Fill In The Blanks[1 Marks ]

Question 11 Mark

The line which cuts off equal intercept from the axes and pass through the point (1, -2) is ____.

Answer

The line which cuts off equal intercept from the axes and pass through the point (1, .2) is x + y + 1 = 0.
Solution:
Intercept form of the lines is
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1\ .....(\text{i})$
Given that a = b
$\therefore \frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{a}}=1$
$\Rightarrow \text{x}+\text{y}=\text{a}\ .....\text{ii}$
If the line (i) passes through (1, -2) we get
⇒ 1 - 2 = a
⇒ a = -1
So, the required equation is x + y = -1
⇒ x + y + 1 = 0.
Hence, the value of the filler is x + y + 1 = 0.

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Question 21 Mark

If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through ____.

Answer

If a, b, c are in A.P., then the straight lines ax + by + c = 0 will always pass through (1, -2).
Solution:
Given equation is ax + by + c = 0 .....(i)
Since a, b and c are in A.P.
$\therefore \text{b}=\frac{\text{a}+\text{c}}{2}$
⇒ a + c = 2b
⇒ a - 2b + c = 0 .....(ii)
Comparing eq. (i) with eq. (ii) we get.
x = 1, y = -2
So, the line will pass through (1, -2)
Hence, the value of the filler is (1, -2).

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Question 31 Mark

Equations of the lines through the point (3, 2) and making an angle of 45° with the line x - 2y = 3 are ____.

Answer

Equations of the lines through the point (3, 2) and making an angle of 45° with the line x - 2y = 3 are x + 3y - 9 = 0.
Solution:
Slope of the given line x - 2y = 3 is $\frac{1}{2}.$
Let the slope of the required line be m.
$\therefore \tan 45^\circ = \begin{vmatrix} \frac{\text{m}-\frac{1}{2}}{1+\frac{1}{2}\text{m}} \end{vmatrix}$
$\Rightarrow 1=\pm\frac{2\text{m}-1}{2+\text{m}}$
$\Rightarrow 2\text{m}-1=2+\text{m}$ or $1-2\text{m}=2+\text{m}$
$\Rightarrow \text{m}=3$ or $\text{m}=-\frac{1}{3}$
Also, the required lines passes through the point (3, 2).
So, equation of the line is: y - 2 = 3(x - 3) or $\text{y}-2=-\frac{1}{3}(\text{x}-3)$
$\therefore$ 3x - y - 7 = 0 or x + 3y - 9 = 0

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Question 41 Mark

$A$ point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 3.$ The equation of its locus is ____.

Answer

$A$ point moves so that square of its distance from the point $(3, -2)$ is numerically equal to its distance from the line $5x - 12y = 3.$ The equation of its locus is $13x^2 + 13y^2 - 83x + 64y + 172 = 0.$
Let the moving point be $P(h, k).$
Given point is $A(3, -2).$
$\text{AP}^2=(\text{h}-3)^2+(\text{k}+2)^2=\text{d}_1^2$
Now, distance of the point $(h, k)$ from the line $5x - 12y - 3 = 0$ is
$\text{d}_2\Big|\frac{5\text{h}-12\text{k}-3}{\sqrt{25+144}}\Big|=\Big|\frac{5\text{h}-12\text{k}-3}{13}\Big|$
Given that, $\text{d}_1^2=\text{d}_2$
$\Rightarrow (\text{h}-3)^2+(\text{k}+2)^2=\frac{5\text{h}-12\text{k}-3}{13} ($Taking $+$ ve sig$n)$
$\Rightarrow \text{h}^2-6\text{h}+9+\text{k}^2+4\text{k}+4=\frac{5\text{h}-12\text{k}-3}{13}$
$\Rightarrow 13\text{h}^2+13\text{k}^2-78\text{h}+52\text{k}+169=5\text{h}-12\text{k}-3$
$\Rightarrow 13\text{h}^2+13\text{k}^2-83\text{h}+64\text{k}+172=0$
So$,$ locus of this points is: $13\text{h}^2+13\text{k}^2-83\text{h}+64\text{k}+172=0.$

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Question 51 Mark

The points (3, 4) and (2, -6) are situated on the ____ of the line 3x - 4y - 8 = 0.

Answer

The points (3, 4) and (2, -6) are situated on the 22 > 0 of the line 3x - 4y - 8 = 0.
Solution:
Given line is 3x - 4y - 8 = 0 .....(i)
and given points are (3, 4) and (2, -6).
For point (3, 4) line becomes = 3(3) - 4(4) - 8
= 9 - 16 - 8
= 9 - 24 = -15 < 0
For the point (2, -6), line becomes = 3(2) - 4(-6) - 8
= 6 + 24 - 8 = 30 - 8
= 22 > 0
So, the points (3, 4) and (2, -6) are situated on the opposite sides of 3x - 4y - 8 = 0.
Hence, the value of the filler is opposite.

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Question 61 Mark

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 is ____.

Answer

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 is $4x^2y^2 = p^2(x^2 + y^2).$
Line $\text{x}\sin\theta+\text{y}\cos\theta=\text{p}$ meets axes at $\text{A}\Big(\frac{\text{p}}{\sin\theta},0\Big)$ and $\text{B}\Big(0,\frac{\text{p}}{\cos\theta}\Big).$
Let $\text{P}(h, k)$ be the mid$-$point of $AB$.
$\therefore \text{h}=\frac{\text{p}}{2\sin\theta}$ and $\text{k}=\frac{\text{p}}{2\cos\theta}$
$\therefore \sin\theta=\frac{\text{p}}{2\text{h}}$ and $\cos\theta=\frac{\text{p}}{2\text{k}}$
Squaring and adding$,$ we get
$\sin^2\theta+\cos^2\theta=\frac{\text{p}^2}{4\text{h}^2}+\frac{\text{p}^2}{4\text{k}^2}$ or $1=\frac{\text{p}^2}{4\text{x}^2}+\frac{\text{p}^2}{4\text{y}^2}$
or $4\text{x}^2\text{y}^2=\text{p}^2(\text{x}^2+\text{y}^2).$

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