True False[1 Marks ]

Question 11 Mark

The straight line 5x + 4y = 0 passes through the point of intersection of the straight lines x + 2y - 10 = 0 and 2x + y + 5 = 0.

Answer

True.
Solution:
Given equation are x + 2y - 10 = 0 .....(i)
and 2x + y + 5 = 10 .....(ii)
From eq, (i) x = 10 - 2y .....(iii)
Putting the value of x in eq. (ii) we get
2(10 - 2y) + y + 5 = 0
⇒ 20 - 4y + y + 5 = 0
⇒ -3y + 25 = 0
$\Rightarrow \text{y}=\frac{25}{3}$
Putting the value of y in eq. (iii) we get
$=\frac{30-50}{3}=\frac{-20}{3}$
$\therefore \text{Point}=\Big(\frac{-20}{3},\frac{25}{3}\Big)$
If the given line 5x + 4y = 0 passes through the point $\Big(\frac{-20}{3},\frac{25}{3}\Big)$ then
$5\Big(\frac{-20}{3}\Big)+4\Big(\frac{25}{3}\Big)=0$
$\Rightarrow \frac{-100}{3}+\frac{100}{3}=0$
$\Rightarrow 0=0$ satisfied.
So, the given line passes through the point of intersection of the given lines.
Hence, the given statement is True.

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Question 21 Mark

The line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$ moves in such a way that $\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}^2},$ where c is a constant. The locus of the foot of the perpendicular from the origin on the given line is $x^2 + y^2 = c^2.$

Answer

True.
The given equation is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\ .....(\text{i})$
Equation of the line passing through $(0, 0)$ and perpendicular to eq. $(i)$ is
$\frac{\text{x}}{\text{b}}-\frac{\text{y}}{\text{a}}=0 \ .....(\text{ii})$
Squaring and adding eq. $(i)$ and $(ii)$ we get
$\Big(\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}\Big)^2+\Big(\frac{\text{x}}{\text{b}}-\frac{\text{y}}{\text{a}}\Big)^2=1+0$
$\Rightarrow \frac{\text{x}^2}{\text{a}^2}+\frac{\text{y}^2}{\text{b}^2}+\frac{2\text{xy}}{\text{ab}}+\frac{\text{x}^2}{\text{b}^2}+\frac{\text{y}^2}{\text{a}^2}-\frac{2\text{xy}}{\text{ab}}=1$
$\Rightarrow \text{x}^2\Big(\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}\Big)+\text{y}^2\Big(\frac{1}{\text{b}^2}+\frac{1}{\text{a}^2}\Big)=1$
$\Rightarrow (\text{x}^2+\text{y}^2)\Big(\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}\Big)=1$
$\Rightarrow (\text{x}^2+\text{y}^2)\Big(\frac{1}{\text{c}^2}\Big)=1\Big[\therefore \frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}=\frac{1}{\text{c}^2}\Big]$
$\Rightarrow \text{x}^2+\text{y}^2=\text{c}^2$
Hence$,$
the given statement is True.

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Question 31 Mark

Line joining the points (3, -4) and (-2, 6) is perpendicular to the line joining the points (-3, 6) and (9, -18).

Answer

False.
Solution:
Given points are A(3, -4), B(-2, 6), C(-3, 6) and D(9, -18).
Now, slope of $\text{AB}=\frac{6+4}{-2-3}=-2$
And slope of $\text{CD}=\frac{-18-6}{9+3}=-2$
So, line AB is paralle to line CD.

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Question 41 Mark

The vertex of an equilateral triangle is (2, 3) and the equation of the opposite side is x + y = 2. Then the other two sides are $\text{y}-3=(2\pm\sqrt{3})(\text{x}-2).$

Answer

True.
Solution:
Let ABC be an equilateral with vertex A(2, 3)
Equation of BC is x + y = 2.
Slope of BC = -1.
Let slope of line AB be m.
Noe, the angle between line AB abd BC is 60°.
$\therefore \tan 60^\circ=\Big|\frac{\text{m}+1}{1-\text{m}}\Big|$
$\Rightarrow \frac{\text{m}+1}{1-\text{m}}=\pm\sqrt{3}$
$\Rightarrow \text{m}=2\pm\sqrt{3}$
Equation of other two sides is $\text{y}-3=(2\pm\sqrt{3})(\text{x}-2).$

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Question 51 Mark

The points A (-2, 1), B (0, 5), C (-1, 2) are collinear.

Answer

False.
Solution:
Given points are A(-2, 1), B(0, 5) and C(-1, 2) are collinear.
Slope of $\text{AB}=\frac{5-1}{0+2}=2$
Slope of $\text{BC}=\frac{2-5}{-1-0}=3$
Since the slopes are different, A, b and C are not collinear.

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Question 61 Mark

If the vertices of a triangle have integral coordinates, then the triangle can not be equilateral.

Answer

True.
Solution:
We know that if the vertices of triangle has integral coordinates, then the triangle cannot be equilateral. So, the given statement is True.

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Question 71 Mark

The lines ax + 2y + 1 = 0, bx + 3y + 1 = 0 and cx + 4y + 1 = 0 are concurrent if a, b, c are in G.P.

Answer

False.
Solution:
Given lines are
ax + 2y + 1 = 0 .....(i)
bx + 3y + 1 = 0 .....(ii)
cx + 4y + 1 = 0 .....(iii)
Soliving (i) and (ii) by cross-multiplication method, we get
$\frac{\text{x}}{2-3}=\frac{-\text{y}}{\text{a}-\text{b}}=\frac{1}{3\text{a}-2\text{b}}$
So, the point of intersection is $\Big(\frac{1}{2\text{b}-3\text{a}},\frac{\text{a}-\text{b}}{2\text{b}-3\text{a}}\Big)$
Since, this point lies on cx + 4y + 1 = 0, then
$\frac{\text{c}}{2\text{b}-3\text{a}}+\frac{4(\text{a}-\text{b})}{2\text{b}-\text{a}}+1=0$
$\Rightarrow \text{c}+4\text{a}-4\text{b}+2\text{b}-3\text{a}=0$
$\Rightarrow 2\text{b}=\text{a}+\text{c}$
Hence, a, b, c are in A.P.

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Question 81 Mark

Equation of the line passing through the point $(\text{a} \cos^3\theta, \text{a} \sin^3 \theta)$ and perpendicular to the line $\text{x}\sec\theta+\text{y cosec}\ \theta=\text{a}$ is $\text{x}\cos\theta-\text{y}\sin\theta=\text{a}\sin2\theta.$

Answer

False.
Solution:
Equation of any line perpendicular to $\text{x}\sec\theta+\text{y cosec}\ \theta+\text{y cosec} \theta = \text{a}$ is $\text{x cosec}\ \theta - \text{y} \sec \theta = \text{k}\ .....(\text{i})$
If eq. (i) passes through $(\text{a}\sin^3\theta,\text{a}\sin^3\theta)$ then
$\text{a}\cos^3\theta.\text{cosec} \theta-\text{a}\sin^3\theta.\sec\theta=\text{k}$
$\Rightarrow \frac{\text{a}\cos^3\theta}{\sin\theta}-\frac{\text{a}\sin^3\theta}{\cos\theta}=\text{k}$
$\therefore$ Required equation is
$\text{x cosec} \ \theta-\text{y} \sec\theta=\frac{\text{a}\cos^3\theta}{\sin\theta}-\frac{\text{a}\sin^3\theta}{\cos\theta}$
$\Rightarrow \frac{\text{x}}{\sin\theta}-\frac{\text{y}}{\sin\theta}=\text{a}\Big[\frac{\cos^4\theta-\sin^4\theta}{\sin\theta\cos\theta}\Big]$
$\Rightarrow \frac{\text{x}\cos\theta-\text{y}\sin\theta}{\sin\theta\cos\theta}=\text{a}\Big[\frac{(\cos^2\theta+\sin^2\theta)(\cos^2\theta-\sin^2\theta)}{\sin\theta\cos\theta}\Big]$
$\Rightarrow \text{x}\cos\theta-\text{y}\sin\theta=\text{a}(\cos^2\theta-\sin^2\theta)$
$\Rightarrow \text{x}\cos\theta-\text{y}\sin\theta=\text{a}\cos 2\theta$
Hence, the given statement is false.

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Question 91 Mark

The equation of the line joining the point $(3, 5)$ to the point of intersection of the lines $4x + y - 1 = 0$ and$ 7x - 3y - 35 = 0$ is equidistant from the points $(0, 0)$ and $(8, 34).$

Answer

True.
Given equation of lines are $4x + y - 1 = 0$ and $7x - 3y - 35 = 0.$
Lines intersect at $(2, -7)$
Now$,$ the equation of a line passing through $(3, 5)$ and $(2, -7)$ is:
$\Rightarrow \text{y}-5=\frac{-7-5}{2-3}(\text{x}-3)$
$\Rightarrow y - 5 = 12(x - 3)$
$\Rightarrow 12x - y - 31 = 0 .....(i)$
Distance from $(0, 0)$ to the line $(i), \text{d}_1=\frac{|-31|}{\sqrt{144+1}}=\frac{31}{\sqrt{145}}$
$\therefore$ Distance from $(8, 34)$ to the line $(i), \text{d}_2=\frac{|96-34-34|}{\sqrt{145}}=\frac{31}{\sqrt{145}}$
So$, d_1 = d_2$

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Maths True False[1 Marks ]

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