Straight Lines - Class 11 Mathematics - Chapter 9 - Notes, NCERT Solutions & Extra Questions
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Extra Questions - Straight Lines | NCERT | Mathematics | Class 11
The lines joining the points of intersection of the line $x+y=1$ and curve $x^{2}+y^{2}-2y+\lambda=0$ to the origin are perpendicular. Then the value of $\lambda$ will be:
A) $1/2$
B) $-1/2$
C) $1/\sqrt{2}$
D) $0
The correct option is D (0).
Considering the given line $x + y = 1$, we can transform the curve equation $x^2 + y^2 - 2y + \lambda = 0$ into a homogeneous equation by substituting $y = 1 - x$. The transformed equation becomes: $$ x^{2} + (1-x)^{2} - 2(1-x) + \lambda(1)^{2} = 0 \ \Rightarrow x^{2} + (1 - 2x + x^{2}) - 2 + 2x + \lambda = 0 \ \Rightarrow 2x^{2} - 2 + \lambda = 0. $$
Homogenization: Convert it by expanding and collecting like terms: $$ x^2 + y^2 - 2y(x + y) + \lambda(x + y)^2 = 0 \ \Rightarrow x^2 (1 + \lambda) + y^2 (1 - \lambda) - 2xy = 0. $$
For the lines to be perpendicular (originating from the intersection of the given line and curve and meeting at the origin), the sum $A + B$ in the general form $Ax^2 + By^2 + Cxy = 0$ should be zero:
$$ 1 + \lambda - 1 + \lambda = 0 \Rightarrow 2\lambda = 0 \Rightarrow \lambda = 0. $$
Therefore, $\boldsymbol{\lambda = 0}$ is the correct value.
A line graph can be:
A. A dotted line
B. A whole unbroken line
C. A zigzag disconnected line
D. None of the above
The correct answer is B. A whole unbroken line.
A line graph primarily represents data through data points, which are often referred to as 'markers'. These markers are connected by straight line segments. It’s a common visualization tool in various fields, effectively representing changes and trends over time. Thus, it involves a continuous, unbroken line, rather than being dotted or disconnected.
The railway tracks $\qquad$ when seen from a distance are very far.
A appears to converge
B converge
C cut each other
D appears to diverge
The correct answer is A appears to converge.
Explanation: Though the railway tracks are actually parallel and never meet, due to a visual perception phenomenon known as perspective, the tracks seem to converge as they recede into the distance. This optical illusion makes the distance between the tracks appear to decrease, giving the impression that they meet at a point on the horizon.
Which of the following statements is true for the equation of a line $x - y = 0$?
A. $|$ to the $x$-axis
B. $|$ to the $y$-axis
C. Passing through the origin
D. Passing through $(1, -1)`
The correct choice is C. Passing through the origin.
The equation given is: $$ x - y = 0. $$ This can be rewritten as: $$ y = x. $$ This equation describes a line where the value of $y$ is equal to $x$ at every point on the line, indicating it is a diagonal line through the $x$-axis and $y$-axis. To validate it passes through the origin, substitute $(0,0)$ into the equation: $$ 0 - 0 = 0, $$ which holds true. Therefore, the line does indeed pass through the origin.
Contrapositive for the statement, "If the slope of a straight line is 1, then it is equally inclined to both the axes," is:
A. If a straight line is equally inclined to both axes, then its slope is 1.
B. If a straight line is equally inclined to both axes, then its slope is not 1.
C. If a straight line is not equally inclined to both axes, then its slope is not 1.
D. If a straight line is not equally inclined to both axes, then its slope is 1.
Solution
The correct choice is Option C. To determine the contrapositive of a statement, the form used is from $p \implies q$ to $\neg q \implies \neg p$.
For the given statement:
- Let $p$ represent: "the slope of a straight line is 1"
- Let $q$ represent: "it is equally inclined to both the axes"
Thus, the contrapositive of the statement "If the slope of a straight line is 1, then it is equally inclined to both the axes" is transformed to:
- "If a straight line is not equally inclined to both axes, then its slope is not 1."
Therefore, the correct answer is:
- C. If a straight line is not equally inclined to both axes, then its slope is not 1.
Parallel lines never intersect.
A) True
B) False
Solution: The correct answer is A) True.
Parallel lines are defined as lines that remain equally distant from each other at all points. They never meet or cross each other regardless of how far they are extended. This characteristic distinguishes them from intersecting lines, which cross each other at a certain point.
The image below visually represents the concept of parallel lines as well as intersecting lines. Notice how the parallel lines do not meet:
From the illustration, it's clear that when lines intersect, they breach the condition of maintaining a constant distance between each other, which is a requisite for lines to be parallel. Thus, the statement "Parallel lines never intersect" holds true.
What is a sarcomere?
A) Part between two $\mathrm{H}$-lines.
B) Part between two A-lines.
C) Part between two I-bands.
D) Part between two Z-lines.
Correct answer: D) Part between two Z-lines
A sarcomere is the fundamental functional unit of striated muscle tissue. It is specifically defined as the segment between two Z-lines. These Z-lines serve as the boundaries for each sarcomere and play a crucial role in muscle contraction.
-
H-Zone: This is a narrower region inside the A-Band at the center of the sarcomere, where only thick filaments (myosin) are found. This zone varies in size depending on the muscle's state of contraction.
-
The A-Band is that part of the sarcomere that contains the thick filaments along with parts of the thin filaments, and this region does not change in length during muscle contraction.
-
The I-Band consists solely of thin filaments (actin) and is the lighter area on either side of the Z-line. This area shortens during muscle contraction.
The interactions between the thin filaments (actin) and thick filaments (myosin) within the sarcomere's A-Band are central to the mechanism of muscle contraction. Each sarcomere shortening contributes cumulatively to the overall contraction of a muscle fiber.
The slope of a straight line passing through $A(-2,3)$ is $-\frac{4}{3}$. The point(s) on the line that are 10 units away from $A$ is/are:
A) $(8,0)$
B) $(-8,11)$
C) $(4,-5)$
D) $(4,11)$
The correct option is
B $(-8,11)$ and C $(4,-5)$.
To solve this question, we begin with the information given:
- The point $A(-2,3)$ is on the line, and
- The slope of the line is $-\frac{4}{3}$.
This slope can be written as $\tan \theta$, where $\theta$ is the angle the slope makes with the positive direction of the x-axis in the 2nd quadrant. This yields: $$ \tan \theta = -\frac{4}{3} $$
From trigonometric identities, we know: $$ \cos \theta = -\frac{3}{5} \quad \text{and} \quad \sin \theta = \frac{4}{5} $$
Given the point $A = (-2, 3) = (x_1, y_1)$, and distance $r=10$ from $A$ to the points on the line, we apply the distance formula in terms of $\cos \theta$ and $\sin \theta$ to find these unknown coordinates $(x, y)$. We then have:
For the x-coordinate: $$ x = x_1 \pm r \cos \theta = -2 \pm 10\left(-\frac{3}{5}\right) = -2 \pm -6 $$ This results in two possibilities: $$ x = -2 - 6 = -8 \quad \text{or} \quad x = -2 + 6 = 4 $$
For the y-coordinate: $$ y = y_1 \pm r \sin \theta = 3 \pm 10\left(\frac{4}{5}\right)= 3 \pm 8 $$ Which leads to: $$ y = 3 + 8 = 11 \quad \text{or} \quad y = 3 - 8 = -5 $$
Therefore, the required points that lie on the line and are 10 units away from point A (-2, 3) are $(-8, 11)$ and $(4, -5)$.
Which of the following are the perfect examples of two parallel lines?
A. Zebra crossing on the road
B. Railway tracks (except on junctions)
C. Adjacent walls of a room
D. The line segments that make the 'T' shape
The correct options are:
A. Zebra crossing on the road
B. Railway tracks (except on junctions)
Lines that lie in the same plane and do not intersect each other are termed parallel. Below are the explanations for each option:
-
Railway tracks (except on junctions): These tracks are designed to be parallel to each other to maintain the set distance apart, providing a path for trains on designated routes, except at junctions where they may intersect or diverge.
-
Zebra crossing on the road: This consists of several parallel white stripes painted on the road, facilitating pedestrian crossings. These stripes are laid parallel to each other across the breadth of the roadway.
-
Adjacent walls of a room: These are not examples of parallel lines since adjacent walls meet at corners, thus they intersect.
-
The line segments that make the 'T' shape: These represent perpendicular lines, not parallel, as one line is vertical and the other is horizontal, intersecting at a right angle.
Classify the following pairs of lines as coincident, parallel or intersecting:
(i) $2x + y - 1 = 0$ and $3x + 2y + 5 = 0$
(ii) $x - y = 0$ and $3x - 3y + 5 = 0$
(iii) $3x + 2y - 4 = 0$ and $6x + 4y - 8 = 0$.
To classify the pairs of lines as coincident, parallel, or intersecting, let's analyze each pair step-by-step.
Pair (i): $2x + y - 1 = 0$ and $3x + 2y + 5 = 0$
Rewrite each equation in the slope-intercept form $(y = mx + c)$:
For the first equation: $$ y = -2x + 1 $$
For the second equation: $$ 3x + 2y + 5 = 0 \ 2y = -3x - 5 \ y = -\frac{3}{2}x - \frac{5}{2} $$
The slopes (m) are: $$ m_1 = -2, \quad m_2 = -\frac{3}{2} $$
Since $m_1 \neq m_2$ and the product $m_1 \cdot m_2 \neq -1$, the lines are intersecting.
Pair (ii): $x - y = 0$ and $3x - 3y + 5 = 0$
Rewrite each equation in the slope-intercept form $(y = mx + c)$:
For the first equation: $$ y = x $$
For the second equation: $$ 3x - 3y + 5 = 0 \ 3y = 3x + 5 \ y = x + \frac{5}{3} $$
The slopes (m) are: $$ m = 1, \quad m' = 1 $$
Since both lines have equal slopes ($m = 1$), they are parallel.
Pair (iii): $3x + 2y - 4 = 0$ and $6x + 4y - 8 = 0$
Rewrite each equation in the slope-intercept form $(y = mx + c)$:
For the first equation: $$ 3x + 2y - 4 = 0 \ 2y = -3x + 4 \ y = -\frac{3}{2}x + 2 $$
For the second equation: $$ 6x + 4y - 8 = 0 \ 4y = -6x + 8 \ y = -\frac{3}{2}x + 2 $$
Since both equations simplify to the same line: $$ y = -\frac{3}{2}x + 2 $$
Therefore, the lines are coincident because they have the same slope and intercept.
In summary:
Pair (i): Intersecting
Pair (ii): Parallel
Pair (iii): Coincident
Find the ratio in which the line $ax+by+c=0$ divides the line segment joining $P(x_{1}, y_{1})$ and $Q(x_{2}, y_{2})$. You are given $t=\frac{ax_{2}+by_{2}+c}{ax_{1}+by_{1}+c}$.
A $-t$
B $t$
C $-\frac{1}{t}$
D $\frac{1}{t}$
The correct answer is Option C:
$$ -\frac{1}{t} $$
To determine how the line $ax + by + c = 0$ divides the segment joining points $P(x_1, y_1)$ and $Q(x_2, y_2)$, we start by establishing the ratio $\frac{m}{n}$. This ratio is given by: $$ \frac{ax_1 + by_1 + c}{ax_2 + by_2 + c} $$
This can be written as: $$ \frac{\left(a x_{1}+b y_{1}+c\right)}{a x_{2}+b y_{2}+c} = -\frac{1}{t} $$
Let's derive this formula step-by-step:
Assume the ratio is $K:1$. Suppose the line divides the segment at point $R$.
Find coordinates of $R$ using $K$. The coordinates $R$ can be expressed in terms of $K$.
Substitute coordinates of $R$ in the line equation $ax + by + c = 0$. This will give us the value of $K$.
Remember, you can utilize this derivation approach to solve similar problems effectively. Additionally, it's beneficial to memorize the formula for quick reference: $$ \frac{m}{n} = -\frac{1}{t} $$
Identify the points which lie on the line 3x + 4y = 7.
Option 1) $(1, 1)$
Option 2) $(0, \frac{7}{4})$
Option 3) $(0, 7)$
Option 4) $(\frac{7}{3}, 0)$
To determine which points lie on the line represented by the equation ( 3x + 4y = 7 ), we need to verify if the coordinates of the points provided in the options satisfy this equation.
Here’s how we'll verify each point:
Option 1:
Point: ( (1, 1) )
Substitute into the equation ( 3x + 4y = 7 ):
[ 3(1) + 4(1) = 3 + 4 = 7 ]
Since ( 7 = 7 ), this point lies on the line.
Option 2:
Point: $ \left(0, \frac{7}{4}\right) $
Substitute into the equation $ 3x + 4y = 7 $:
[ 3(0) + 4\left(\frac{7}{4}\right) = 0 + 7 = 7 ]
Since ( 7 = 7 ), this point lies on the line.
Option 3:
Point: ( (0, 7) )
Substitute into the equation $ 3x + 4y = 7 $:
[ 3(0) + 4(7) = 0 + 28 \neq 7 ]
Since $ 28 \neq 7 $, this point does not lie on the line.
Option 4:
Point: $ \left(\frac{7}{3}, 0\right) $
Substitute into the equation $ 3x + 4y = 7 $:
$$ 3\left(\frac{7}{3}\right) + 4(0) = 7 + 0 = 7 $$
Since ( 7 = 7 ), this point lies on the line.
Based on the evaluations, the points that lie on the line ( 3x + 4y = 7 ) correspond to the equations:
Option 1:
Option 2:
Option 4:
Therefore, the correct options are 1, 2, and 4.
The distance of the point (1, 2) from the line ( 3x + 4y - 5 = 0 ) is __________ units.
A) ( \frac{6}{5} )
B) ( \frac{5}{6} )
C) ( \frac{16}{5} )
The correct option is $\mathbf{A} \frac{6}{5}$
The distance of a point $\left(x_{1}, y_{1}\right)$ from a line given by $ax + by + c = 0$ is calculated using the formula:
$$ \text{Distance} = \frac{\left|a x_{1} + b y_{1} + c\right|}{\sqrt{a^{2} + b^{2}}} $$
Given the point $(1, 2)$ and the line equation $3x + 4y - 5 = 0$, we can substitute into the formula as follows:
$$ d = \frac{|3 \cdot 1 + 4 \cdot 2 - 5|}{\sqrt{3^2 + 4^2}} $$
Breaking it down step-by-step:
$$ d = \frac{|3 + 8 - 5|}{\sqrt{9 + 16}} $$
Simplify the numerator and the denominator:
$$ d = \frac{|6|}{\sqrt{25}} $$
Since $|6| = 6$ and $\sqrt{25} = 5$, the distance becomes:
$$ d = \frac{6}{5} \text{ units} $$
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