1 Marks Question - MATHS STD 11 Science Questions - Vidyadip

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

Question 11 Mark

If the centroid of a triangle formed by the points (0, 0), $(\cos \theta, \sin \theta)$ and $\sin \theta, - \cos \theta$ lies on the line y = 2x, then write the value of $\tan\theta.$

Answer

The centroid of a triangle with vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given below:
$\Big(\frac{\text{x}_1+\text{x}_2+\text{x}_3}{3},\frac{\text{y}_1+\text{y}_2+\text{y}_3}{3}\Big)$
Therefore, the centre of the triangle having vertices (0, 0), $(\cos \theta, \sin \theta)$ and $\sin \theta, - \cos \theta$ is
$\Big(\frac{0+\cos\theta+\sin\theta}{3},\frac{0+\sin\theta-\cos\theta}{3}\Big)\equiv\Big(\frac{\cos\theta+\sin\theta}{3},\frac{\sin\theta-\cos\theta}{3}\Big)$
This point lies on the line y = 2x
$\frac{\sin\theta-\cos\theta}{3}=2\times\frac{\cos\theta+\sin\theta}{3}$
$\Rightarrow\sin\theta-\cos\theta=2\cos\theta+2\sin\theta$
$\Rightarrow\tan\theta=-3$
$\therefore\tan\theta=-3$

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Question 21 Mark

Find the slopes of the lines which make the following angles with the positive direction of x-axis:
$\frac{3\pi}{4}$

Answer

Angle made with positive x axis is $\frac{3\pi}{4}$
$\therefore\text{m}=\tan\theta=\tan\Big(\frac{3\pi}{4}\Big)=\tan\Big(\pi-\frac{\pi}{4}\Big)=-1$

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Question 31 Mark

Find the slope of a line passing through the following points:
(3, 5), and (1, 2)

Answer

(3, 5), and (1, 2)
slope of line $=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{2-(-5)}{1-3}=\frac{7}{-2}=\frac{-7}{2}$

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Question 41 Mark

Write the coordinates of the orthocentre of the triangle formed by the lines x² - y² = 0 and x + 6y = 18.

Answer

The equation x² - y² = 0 represents a pair of straight line, which can be written in the following way:
(x + y)(x - y) = 0
So, the lines can be written separately in the following manner:
x + y = 0 ...(1)
x - y = 0 ...(2)
The third line is
x + 6y = 18 ... (3)
Lines (1) and (2) are perpendicular to each other as their slopes are -1 and 1, respectively
⇒ -1 × 1 = −1
Therefore, the triangle formed by the lines (1), (2) and (3) is a right-angled triangle.
Thus, the orthocentre of the triangle formed by the given lines is the intersection of x + y= 0 and x - y = 0, which is (0, 0).

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Question 51 Mark

Write the locus of a point the sum of whose distances from the coordinates axes is unity.

Answer

Let (h, k) be the locus.
It is given that the sum of distances of (h, k) from the coordinate axis is unity.
$\therefore$ |h| + |k| = 1
Taking locus of (h, k), we get:
|x| + |y| = 1
This represents a square.

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Question 61 Mark

Determine the distance between the following pair of parallel lines:
4x - 3y - 9 = 0 and 4x - 3y - 24 = 0

Answer

Determine between parallel lines
ax + by +c₁ = 0 and ax + by + c₂ = 0 is
$\Big|\frac{\text{c}_2-\text{c}_1}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$
4x - 3y - 9 = 0 and 4x - 3y - 24 = 0
Distance between the two parallel lines is
$\Big|\frac{-24-(-9)}{\sqrt{4^2+3^2}}\Big|=\Big|\frac{-24+9}{5}\Big|$
= 3 units.

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Question 71 Mark

Write the integral values of m for which the x-coordinate of the point of intersection of the lines y = mx + 1 and 3x + 4y = 9 is an integer.

Answer

The given lines can be written as
mx - y + 1 = 0 ...(1)
3x + 4y - 9 = 0 ...(2)
Solving (1) and (2) by cross multiplication, we get:
$\frac{\text{x}}{9-4}=\frac{\text{y}}{3+9\text{m}}=\frac{1}{4\text{m}+3}$
$\Rightarrow\text{x}=\frac{5}{4\text{m}+3},\text{y}=\frac{9\text{m}+3}{4\text{m}+3}$
For x to be integer we have, 4m + 3 = 1,-1, 5 and -5
$\Rightarrow \ \text{m}=\frac{-1}{2},-1,\frac{1}{2}$ and $-2.$
Hence, the integral values of m are -1 and -2.

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Question 81 Mark

If the diagonals of the quadrilateral formed by the lines l₁x + m₁y + n₁ = 0, l₂x + m₂y + n₂= 0, l₁x + m₁y + n₁' = 0 and l₂x + m₂y + n₂' = 0 are perpendicular, then write the value of l₁² - l₂² + m₁² - m₂².

Answer

The given lines are

l₁x + m₁y + n₁ = 0 ...(1)

l₂x + m₂y + n₂ = 0 ...(2)

l₁x + m₁y + n₁' = 0 ...(3)

l₂x + m₂y + n₂' = 0 ...(4)

Let (1), (2), (3) and (4) represent the sides AB, BC, CD and DA, respectively.

The equation of diagonal AC passing through the intersection of (2) and (3) is given by

$\text{l}_1\text{x} + \text{m}_1\text{y} + \text{n}_1' +\lambda(\text{l}_2\text{x} + \text{m}_2\text{y} + \text{n}_2) = 0$

$\Rightarrow(\text{l}_2+\lambda\text{l}_2)\text{x}+(\text{m}_1+\lambda\text{m}_2)\text{y}+(\text{n}_1'+\lambda\text{n}_2)=0$

⇒ Slope of diagonal $\text{AC}=-\Big(\frac{\text{l}_1+\lambda\text{l}_2}{\text{m}_1+\lambda\text{m}_2}\Big)$

Also, the equation of diagonal BD, passing through the intersection of (1) and (2), is given by

$\text{l}_1\text{x}+\text{m}_1\text{y}+\text{n}_1+\mu(\text{l}_2\text{x}+\text{m}_2\text{y}+\text{n}_2)=0$

$\Rightarrow(\text{l}_1+\mu\text{l}_2)\text{x}+(\text{m}_1+\mu\text{m}_2)\text{y}+(\text{n}_1+\mu\text{n}_2)=0$

⇒ Slope of diagonal $\text{BD}=-\Big(\frac{\text{l}_1+\mu\text{l}_2}{\text{m}_1+\mu\text{m}_2}\Big)$

The diagonals are perpendicular to each other.

$\therefore\Big(\frac{\text{l}_1+\lambda\text{l}_2}{\text{m}_1+\lambda\text{m}_2}\Big)\Big(\frac{\text{l}_1+\mu\text{l}_2}{\text{m}_1+\mu\text{m}_2}\Big)=-1$

$\Rightarrow(\text{l}_1-\text{l}_2)(\text{l}_1+\text{l}_2)=-(\text{m}_1-\text{m}_2)(\text{m}_1+\text{m}_2)$

$\Rightarrow\Big(\text{l}_1^2-\text{l}_2^2\Big)=-\Big(\text{m}_1^2-\text{m}_2^2\Big)$

$\Rightarrow\Big(\text{l}_1^2-\text{l}_2^2\Big)+\Big(\text{m}_1^2-\text{m}_2^2\Big)=0$

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Question 91 Mark

Write the distance between the lines 4x + 3y - 11 = 0 and 8x + 6y - 15 = 0.

Answer

The distance between the two parallel lines ax + by + c₁= 0 and ax + by + c₂= 0 is $\Big|\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$
The given lines can be written as
$4\text{x} + 3\text{y} - 11 = 0 \ ...(1)$
$8\text{x} + 6\text{y} - 15 = 0 ⇒ 4\text{x} + 3\text{y} - \frac{15}{2}=0 \ ...(2)$
Let d be the distance between the lines (1) and (2)
$\text{d}=\Big|\frac{-11-\big(-\frac{15}{2}\big)}{\sqrt{4^2+3^2}}\Big|=\frac{7}{10} \text{units}$

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Question 101 Mark

Write the value of $\theta\in\Big(0,\frac{\pi}{2}\Big)$ for which area of the triangle formed by points $\text{O}(0,0),\text{A} (\text{a} \cos \theta, \text{b} \sin \theta)$ and $\text{B} (\text{a} \cos \theta, − \text{b} \sin \theta)$ is maximum.

Answer

Let A be the area of the triangle formed by the points $\text{O}(0,0),\text{A} (\text{a} \cos \theta, \text{b} \sin \theta)$ and $\text{B} (\text{a} \cos \theta, − \text{b} \sin \theta)$
$\text{A}=\frac{1}{2}\begin{vmatrix}0&0&1\\\text{a}\cos\theta&\text{b}\sin\theta&1\\\text{a}\cos\theta&-\text{b}\sin\theta&1\end{vmatrix}$
$\Rightarrow\text{A}=\frac{1}{2}|(-\text{ab}\sin\theta\cos\theta-\text{ab}\sin\theta\cos\theta)|$
$\Rightarrow\text{A}=\frac{1}{2}=\text{ab}\sin\theta\cos\theta=\frac{1}{2}\sin2\theta$
Now,
$\therefore \ \text{A}_\text{max}=\frac{1}{2},$ when $\sin2\theta=1$
$\Rightarrow\therefore\text{A}_\text{max}=\frac{1}{2},$ when $2\theta=\frac{\pi}{2}\Rightarrow\theta=\frac{\pi}{4}$
Hence, the area of the triangle formed by the given points is maximum when $\theta=\frac{\pi}{4}.$

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Question 111 Mark

Write the area of the triangle formed by the coordinate axes and the line $(\sec \theta - \tan \theta) \text{x}+ (\sec \theta + \tan \theta) \text{y} = 2.$

Answer

The point of intersection of the coordinate axes is (0, 0).
Let us find the intersection of the line $(\sec \theta - \tan \theta) \text{x}+ (\sec \theta + \tan \theta) \text{y} = 2.$ and the coordinate axis.
For x-axis:
$\text{y}=0,\text{x}=\frac{2}{\sec\theta-\tan\theta}$
For x-axis:
$\text{x}=0,\text{y}=\frac{2}{\sec\theta+\tan\theta}$
Thus, the coordinates of the triangle formed by the coordinate axis and the line $(\sec \theta - \tan \theta) \text{x}+ (\sec \theta + \tan \theta) \text{y} = 2.$ are $(0, 0),$ $\Big(\frac{2}{\sec\theta-\tan\theta},0\Big)$ and $\Big(0,\frac{2}{\sec\theta+\tan\theta}\Big)$
Let A be the area of the required triangle.
$\therefore\text{A}=\frac{1}{2}\begin{vmatrix}0&0&1\\\frac{2}{\sec\theta-\tan\theta}&0&1\\0&\frac{2}{\sec\theta+\tan\theta}&1\end{vmatrix}$
$\Rightarrow\text{A}=\frac{1}{2}\times\frac{2}{\sec\theta-\tan\theta}\times\frac{2}{\sec\theta+\tan\theta}$
$\Rightarrow\text{A}=\frac{2}{(\sec\theta-\tan\theta)(\sec\theta+\tan\theta)}=\frac{2}{(\sec^2\theta+\tan^2\theta)}=2$
Hence, the area of the triangle is 2 square units.

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Question 121 Mark

Find the slopes of the lines which make the following angles with the positive direction of x-axis:
$\frac{2\pi}{3}$

Answer

Angle made with positive x axis is $\frac{2\pi}{3}$
$\therefore \text{m}=\tan\theta=\tan\Big(\frac{2\pi}{3}\Big)=-\sqrt3$

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Question 131 Mark

State whether the two lines in the following are parallel, perpendicular or neither.
Through (3, 15) and (16, 6); through (-5, 3) and (8, 2)

Answer

Slope of line joining (3, 15) and (16, 6)

$\text{m}_1=\frac{6-5}{16-3}=\frac{-9}{13}$

Slope of the line joining (-5, 3) and (8, 2)

f$\text{m}_2=\frac{2-3}{8-(-5)}=\frac{-1}{13}$

Here, neither m₁ = m₂ nor m₁ × m₂ = -1

$\therefore$ The lines are neither parallel nor perpendicular.

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Question 141 Mark

Write the coordinates of the orthocentre of the triangle formed by the lines xy = 0 and x + y = 1.

Answer

The equation xy = 0 represents a pair of straight lines.
The lines can be written separately in the following way:
x = 0 ...(1)
y = 0 ...(2)
The third line is
= x + y = 1 ...(3)
Lines (1) and (2) are perpendicular to each other as they are coordinate axes.
Therefore, the triangle formed by the lines (1), (2) and (3) is a right-angled triangle.
Thus, the orthocentre of the triangle formed by the given lines is the intersection of x = 0 and y = 0, which is (0, 0)

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Question 151 Mark

Determine the distance between the following pair of parallel lines:

y = mx + c and y = mx + d

Answer

Determine between parallel lines

ax + by +c₁ = 0 and ax + by + c₂ = 0 is

$\Big|\frac{\text{c}_2-\text{c}_1}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$

y = mx + c and y = mx + d

Distance between the two parallel lines is

$\Big|\frac{\text{c}-\text{d}}{\sqrt{\text{m}^2+1}}\Big|$

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Question 161 Mark

Write the equation of the line passing through the point (1, -2) and cutting off equal intercepts from the axes.

Answer

Let the equation of the required line be
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}} \ \big[\because$ the line has equal intercepts$\big]$
Now, it is passing through (1, -2)
$\therefore \ \frac{1}{\text{a}}-\frac{2}{\text{a}}=1$
$\Rightarrow\text{a}=-1$
Hence, the required equation is given by
$\frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1$
$\Rightarrow\text{x}+\text{y}+1=0$

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Question 171 Mark

State whether the two lines in the following are parallel, perpendicular or neither.
Through (9, 5) and (-1, 1); through (3, -5) and (8, -3)

Answer

Slope of line joining (-1, 1) and (9, 5)
$\text{m}_1=\frac{5-1}{9-(-1)}=\frac{4}{10}=\frac{2}{5}$
Slope of line joining (3, -5) and (9, 5)
$\text{m}_2=\frac{-3-(-5)}{8-3}=\frac{-3+5}{5}=\frac{2}{5}$
Here m₁ = m₂
$\therefore$ The two lines are parallel.

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Question 181 Mark

Write the area of the figure formed by the lines a |x| + b |y| + c = 0.

Answer

The given lines can be written separately in the following way:
$\text{ax}+\text{by}+\text{c}=0; \ \text{x},\text{y}\geq0 \ ...(1)$
$-\text{ax}+\text{by}+\text{c}=0; \ \text{x}<\text{y}\geq0 \ ...(2)$
$-\text{ax}-\text{by}+\text{c}=0; \ \text{x}<\text{y}<0 \ ...(3)$
$\text{ax}-\text{by}+\text{c}=0; \ \text{x}\geq0 \ \text{y}<0 \ ...(4)$
The lines and the region enclosed between them is shown below.

So, the area of the figures formed by the lines a|x| + b|y| + c = 0 is
$4\times\frac{1}{2}\big|\frac{\text{c}}{\text{a}}\times\frac{\text{c}}{\text{b}}\big|=\frac{2\text{c}^2}{|\text{ab}|}\text{squnits}$

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Question 191 Mark

The equations of two sides of a square are 5x - 12y - 65 = 0 and 5x - 12y + 26 = 0. Find the area of the square.

Answer

The two sides of square are
5x - 12y - 65 = 0 and 5x - 12y + 26 = 0
The distance between these two parallel sides $\Big($as both have slpoe $\frac{5}{12}\Big)$ is
$\Big|\frac{-65-26}{\sqrt{5^2+12^2}}\Big|=\Big|\frac{-91}{13}\Big|$
= 7 units.
And all sides of square are equal.
$\therefore$ Area of the square is 7 × 7 = 49 sq units.

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Question 201 Mark

Find the slope of a line passing through the following points:
$\big(\text{at}_1^2,2\text{at}_1\big)$ and $\big(\text{at}_2^2,2\text{at}_2\big)$

Answer

$\big(\text{at}_1^2,2\text{at}_1\big)$ and $\big(\text{at}_2^2,2\text{at}_2\big)$
slope of line $=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{2\text{at}_2-2\text{at}_1}{\text{at}_2^2-\text{at}_1^2}=\frac{2}{\text{t}_2+\text{t}_1}$

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Question 211 Mark

If a, b, c are in A.P., then the line ax + by + c = 0 passes through a fixed point. Write the coordinates of that point.

Answer

If, a, b, c are in A.P, then
a + c = 2b
⇒ a - 2b + c = 0
Comparing the coefficient of ax + by + c = 0 and a - 2b + c = 0, we get
x = 1 and y = -2
So, the the coordinates of that point is (1, -2)

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Question 221 Mark

If a, b, c are in G.P. write the area of the triangle formed by the line ax + by + c = 0 with the coordinates axes.

Answer

The point of intersection of the line ax + by + c = 0 with the coordinate axis are $\Big(-\frac{\text{c}}{\text{a}},0\Big)$ and $\Big(0,-\frac{\text{c}}{\text{b}}\Big)$
So, the vertices of the triangle are (0, 0), $\Big(-\frac{\text{c}}{\text{a}},0\Big)$ and $\Big(0,-\frac{\text{c}}{\text{b}}\Big)$
Let A be the area of the required triangle.
$\text{A}=\frac{1}{2}\begin{vmatrix}0&0&1\\\frac{\text{-c}}{\text{a}}&0&1\\0&\frac{\text{-c}}{\text{b}}&1\end{vmatrix}$
$\text{A}=\frac{1}{2}\big|-\frac{\text{c}}{\text{a}}\times\frac{\text{-c}}{\text{a}}\big|=\frac{1}{2}\big|\frac{\text{c}^2}{\text{ab}}\big|$
It is given that a, b and c are in GP.
$\therefore \ \text{b}^2=\text{ac}$
$\Rightarrow\text{A}=\frac{1}{2}\big|\frac{\text{b}^4}{\text{a}^2\times\text{ab}}\big|=\frac{1}{2}\big|\frac{\text{b}}{\text{a}}\big|^3 \ \text{sq units}$

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Question 231 Mark

Determine the distance between the following pair of parallel lines:
4x + 3y - 11 = 0 and 8x + 6y = 15

Answer

Determine between parallel lines

ax + by +c₁ = 0 and ax + by + c₂ = 0 is

$\Big|\frac{\text{c}_2-\text{c}_1}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$

4x + 3y - 11 = 0 and 8x + 6y = 15

Distance between the two parallel lines is

$\Big|\frac{-11-15}{\sqrt{4^2+3^2}}\Big|=\frac{7}{10}\text{units}$

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Question 241 Mark

Find the locus of the mid-points of the portion of the line $\text{x}\sin\theta+ \text{y}\cos\theta =\text{p}$ intercepted between the axes.

Answer

We have $\text{x}\sin\theta+ \text{y}\cos\theta =\text{p}$
$\Rightarrow\frac{\text{x}}{\frac{\text{p}}{\sin\theta}}+\frac{\text{y}}{\frac{\text{p}}{\cos\theta}}=1$
So, the x and y intercepts are given by
$\Big(\frac{\text{p}}{\sin\theta},0\Big)$ and $\Big(0,\frac{\text{p}}{\cos\theta}\Big)$
Now, let the coordinates of the mid point be (h, k)
$\therefore\text{h}=\frac{\frac{\text{p}}{\sin\theta}+0}{2}$ and $\text{k}=\frac{0+\frac{\text{p}}{\cos\theta}}{2}$
$\Rightarrow\text{h}=\frac{\text{p}}{2\sin\theta}$ and $\text{k}=\frac{\text{p}}{2\cos\theta}$
$\Rightarrow\sin\theta=\frac{\text{p}}{2\text{h}}$ and $\cos\theta=\frac{\text{p}}{2\text{k}}$
$\Rightarrow\sin^2\theta=\frac{\text{p}^2}{2\text{h}^2}$ and $\cos^2\theta=\frac{\text{p}^2}{2\text{k}^2}$
Now, squaring and adding, we get
$\sin^2\theta+\cos^2\theta=\frac{\text{p}^2}{4\text{h}^2}+\frac{\text{p}^2}{4\text{k}^2}$
$\Rightarrow1=\frac{\text{p}^2}{4\text{h}^2}+\frac{\text{p}^2}{4\text{k}^2}$
$\Rightarrow\frac{4}{\text{p}^2}=\frac{1}{\text{h}^2}+\frac{1}{\text{k}^2}$
since, (h, k) is the mid point, so it will also pass through xsinθ+ ycosθ = p.
Hence, the given equation of locus can also be written as:
$\frac{4}{\text{p}^2}=\frac{1}{\text{x}^2}+\frac{1}{\text{y}^2}$

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Question 251 Mark

Find the slopes of the lines which make the following angles with the positive direction of x-axis:
$-\frac{\pi}{4}$

Answer

Angle made with positive x axis is $-\frac{\pi}{4}.$
$\therefore\text{m}=\tan\theta=\tan\Big(-\frac{\pi}{4}\Big)=-1$

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Question 261 Mark

State whether the two lines in the following are parallel, perpendicular or neither.
Through (6, 3) and (1, 1); through (-2, 5) and (2, -5)

Answer

Slope of line joining (6, 3) and (1, 1)
$\text{m}_1 =\frac{1-3}{1-6}=\frac{-2}{-5}=\frac{2}{5}$
Slope of line joining (-2, 5) and (2, -5)
$\text{m}_2=\frac{\text{5}-{5}}{2-(-2)}=\frac{-10}{4}=\frac{-5}{2}$
Here $\text{m}_1\times\text{m}_2=\frac{2}{5}\times\frac{-5}{2}=-1$
$\therefore$ The lines are perpendicular to each other.

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Question 271 Mark

Write the coordinates of the image of the point (3, 8) in the line x + 3y - 7 = 0.

Answer

Let the given point be A(3, 8) and its image in the line x + 3y - 7 = 0 is B(h, k).
The midpoint of AB is $\Big(\frac{3+\text{h}}{2},\frac{8+\text{k}}{2}\Big)$ that lies on the line x + 3y - 7 = 0
$\therefore\frac{ 3+\text{h}}{2}+3\times\frac{8+\text{k}}{2}-7=0$
$\text{h}+3\text{k}+13=0 \ ...(1)$
AB and the line x + 3y - 7 = 0 are perpendicular.
$\therefore$ Slope of AB × Slope of the line = -
$\Rightarrow\frac{\text{k}-8}{\text{h}-3}\times\frac{-1}{3}=-1$
$\Rightarrow3\text{h}-\text{k}-1 \ ...(2)$
Solving (1) and (2), we get:
(h, k) = (-1, -4)
Hence, the image of the point (3, 8) in the line x + 3y - 7 = 0 is (-1, -4).

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Question 281 Mark

Determine the distance between the following pair of parallel lines:
8x + 15y - 34 = 0 and 8x + 15y + 31 = 0

Answer

Determine between parallel lines
ax + by +c₁ = 0 and ax + by + c₂ = 0 is
$\Big|\frac{\text{c}_2-\text{c}_1}{\sqrt{\text{a}^2+\text{b}^2}}\Big|$
8x + 15y - 34 = 0 and 8x + 15y + 31 = 0
Distance between the two parallel lines is
$\Big|\frac{-34-31}{\sqrt{8^2+15^2}}\Big|=\frac{65}{17}\text{units}$

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Question 291 Mark

If a ≠ b ≠ c, write the condition for which the equations (b - c) x + (c - a) y + (a - b) = 0 and (b³ - c³)x + (c³ - a³)y + (a³ - b³) = 0 represent the same line.

Answer

The given lines are
(b - c)x + (c - a)y + (a - b) = 0 ...(1)
(b³ - c³)x + (c³ - a³)y + (a³ - b³) = 0 ...(2)
The lines (1) and (2) will represent the same lines if
$\frac{\text{b}-\text{c}}{\text{b}^3-\text{c}^3}=\frac{\text{c}-\text{a}}{\text{c}^3-\text{a}^3}=\frac{\text{a}-\text{b}}{\text{a}^3-\text{b}^3}$
$\Rightarrow\frac{\text{b}-\text{c}}{(\text{b}-\text{c})(\text{b}^2+\text{bc}+\text{c}^2)}=\frac{\text{c}-\text{a}}{(\text{c}-\text{a})(\text{c}^2+\text{ac}+\text{a}^2)}=\frac{\text{a}-\text{b}}{(\text{a}-\text{b})(\text{a}^2+\text{ab}+\text{b}^2)}$
$\Rightarrow\frac{1}{(\text{b}^2+\text{bc}+\text{c}^2)}=\frac{1}{(\text{c}^2+\text{ac}+\text{a}^2)}=\frac{1}{(\text{a}^2+\text{ab}+\text{b}^2)} \ (\because\text{a}\neq\text{b}\neq \text{c})$
$\Rightarrow\text{b}^2+\text{bc}+\text{c}^2=\text{c}^2+\text{ac}+\text{a}^2$ and $\text{c}^2+\text{ac}+\text{a}^2=\text{a}^2+\text{ab}+\text{b}^2$
$\Rightarrow(\text{a}-\text{b})(\text{a}+\text{b}+\text{c})=0$ and $(\text{b}-\text{c})(\text{b}+\text{c}+\text{a})=0$
$\Rightarrow\text{a}+\text{b}+\text{c} \ (\because\text{a}\neq\text{b}\neq\text{c})$
Hence, the given lines will represent the same lines if a + b + c = 0.

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Question 301 Mark

Find the slope of a line passing through the following points:
(-3, 2) and (1, 4)

Answer

(-3, 2) and (1, 4)
slope of line $=\frac{\text{y}_2-\text{y}_1}{\text{x}_2-\text{x}_1}=\frac{4-2}{1-(-3)}=\frac{2}{3}=\frac{1}{2}$

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Question 311 Mark

If the lines x + ay + a = 0, bx + y + b = 0 and cx + cy + 1 = 0 are concurrent, then write the value of 2abc - ab - bc - ca.

Answer

The given lines are

x + ay + a = 0 ...(1)

bx + y + b = 0 ...(2)

cx + cy + 1 = 0 ...(3)

It is given that the lines (1), (2) and (3) are concurrent.

$\therefore\begin{vmatrix}1&\text{a}&\text{a} \\ \text{b}&1&\text{b}\\\text{c}&\text{c}&1\end{vmatrix}=0$

$\Rightarrow(1-\text{bc})-\text{a}(\text{b}-\text{bc})+\text{a}(\text{bc}-\text{c})=0$

$\Rightarrow1-\text{bc}-\text{ab}+\text{abc}+\text{abc}-\text{ac}=0$

$\Rightarrow2\text{abc}-\text{ab}-\text{bc}-\text{ca}=-1$

Hence, the value of 2abc - ab - bc - ca is -1

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Question 321 Mark

Find the slopes of the lines which make the following angles with the positive direction of x-axis:
$\frac{\pi}{3}$

Answer

Angle made with positive x axis is $\frac{\pi}{3} $
$\therefore\text{m}=\tan\theta=\tan\Big(\frac{\pi}{3}\Big)=\sqrt3$

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Question 331 Mark

Find the equation of the straight line which passes through (1, -2) and cuts off equal intercepts on the axes.

Answer

The equation of straight line in the intercepts form is

$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1 \ ...(\text{i})$

If (i) passes through point (1, -2) ans has equal intercepts (a = b = k), we get,

$\frac{1}{\text{k}}+\frac{(-2)}{\text{k}}=1$

$\frac{-1}{\text{k}}-\frac{2}{\text{k}}=1$

$1-2 =\text{k}$

$\text{k}=-1$

$\Rightarrow\text{a}=\text{b}=-1$

Putting in (i)

$\frac{\text{x}}{-1}+\frac{\text{y}}{-1}=1$

$\text{x}+\text{y}=-1$

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Question 341 Mark

Find the equation to the straight line:

Cutting off intercepts -5 and 6 from the axes.

Answer

If (a, 0) and (0, b) are the intercepts of a line then the intercept form of equation is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$

Here, a = -5, b = 6

$\therefore$ The required equation is

$\frac{\text{x}}{-5}+\frac{\text{y}}{6}=1$

$\Rightarrow6\text{x}-5\text{y}=-30$

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Question 351 Mark

Find the equation to the straight line:
Cutting off intercepts 3 and 2 from the axes.

Answer

If (a, 0) and (b, 0) are the intercepts of the line then the intercept form the equation is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$

Here, a = 3, b = 2

$\therefore$ The required equation is

$\frac{\text{x}}{\text{3}}+\frac{\text{y}}{2}=1$

$\Rightarrow2\text{x}+3\text{y}=6$

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Question 361 Mark

Write an equation representing a pair of lines through the point (a, b) and parallel to the coordinate axes.

Answer

The lines passing through (a, b) and parallel to the x-axis and y-axis are y = b and x = a respectively.

Therefore, their combined equation is given below:

(x - a)(y - b) = 0

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