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MATHS MCQ

STRAIGHT LINES · Gujarat Board (GSEB) STD 11 Science (GSEB - English Medium). 170 questions.

Questions · Page 2 of 4

MCQ

MCQ 511 Mark
Distance between the lines 5x + 3y - 7 = 0 and 15x + 9y + 14 = 0 is:
  • A
    $\frac{35}{\sqrt{34}}$
  • B
    $\frac{1}{3\sqrt{34}}$
  • ✓
    $\frac{35}{3\sqrt{34}}$
  • D
    $\frac{35}{2\sqrt{34}}$
Answer
Correct option: C.
$\frac{35}{3\sqrt{34}}$
The given lines can be written as
$5\text{x}+3\text{y}-7=0 \ ...(1)$
$5\text{x}+3\text{y}+\frac{14}{3}=0 \ ...(2)$
Let d be the distance between the lines 5x + 3y - 7 = 0 and 15x + 9y + 14 = 0
Then, $\text{d}=\Bigg|\frac{-7-\frac{14}{3}}{\sqrt{5^2+3^2}}\Bigg|$
$\Rightarrow\text{d}=\frac{35}{3\sqrt{34}}$
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MCQ 521 Mark
Choose the correct answer. For specifying a straight line, how many geometrical parameters should be known?
  • A
    1
  • ✓
    2
  • C
    4
  • D
    3
Answer
Correct option: B.
2
  1. 2
Solution:
Different form of equation of straight line are slope intercept form, y = mx + c, Paramerer = 2
Intercept form, $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ parameter = 2
One-point from, $y - y_1 = m(x - x_1)$, parameter = 2
Normal form, $\text{x}\cos \text{w}+\text{y}\sin\text{w}=\text{P},$ Parameter = 2
Hence, the correct option is (b).
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MCQ 531 Mark
A line passes through P (1, 2) such that its intercept between the axes is bisected at P. The equation of the line is:
  • A
    x + 2y = 5
  • B
    x - y + 1 = 0
  • C
    x + y - 3 = 0
  • ✓
    2x + y - 4 = 0
Answer
Correct option: D.
2x + y - 4 = 0
  1. 2x + y - 4 = 0
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MCQ 541 Mark
Find the equation of line parallel to x-axis and passing through (3, 4):
  • A
    x = 3
  • B
    x = 4
  • ✓
    y = 4
  • D
    y = 3
Answer
Correct option: C.
y = 4
Let general equation of line be y = m × x + c.
Since line is parallel to x-axis so, m = 0.
⇒ y = c
⇒ y = 4 by substituting the point (3, 4).
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MCQ 551 Mark
If -40°F is equal to -40°C and 0°C is equal to 32°F then find the value of 40°C:
  • ✓
    104°F
  • B
    112°F
  • C
    86°F
  • D
    92°F
Answer
Correct option: A.
104°F
Let general equation be F = m × c + k
-40 = -40m + k
and 32 = 0 + k
⇒ -40 = -40m + 32
$\text{ m}=\frac{72}{40} = \frac{18}{10}$
$\text{F}=\frac{18}{10} \times 40 + 32$
= 72 + 32 = 104.
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MCQ 561 Mark
The area of a triangle with vertices at (-4, -1), (1, 2) and (4, -3) is:
  • ✓
    17
  • B
    16
  • C
    15
  • D
    None of these.
Answer
Correct option: A.
17
Let A be the area of the triangle formed by the points (-4, -1), (1, 2) and (4, -3).
$\therefore\text{A}=\frac{1}{2}\big|\{\text{x}_1(\text{y}_2-\text{y}_3)+\text{x}_2(\text{y}_3-\text{y}_1)+\text{x}_3(\text{y}_1-\text{y}_2)\}\big|$
$\Rightarrow\text{A}=\frac{1}{2}\big|\{-4(2+3)+1(-3+1)+4(-1-2)\}\big|$
$\Rightarrow\text{A}=17$
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MCQ 571 Mark
If equation of line is y = 5x + 10 then find the value of x-intercept made by the line:
  • A
    2
  • B
    $\frac{1}{2}$
  • C
    $\frac{-1}{2}$
  • ✓
    -2
Answer
Correct option: D.
-2
Given, equation is y = 5x + 10. X-intercept means value of x when y is zero 0 = 5x + 10
⇒ x = -2
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MCQ 581 Mark
The distance between M (-1, 5) and N (x, 5) is 8 units. The value of x is:
  • A
    -9 or 9
  • B
    -7 or 9
  • ✓
    -9 or 7
  • D
    -7 or -9
Answer
Correct option: C.
-9 or 7
$\sqrt{[\text{x}-(-1)^{2}] + (5-5)^{2}} = 8$
$\Rightarrow(\text{x+1})^2=8^2$
$\Rightarrow\text{x}+1=\underline{+}8$
$\therefore\text{x}=-9,7$
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MCQ 591 Mark
The equation of the straight line which passes through the point (-4, 3) such that the portion of the line between the axes is divided internally by the point in the ratio 5 : 3 is:
  • ✓
    9x - 20y + 96 = 0
  • B
    9x + 20y = 24
  • C
    20x + 9y + 53 = 0
  • D
    none of these.
Answer
Correct option: A.
9x - 20y + 96 = 0
Let the required line intersects the coordinate axis at (a, 0) and (0, b).

The point (−4, 3) divides the required line in the ratio 5 : 3
$\therefore \ -4=\frac{5\times0+3\times\text{a}}{5+3}$ and $3=\frac{5\times\text{b}+3\times0}{5+3}$
$\Rightarrow\text{a}=\frac{ -32}{3}$ and $\text{b}=\frac{ 24}{5}$
Hence, The equation of the required line is given below:
$\frac{\text{x}}{\frac{-32}{3}}+\frac{\text{y}}{\frac{24}{5}}=1$
$\Rightarrow\frac{-3\text{x}}{32}+\frac{5\text{y}}{24}=1$
$\Rightarrow-9\text{x}+20\text{y}=96$
$\Rightarrow9\text{x}-20\text{y}+96=0$
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MCQ 601 Mark
The slope of a line ax + by + c = 0 is:
  • A
    $\frac{\text{a}}{\text{b}}$
  • ✓
    $\frac{\text{-a}}{\text{b}}$
  • C
    $\frac{\text{c}}{\text{b}}$
  • D
    $\frac{\text{-c}}{\text{b}}$
Answer
Correct option: B.
$\frac{\text{-a}}{\text{b}}$
We know that the general equation of a line is ax + by + c = 0.
Rearranging the equation, we get
⇒ by = -ax - c
$\Rightarrow\text{y} =\big(\frac{\text{-a}}{\text{b}})\text{ x}-\big(\frac{\text{-c}}{\text{b}}) ...(1)$
This is of the form, y = mx + c … (2)
By comparing (1) and (2), we get
Slope, $\text{m} = \frac{\text{-a}}{\text{b}}$
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MCQ 611 Mark
Two vertices of a triangle are (-2, -1) and (3, 2) and third vertex lies on the line x + y = 5. If the area of the triangle is 4 square units, then the third vertex is:
  • A
    (0, 5) or, (4, 1)
  • ✓
    (5, 0) or, (1, 4)
  • C
    (5, 0) or, (4, 1)
  • D
    (0, 5) or, (1, 4)
Answer
Correct option: B.
(5, 0) or, (1, 4)
Let (h, k) be the third vertex of the triangle.
It is given that the area of the triangle with vertices (h, k), (-2, -1) and (3, 2) is 4 square units.
$\frac{1}{2}\big|\text{h}(-1-2)-3(-1-\text{K})-2(2-\text{K})\big|=4$
$\Rightarrow3\text{h}-5\text{k}+1=\pm 8$
Taking positive sign, we get,
3h - 5k + 1 = 8
3h - 5k - 7 = 0 ...(1)
Taking negative sign, we get,
3h - 5k + 9 = 0 ...(2)
The vertex (h, k) lies on the line x + y = 5.
h + k - 5 = 0 ...(3)
On solving (1) and (3), we find (4, 1) to be the coordinates of the third vertex.
Similarly, on solving (2) and (3), we find (2, 3) to be the coordinates of the third vertex.
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MCQ 621 Mark
If x-intercept of a line is 4 and its y-intercept is 2 then find the equation of line:
  • A
    2x + y - 4 = 0.
  • ✓
    x + 2y - 4 = 0.
  • C
    2x + y + 4 = 0.
  • D
    x + 2y + 4 = 0.
Answer
Correct option: B.
x + 2y - 4 = 0.
If x-intercept of a line is a and y-intercept of line is b so, equation of line is $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}} = 1$
Equation of line is $\frac{\text{x}}{4}+\frac{\text{y}}{2} = 1$
⇒ x + 2y - 4 = 0.
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MCQ 631 Mark
Find the distance between 2x + y + 4 = 0 and 2x + y + 8 = 0:
  • ✓
    $\frac{4}{\sqrt5}$
  • B
    $\frac{3}{\sqrt5}$
  • C
    $\frac{9}{\sqrt5}$
  • D
    $\frac{3}{\sqrt5}$
Answer
Correct option: A.
$\frac{4}{\sqrt5}$
  1. $\frac{4}{\sqrt5}$
Solution:
Distance between parallel lines $ax + by + c_1 = 0$ and $ax + by + c_2 = 0$ is $\mid\frac{\text{c}_{1}-\text{c}_{2}}{\sqrt{\text{a}^2+\text{b}^2}}\mid$
So, distance 2x + y + 4 = 0 and 2x + y + 8 = 0 is $\mid\frac{8-4}{\sqrt2^2+1^2}\mid=\frac{4}{\sqrt5}$
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MCQ 641 Mark
If equation of a line is y = 3x - 4 then find the slope of line:
  • ✓
    3
  • B
    -3
  • C
    4
  • D
    -4
Answer
Correct option: A.
3
Comparing the above equation with general equation y = m × x + c,
m = 3 which is the slope of line.
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MCQ 651 Mark
Choose the correct answer. The distance between the lines $y = mx + c_1$ and $y = mx + c_2$ is:
  • A
    $\frac{\text{c}_1-\text{c}_2}{\sqrt{\text{m}^2+1}}$
  • ✓
    $\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
  • C
    $\frac{c_2 - c_1}{\sqrt{1+m^2}}$
  • D
    $0$
Answer
Correct option: B.
$\frac{|\text{c}_1-\text{c}_2|}{\sqrt{1+\text{m}^2}}$
Let any point on the line y $= mx + c_1$  be $P(x_1, y_1).$
The equation of the other line is: $y = mx + c_2$
$\Rightarrow mx - y + c_2 = 0$
Distance of point P from this line, $\text{d}=\frac{|\text{mx}_1-\text{y}_1+\text{c}_2|}{\sqrt{\text{m}^2+1}}$
Since P line on the first line, we get
$\Rightarrow y_1 = mx_1 + c_1$
$\Rightarrow mx_1 - y_1 = -c_1$
$\therefore \text{d}=\frac{|\text{c}_1-\text{c}_2|}{\sqrt{\text{m}^2+1}}$
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MCQ 661 Mark
Angle made by line with measured anticlockwise is called inclination of the line:
  • ✓
    Positive x-axis
  • B
    Negative x-axis
  • C
    Positive y-axis
  • D
    Negative y-axis
Answer
Correct option: A.
Positive x-axis
We know, inclination of line is always measured with positive x-axis in anticlockwise direction.
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MCQ 671 Mark
Choose the correct answer. If the coordinates of the middle point of the portion of a line intercepted between the coordinate axes is (3, 2), then the equation of the line will be:
  • ✓
    2x + 3y = 12
  • B
    3x + 2y = 12
  • C
    4x - 3y = 6
  • D
    5x - 2y = 10
Answer
Correct option: A.
2x + 3y = 12
Let the given the line meets the axes at A(a, 0) and B(0, b).
Given that C(3, 2) is the mid-point of AB
$\therefore 3=\frac{\text{a}+0}{2}\Rightarrow \text{a}=6$
and $2=\frac{0+\text{b}}{2}\Rightarrow \text{b}=4$

Intercept form of the line AB
$\Rightarrow \frac{\text{x}}{6}+\frac{\text{y}}{4}=1$
$\Rightarrow 2\text{x}+3\text{y}=12$
Hence, the correct option is (a).
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MCQ 681 Mark
If slope of a line is negative then its inclination is:
  • A
    Right angle.
  • B
    Acute angle.
  • ✓
    Obtuse angle.
  • D
    Zero.
Answer
Correct option: C.
Obtuse angle.
  1. Obtuse angle.
Solution:
If inclination is $\alpha$ slope is given by $\tan\alpha$ Given that slope of line is negative which means $\tan\alpha$ is negative. We know, $\tan\alpha$ is negative in $2^{nd}$ quadrant i.e. $\alpha$ should be obtuse angle.
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MCQ 691 Mark
Choose the correct answer. Equations of diagonals of the square formed by the lines x = 0, y = 0, x = 1 and y = 1 are:
  • ✓
    y = x, y + x = 1
  • B
    y = x, x + y = 2
  • C
    $2\text{y} = \text{x}, \text{y} + \text{x} = \frac{1}{3}$
  • D
    y = 2x, y + 2x = 1
Answer
Correct option: A.
y = x, y + x = 1
Given lines are plotted on coordinate plane as shown in the adjacent figure.
From the figure, equation of diagonal OB is y = x.
Equation of the diagonal AC is x + y = 1 (using intercept form).
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MCQ 701 Mark
If p be the length of the perpendicular from the origin on the line $\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1,$ then:
  • A
    $\text{p}^2=\text{a}^2+\text{b}^2$
  • B
    $\text{p}^2=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
  • ✓
    $\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
  • D
    None of these.
Answer
Correct option: C.
$\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
It is given that p is the length of the perpendicular from the origin on the line
$\frac{\text{x}}{\text{a}}+\frac{\text{y}}{\text{b}}=1$
$\frac{1}{\text{a}}\text{x}+\frac{1}{\text{b}}\text{y}-1=0$
$\therefore\text{p}=\begin{vmatrix}\frac{0+0+1}{\sqrt{\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}}} \end{vmatrix}$
Squaring both sides,
$\Rightarrow\frac{1}{\text{p}^2}=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}$
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MCQ 711 Mark
If equation of line is y = 5x + 10 then find the value of x-intercept made by the line:
  • A
    $2$
  • B
    $\frac{1}{2}$
  • C
    $\frac{-1}{2}$
  • ✓
    $-2$
Answer
Correct option: D.
$-2$
Given, equation is y = 5x + 10. X-intercept means value of x when y is zero. 0 = 5x + 10
⇒ x = -2
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MCQ 721 Mark
Slope of a line is given by if inclination of line is $\alpha$:
  • A
    $\sin\alpha$
  • B
    $\cos\alpha$
  • ✓
    $\tan\alpha$
  • D
    $\cot\alpha$
Answer
Correct option: C.
$\tan\alpha$
Slope of a line is given by $\tan\alpha$ if inclination of line is $\alpha$ Slope is denoted by tangent of the inclination angle.
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MCQ 731 Mark
The ratio in which the line 3x + 4y + 2 = 0 divides the distance between the line 3x + 4y + 5 = 0 and 3x + 4y - 5 = 0 is:
  • A
    1 : 2
  • ✓
    3 : 7
  • C
    2 : 3
  • D
    2 : 5
Answer
Correct option: B.
3 : 7
Here, in all equations the coefficient of x is same.
It means all the lines have same slope
So, all the lines are parallel.
Now, the distance between the line 3x + 4y + 2 = 0 and 3x + 4y + 5 = 0 is given by
$\frac{|2-5|}{\sqrt{3^2+4^2}}$
$=\frac{3}{\sqrt{25}}=\frac{3}{5}$
Hence, the ratio is given by
$\frac{3}{5}:\frac{7}{5}$
$=3:7$
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MCQ 741 Mark
If a line makes an angle a with the positive direction of x-axis, then the slope of the line is given by:
  • A
    $\text{m} = \sin\text{a}$
  • B
    $\text{m} = \cos\text{a}$
  • ✓
    $\text{m} = \tan\text{a}$
  • D
    $\text{m} = \sec\text{a}$
Answer
Correct option: C.
$\text{m} = \tan\text{a}$
  1. $\text{m} = \tan\text{a}$
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MCQ 751 Mark
The equation of a straight line that passes through the point (3, 4) and perpendicular to the line 3x + 2y + 5 = 0 is:
  • ✓
    2x - 3y + 6 = 0
  • B
    2x + 3y + 6 = 0
  • C
    2x - 3y - 6 = 0
  • D
    2x + 3y - 6 = 0
Answer
Correct option: A.
2x - 3y + 6 = 0
The equation of a straight line perpendicular to 3x + 2y + 5 = 0 is $2\text{x}-3\text{y}+\lambda=0 …(1)$
This passes through the point (3, 4).
Now, substitute in equation (1), we get
$2(2) - 3(4) +\lambda = 0$
$4-12+\lambda =0$
$-6+\lambda =0$
$\lambda=6$
Substituting $\lambda=6 $ in (1), we get 2x - 3y + 6 = 0, which is the required equation.
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MCQ 761 Mark
The inclination of the straight line passing through the point (-3, 6) and the mid-point of the line joining the point (4, -5) and (-2, 9) is:
  • A
    $\frac{\pi}{4}$
  • B
    $\frac{\pi}{6}$
  • C
    $\frac{\pi}{3}$
  • ✓
    $\frac{3\pi}{4}$
Answer
Correct option: D.
$\frac{3\pi}{4}$
The midpoint of the line joining the points (4, -5) and (-2, 9) is (1, 2).
Let $\theta$ be the inclination of the straight line passing through the points (-3, 6) and (1, 2).
Then, $\tan\theta=\frac{ 2-6}{1+3}=-1$
$\Rightarrow\theta=\frac{3\pi}{4}$
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MCQ 771 Mark
If slope of a line is 4 and x-intercept made by the line is 2 then the equation of line will be:
  • ✓
    y = 4x - 8
  • B
    y = 4x + 8
  • C
    y = 2x + 4
  • D
    y = 2x - 4
Answer
Correct option: A.
y = 4x - 8
Let general equation of line be y = m × x + c.
Given, m = 4 and value of x when y = 0 is 2.
C = -m × 2 = -4 × 2 = -8.
⇒ y = 4x - 8
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MCQ 781 Mark
The angle between the lines 2x - y + 3 = 0 and x + 2y + 3 = 0 is:
  • ✓
    90°
  • B
    60°
  • C
    45°
  • D
    30°
Answer
Correct option: A.
90°
  1. 90°
Solution:
Let $m_1$ and $m_2$ be the slope of the lines 2x - y + 3 = 0 and x + 2y + 3 = 0, respectively.
Let $\theta$ be the angle between them.
Here, $m_1 = 2$ and $\text{m}_2=-\frac{1}{2}$
$\because\text{m}_1\text{m}_2=-1$
Therefore, the angle between the given lines is 90°.
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MCQ 791 Mark
If A (6, 4) and B (2, 12) are the two points, then the slope of a line perpendicular to line AB is:
  • A
    $-2$
  • B
    $2$
  • ✓
    $\frac{1}{2}$
  • D
    $\frac{-1}{2}$
Answer
Correct option: C.
$\frac{1}{2}$
  1. $\frac{1}{2}$
Solution:
Given points: $A(6,4)=\left(x_1, y_1\right)$
$B(2,12)=\left(x_2, y_2\right)$
We know that the slope of a line passing through two points $\left(x_1, y_1\right)$ and $\left(x_2, y_2\right)$ is $\frac{\left(y_2-y_1\right)}{\left(x_2-x_1\right)}$
$\mathrm{m}=\frac{(12-4)}{(2-6)}=\frac{8}{-4}=-2 .$
We know that the slope of two perpendicular lines $\mathrm{m}_{1,} \mathrm{~m}_2=-1$.
The slope of a line perpendicular to line $A B$ is $\frac{-1}{\mathrm{~m}}=\frac{-1}{-2}=\frac{1}{2}$
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MCQ 801 Mark
A(6, 3), B(-3, 5), C(4, -2) and D(x, 3x) are four points. If $\triangle\text{DBC} : \triangle\text{ABC}= 1 : 2,$ then x is equal to:
  • ✓
    $\frac{11}{8}$
  • B
    $\frac{8}{11}$
  • C
    3
  • D
    None of these
Answer
Correct option: A.
$\frac{11}{8}$
The area of a triangle with vertices D(x, 3x), B(-3, 5) and C(4, -2) is given below:
Area of $\triangle\text{DBC}=\frac{1}{2}\{\text{x}(5+2)-3(-2-3\text{x})+4(3\text{x}-5)\}$
⇒ Area of $\triangle\text{DBC}=(14\text{x}-7)\text{sq units}$
Similarly, the area of a triangle with vertices A(6, 3), B(-3, 5) and C(4, -2) is given below:
$\triangle\text{ABC}=\frac{1}{2}\{6(5+2)-3(-2-3)+4(3-5)\}$
$\triangle\text{ABC}=\frac{49}{2}\text{sq units}$
Given:
$\triangle\text{DBC}:\triangle\text{ABC}=1:2$
$\frac{2(14\text{x}-7)}{49}=\frac{1}{2}$
$\Rightarrow8\text{x}-4=7$
$\Rightarrow\text{x}=\frac{11}{8}$
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MCQ 811 Mark
The equation $\frac{(2 \text{x})}{2} - \frac{(\text{y}{ 2})}{2} = -3,$ if x = -2, then y is equal to:
  • A
    -2
  • B
    -7
  • ✓
    8
  • D
    5
Answer
Correct option: C.
8
  1. 8
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MCQ 821 Mark
Equation of vertical line to the left of y-axis at 5 units from y-axis is:
  • A
    x = 5
  • ✓
    x = -5
  • C
    y = 5
  • D
    y = -5
Answer
Correct option: B.
x = -5
Equation of y-axis is x = 0. Vertical line is parallel to y-axis and to the left by 5 units so, equation of line is x = -5.
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MCQ 831 Mark
Find the equation perpendicular to 2x - y = 4 and pass through (2, 4):
  • A
    2x + y - 10 = 0
  • B
    x + 2y + 10 = 0
  • ✓
    x + 2y - 10 = 0
  • D
    x + y - 10 = 0
Answer
Correct option: C.
x + 2y - 10 = 0
Line 2x - y = 4 has slope 2. Line perpendicular to given line has slope $\frac{-1}{2}$
equation is $(\text{y}-4) = \big(\frac{-1}{2}\big) (\text{x}-2)$
2y - 8 = -x + 2
⇒ x + 2y - 10 = 0
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MCQ 841 Mark
Which of the following lines is farthest from the origin?
  • A
    x - y + 1 = 0
  • B
    2x - y + 3 = 0
  • C
    x + 2y - 2 = 0
  • ✓
    x + y - 2 = 0
Answer
Correct option: D.
x + y - 2 = 0
  1. x + y - 2 = 0
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MCQ 851 Mark
Equation of horizontal line above x-axis at 5 units from x-axis is:
  • A
    x = 5
  • B
    x = -5
  • ✓
    y = 5
  • D
    y = -5
Answer
Correct option: C.
y = 5
Equation of x-axis is y = 0. Horizontal line is parallel to x-axis and above it by 5 units so, equation of line is y = 5.
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MCQ 861 Mark
The inclination of the line x - y + 3 = 0 with the positive direction of x-axis is:
  • ✓
    45°
  • B
    135°
  • C
    -45°
  • D
    -135°
Answer
Correct option: A.
45°
  1. 45°
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MCQ 871 Mark
What is the inclination of a line which is parallel to y-axis?
  • A
    0°
  • B
    180°
  • C
    45°
  • ✓
    90°
Answer
Correct option: D.
90°
If a line is parallel to y-axis then angle formed by it with x-axis is 90°. So, its inclination is 90°.
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MCQ 881 Mark
Find the equation of line parallel to x-axis and passing through (3, 4):
  • A
    x = 3
  • B
    x = 4
  • ✓
    y = 4
  • D
    y = 3
Answer
Correct option: C.
y = 4
Let general equation of line be y = m × x + c. Since line is parallel to x-axis so, m = 0.
⇒ y = c
⇒ y = 4 by substituting the point (3, 4).
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MCQ 891 Mark
Equation of horizontal line above x-axis at 5 units from x-axis is:
  • A
    x = 5
  • B
    x = -5
  • ✓
    y = 5
  • D
    y = -5
Answer
Correct option: C.
y = 5
Equation of x-axis is y = 0. Horizontal line is parallel to x-axis and above it by 5 units so, equation of line is y = 5
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MCQ 901 Mark
The equations of the sides AB, BC and CA of $\triangle \text{ABC}$ are y - x = 2, x + 2y = 1 and 3x + y + 5 = 0 respectively. The equation of the altitude through B is:
  • A
    x - 3y + 1 = 0
  • ✓
    x - 3y + 4 = 0
  • C
    3x - y + 2 = 0
  • D
    None of these.
Answer
Correct option: B.
x - 3y + 4 = 0
The equation of the sides AB, BC and CA of $\triangle \text{ABC}$ are y - x = 2, x + 2y = 1 and 3x + y + 5 = 0, respectively.
Solving the equations of AB and BC, i.e. y - x = 2 and x + 2y = 1, we get:
x = -1, y = 1
So, the coordinates of B are (-1, 1).
The altitude through B is perpendicular to AC.
$\therefore$ Slope of AC = -3
Thus, slope of the altitude through B is 13. Thus, slope of the altitude through B is $\frac{1}{3}.$
Equation of the required altitude is given below:
$\text{y}-1+\frac{1}{3}(\text{x}+1)$
$\Rightarrow\text{x}-3\text{y}+4=0$
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MCQ 911 Mark
For specifying a straight line, how many geometrical parameters should be known:
  • A
    1
  • ✓
    2
  • C
    4
  • D
    3
Answer
Correct option: B.
2
  1. 2
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MCQ 921 Mark
Two lines are said to be parallel if the difference of their slope is:
  • A
    -1
  • ✓
    0
  • C
    1
  • D
    None of these
Answer
Correct option: B.
0
  1. 0
Solution:
We know that two lines are said to be parallel if their slope is equal. If $m_1$ and $m_2$ are the slopes of two parallel lines, then it is represented as $m_1 = m_2$.
The difference of their slope should be $m_1- m_2 = 0$.
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MCQ 931 Mark
The distance between the lines 3x + 4y = 9 and 6x + 8y = 15 is:
  • A
    $\frac{3}{2}$
  • ✓
    $\frac{3}{10}$
  • C
    $6$
  • D
    $\frac{9}{4}$
Answer
Correct option: B.
$\frac{3}{10}$
  1. $\frac{3}{10}$
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MCQ 941 Mark
If equation of line is x + y = 2 then find the angle made by line with x-axis:
  • ✓
    45°
  • B
    60°
  • C
    30°
  • D
    75°
Answer
Correct option: A.
45°
Given, equation is x + y = 2. Reducing the above equation to normal form
$\frac{(\text{x + y})}{\sqrt2} ={\sqrt2}.$
$\text{x} \cos⁡45° + \text{y} \sin 45^\circ = {\sqrt2}$
Angle made with x-axis is 45°.
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MCQ 951 Mark
Angle made by line with measured anticlockwise is called inclination of the line:
  • ✓
    Positive x-axis.
  • B
    Negative x-axis.
  • C
    Positive y-axis.
  • D
    Negative y-axis.
Answer
Correct option: A.
Positive x-axis.
We know, inclination of line is always measured with positive x-axis in anticlockwise direction.
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MCQ 961 Mark
If a + b + c = 0, then the family of lines 3ax + by + 2c = 0 pass through fixed point:
  • A
    $\Big(2,\frac{2}{3}\Big)$
  • ✓
    $\Big(\frac{2}{3},2\Big).$
  • C
    $\Big(-2,\frac{2}{3}\Big)$
  • D
    None of these.
Answer
Correct option: B.
$\Big(\frac{2}{3},2\Big).$
  1. $\Big(\frac{2}{3},2\Big).$
Solution:
Given:
a + b + c = 0
Substituting c = -a - b in 3ax + by + 2c = 0, we get:
$3\text{ax}+\text{by}-2\text{a}-2\text{b}=0$
$\Rightarrow\text{a}(3\text{x}-2)+\text{b}(\text{y}-2)=0$
$\Rightarrow(3\text{x}-2)+\frac{\text{b}}{\text{a}}(\text{y}-2)=0$
This line is of the form $\text{L}_1+\lambda\text{L}_2=0,$ which passes through the intersection of the lines $L_1$ and $L_2$ i.e. 3x - 2 = 0 and y - 2 = 0.
Solving 3x - 2 = 0 and y - 2 = 0, we get
$\text{x}=\frac{2}{3},\text{y}=2$
Hence, the required fixed point is $\Big(\frac{2}{3},2\Big).$
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MCQ 971 Mark
Equation of the straight line making equal intercepts on the axes and passing through the point (2, 4) is:
  • A
    4x - y - 4 = 0
  • B
    2x + y - 8 = 0
  • ✓
    x + y - 6 = 0
  • D
    x + 2y - 10 = 0
Answer
Correct option: C.
x + y - 6 = 0
  1. x + y - 6 = 0
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MCQ 981 Mark
Equation of vertical line to the right of y-axis at 5 units from y-axis is:
  • ✓
    x = 5
  • B
    x = -5
  • C
    y = 5
  • D
    y = -5
Answer
Correct option: A.
x = 5
Equation of y-axis is x = 0. Vertical line is parallel to y-axis and to the right by 5 units so, equation of line is x = 5.
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MCQ 991 Mark
What can be said regarding a line if its slope is negative?
  • A
    $\theta$ is an acute angle
  • ✓
    $\theta$ is an obtuse angle
  • C
    Either the line is x-axis or it is parallel to the x-axis
  • D
    None of these
Answer
Correct option: B.
$\theta$ is an obtuse angle
The line with a negative slope makes an obtuse angle with a positive x-axis when measured in the anti-clockwise direction.
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MCQ 1001 Mark
The centroid of the triangle with vertices (2, 6), (-5, 6) and (9,3) is:
  • A
    (2, -3)
  • ✓
    (2, 5)
  • C
    (-2, 3)
  • D
    (-2, -3)
Answer
Correct option: B.
(2, 5)
$\text{G}=\Big(\frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}}{3},\frac{\text{y}_{1}+\text{y}_{2}+\text{y}_{3}}{3}\Big)$
$=\Big(\frac{2 - 5 + 9}{3},\frac{6 + 6 + 6}{3}\Big) = (2.5)$
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