -8
$\Rightarrow \frac{\text{a}}{-8}\text{x}+\frac{\text{b}}{-8}\text{y}=1$
$\Rightarrow \frac{\text{x}}{\frac{-8}{\text{a}}}+\frac{\text{y}}{\frac{-8}{\text{b}}}=1$
So, the intercepts ono the axes are $\frac{-8}{\text{a}}$ and $\frac{-8}{\text{b}}$
From eq. (ii), we get
⇒ 2x - 3y + 6 = 0
⇒ 2x - 3y = -6
$\Rightarrow \frac{2\text{x}}{6}-\frac{3\text{y}}{-6}=1$
$\Rightarrow \frac{\text{x}}{-3}+\frac{\text{y}}{2}=1$
So, the intercepts are -3 and 2.
According to the question
$\Rightarrow \frac{-8}{\text{a}}=+3$
$\Rightarrow \text{a}=-\frac{8}{3}$
and $\frac{-8}{\text{b}}=-2\Rightarrow \text{b}=+4$
Hence, the required values of a and b are $\frac{-8}{3}$ and 4.
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Question 43 Marks
Find the equation of the circle which touches the both axes in first quadrant and whose radius is a.
Answer
Intercept form of straight line
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ where a and b are the intercepts on the axis
Given that a = b
$\therefore \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\ .....(\text{i})$
If eq. (i) passes throught the point (1, -2), we get
$\Rightarrow\frac{1}{\text{a}}-\frac{2}{\text{a}}=1$
$\Rightarrow -\frac{1}{\text{a}}=1$
$\Rightarrow \text{a}=-1$
So, equation of the straight line is
$\Rightarrow\frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1$
$\Rightarrow \text{x}+\text{y}=-1$
$\Rightarrow \text{x}+\text{y}+1=0$
Hence, the required equation is x + y = 1 = 0.
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Question 53 Marks
Find the equation of the line passing through the point of intersection of 2x + y = 5 and x + 3y + 8 = 0 and parallel to the line 3x + 4y = 7.
Answer
Given lines are:
2x + y = 5 .....(i)
x + 3y = -8 .....(ii)
Solving (i) and (ii), we get their point intersection as $\Big(\frac{23}{5},\frac{-21}{5}\Big).$
Slope of line 3x + 4y = 7 is $\frac{-3}{4}.$ So, the line parallel to this line has slope $\frac{-3}{4}.$
Then the equation of the line passing through the point $\Big(\frac{23}{5},\frac{-21}{5}\Big)$ having slope $\frac{-3}{4}$ is:
$\text{y}+\frac{21}{5}=\frac{-3}{4}\Big(\text{x}-\frac{23}{5}\Big)$
$\Rightarrow 4\text{y}+\frac{84}{5}=-3\text{x}+\frac{69}{5}$
$\Rightarrow 3\text{x}+4\text{y}=\frac{84-69}{5}=3$
$\Rightarrow 3\text{x}+4\text{y}+3=0$
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Question 63 Marks
If the sum of the distances of a moving point in a plane from the axes is 1, then find the locus of the point.
[Hint: Given that |x| + |y| = 1, which gives four sides of a square.]
Answer
Let coordinates of a moving point P be (x, y).
Given that the sum of the distance from the axes to the point is always 1

$\therefore$ |x| + |y| = 1
⇒ x + y = 1
⇒ -x - y = 1
⇒ -x + y = 1
⇒ x - y = 1
Hence, these equations gives up the locus of the point P which is a square.
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Question 73 Marks
A variable line passes through a fixed point $P.$ The algebraic sum of the perpendiculars drawn from the points $(2, 0), (0, 2)$ and $(1, 1)$ on the line is zero. Find the coordinates of the point $P.$
$[$Hint: Let the slope of the line be m. Then the equation of the line passing through the fixed point $P (x_1, y_1)$ is $y - y_1 = m (x - x_1)$. Taking the algebraic sum of perpendicular distances equal to zero$,$ we get $y - 1 = m (x - 1).$ Thus $(x_1, y_1)\  is\  (1, 1).]$
Answer
Let the variable line throught the fixed point $P$ is $ax + by + c = 0 .....(i)$
Perpendicular distance from $\text{A}(2,0)=\frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}$
Perpendicular distance from $\text{A}(0,2)=\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}$
Perpendicular distance from $\text{A}(1, 1)=\frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}$
According to the question$,$ we have
$\Rightarrow \frac{2\text{a}+0+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{0+2\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}+\frac{\text{a}+\text{b}+\text{c}}{\sqrt{\text{a}^2+\text{b}^2}}=0$
$\Rightarrow 3a + 3b + 3c = 0 or a + b + c = 0 .....(ii)$
From $(i)$ and $(ii),$ variable line passes through the fixed point $(1, 1).$
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Question 83 Marks
If $p$ is the length of perpendicular from the origin on the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ and $a^2, p^2, b^2$ are in $A.P,$ then show that $a^4 + b^4 = 0.$
Answer
Since $p$ is the length of perpendicular fro m the origin on the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$
We have
$\Rightarrow\text{p}=\frac{0+0-1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}}}=\frac{\text{ab}}{\sqrt{\text{a}^2+\text{b}^2}}$
$\Rightarrow \text{p}^2=\frac{\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}$
Given that, $a^2, p^2$ and $b^2$ are in $A.P.$
$\therefore 2\text{p}^2=\text{a}^2+\text{b}^2$
$\Rightarrow \frac{2\text{a}^2\text{b}^2}{\text{a}^2+\text{b}^2}=\text{a}^2+\text{b}^2$
$\Rightarrow 2\text{a}^2\text{b}^2=(\text{a}^2+\text{b}^2)^2$
$\Rightarrow 2\text{a}^2\text{b}^2=\text{a}^2+\text{b}^4+2\text{a}^2\text{b}^2$
$\Rightarrow \text{a}^4+\text{b}^4=0$
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Question 93 Marks
Show that the tangent of an angle between the lines $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ and $\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=1$ is $\frac{2\text{ab}}{\text{a}^2-\text{b}^2}.$
Answer
Given that: $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\ .....(\text{i})$
and $\frac{\text{x}}{\text{a}}-\frac{\text{y}}{\text{b}}=1\ .....(\text{ii})$
Slope of $eq. (i) m_1 (say) = -\frac{\text{b}}{\text{a}}$
And slope of$ eq. (ii) m_2 (say) = \frac{\text{b}}{\text{a}}$
Let $\theta$ be the angle between the equation $(i)$ and $(ii)$
$\therefore \tan\theta=\bigg|\frac{\text{m}_1-\text{m}_2}{1+\text{m}_1\text{m}_2}\bigg|=\begin{vmatrix} \frac{-\frac{\text{b}}{\text{a}}-\frac{\text{b}}{\text{a}}}{1+\Big(-\frac{\text{b}}{\text{a}}\Big)\Big(\frac{\text{b}}{\text{a}}\Big)} \end{vmatrix}$
$\Rightarrow \tan\theta=\begin{vmatrix} \frac{-\frac{2\text{b}}{\text{a}}}{1-\frac{\text{b}^2}{\text{a}^2}} \end{vmatrix}=\bigg|\frac{-2\text{ab}}{\text{a}^2-\text{b}^2}\bigg|$
$\Rightarrow\tan\theta=\frac{2\text{ab}}{\text{a}^2-\text{b}^2}.$
Hence proved.
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