Class 11 Mathematics Chapter 9 Straight Line

Key Concept: Midpoint Formula

b) (5, 5)

[Solution Description] The midpoint formula is given by $$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$. Substituting the values:

$$\text{Midpoint} = \left(\frac{2 + 8}{2}, \frac{3 + 7}{2}\right) = \left(\frac{10}{2}, \frac{10}{2}\right) = (5, 5)$$

Key Concept: General Equation

a) $Ax + By + C = 0$

[Solution Description] The general form of a linear equation in two variables $x$ and $y$ is given by the formula:

$$Ax + By + C = 0$$

where $A$, $B$, and $C$ are constants.

Key Concept: Perpendicular Slope

c) $-\frac{4}{3}$

[Solution Description] If a line has a slope $m$, then a line perpendicular to it will have a slope of $-\frac{1}{m}$. Given that the slope of the original line is $\frac{3}{4}$, the slope of the perpendicular line is:

$$-\frac{1}{\left(\frac{3}{4}\right)} = -\frac{4}{3}$$

Therefore, the slope of the perpendicular line is $-\frac{4}{3}$.

Key Concept: Parallel Lines

a) $m_1 = 5, m_2 = 5$

[Solution Description] Two non-vertical lines are parallel if their slopes are equal. We compare the slopes provided:

– $$m_1 = 5, m_2 = 5$$: These lines have equal slopes.

– $$m_1 = -3, m_2 = 3$$: These lines do not have equal slopes.

– $$m_1 = 0, m_2 = -1$$: These lines do not have equal slopes.

– $$m_1 = 4, m_2 = -\frac{1}{4}$$: These lines do not have equal slopes.

Therefore, option A with slopes $m_1 = 5$ and $m_2 = 5$ represents parallel lines.

Key Concept: Intermediate Calculation

a) $4x + 6y – 2 = 0$

[Solution Description] Using the two-point form: $$\frac{y – y_1}{y_2 – y_1} = \frac{x – x_1}{x_2 – x_1}$$, we substitute $(-2, 1)$ for $(x_1, y_1)$ and $(4, -3)$ for $(x_2, y_2)$. This gives us:

$$\frac{y – 1}{-3 – 1} = \frac{x + 2}{4 + 2}$$

Simplifying further,

$$\frac{y – 1}{-4} = \frac{x + 2}{6}$$

Cross-multiplying, we get:

$$6(y – 1) = -4(x + 2)$$

Distributing both sides:

$$6y – 6 = -4x – 8$$

Rearranging to standard form:

$$4x + 6y – 2 = 0$$

Therefore, the equation of the line is $4x + 6y – 2 = 0$.

Key Concept: Advanced Line Division, Complex Median Equation

a) Line: $y = 12x – 49$; Not a median

[Solution Description] First, find the point dividing $GH$ in the ratio 2:3 using section formula:

$$P = \left(\frac{2 \times 8 + 3 \times 2}{2+3}, \frac{2 \times 5 + 3 \times 3}{2+3}\right)$$

Calculate:

$$P = \left(\frac{16 + 6}{5}, \frac{10 + 9}{5}\right) = \left(\frac{22}{5}, \frac{19}{5}\right)$$

For the median through $I$, compute midpoint of $GH$:

$$M_{GH} = \left(\frac{2 + 8}{2}, \frac{3 + 5}{2}\right) = (5, 4)$$

Slope of $IP$:

$$m_{IP} = \frac{\frac{19}{5} + 1}{\frac{22}{5} – 4} = \frac{\frac{24}{5}}{\frac{2}{5}} = 12$$

Equation of line $IP$ using point-slope form at $I(4, -1)$:

$$y + 1 = 12(x – 4)$$

Simplify:

$$y = 12x – 49$$

The calculated line cannot be the median through $I$ as it doesn’t pass through midpoint $M_{GH}(5, 4)$, indicating a different role such as internal division but not as a median.

Key Concept: Intercepts Calculation

b) $-2$

[Solution Description] The y-intercept occurs when $x = 0$. Substitute $x = 0$ into the equation $4x – 5y = 10$:

$$4(0) – 5y = 10$$

$$-5y = 10$$

Solving for $y$:

$$y = \frac{10}{-5}$$

$$y = -2$$

Therefore, the y-intercept is -2.

Key Concept: General Equation

b) $3x – 4y + 7 = 0$

[Solution Description] A general linear equation has the form $Ax + By + C = 0$, where $A$, $B$, and $C$ are constants. Option $b)$, $3x – 4y + 7 = 0$, fits this form as it has terms in $Ax$, $By$, and $C$ format.

Key Concept: Basic Equation

a) $Ax + By + C = 0$

[Solution Description] The general form of a line equation is given by $Ax + By + C = 0$, where A and B are not zero simultaneously.

Key Concept: Coordinate Identification

a) (2, 3)

[Solution Description] Use the midpoint formula: $$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$. Here, $x_1 = 0$, $x_2 = 4$, $y_1 = 0$, $y_2 = 6$:

$$\left(\frac{0 + 4}{2}, \frac{0 + 6}{2}\right) = \left(\frac{4}{2}, \frac{6}{2}\right) = (2, 3)$$

Thus, the midpoint is (2, 3).

Key Concept: Slope Identification

a) $\frac{10}{3}$

[Solution Description] To determine the slope $m$, use the slope formula:

$$m = \frac{y_2 – y_1}{x_2 – x_1} = \frac{7 – (-3)}{5 – 2} = \frac{10}{3}$$

Thus, the slope of the line is $\frac{10}{3}$.

Key Concept: Parallel Line

a) $y = 2x – 5$

[Solution Description] The given line has a slope of 2. A parallel line will have the same slope. Using point-slope form:

$$y – 1 = 2(x – 3).$$

Expand the equation:

$$y – 1 = 2x – 6.$$

Add 1 to both sides:

$$y = 2x – 5.$$

The equation of the line parallel to the given line and passing through the point (3, 1) is $y = 2x – 5$.

Key Concept: Basic Angle Calculation

c) $90^\circ$

[Solution Description] The angle $\theta$ between two lines with slopes $m_1$ and $m_2$ is given by the formula:

$$\tan \theta = \left| \frac{m_2 – m_1}{1 + m_1 m_2} \right|$$

Here, $m_1 = 1$ and $m_2 = -1$. Plugging in these values:

$$\tan \theta = \left| \frac{-1 – 1}{1 + (1)(-1)} \right| = \left| \frac{-2}{0} \right|$$

Since division by zero is undefined, this suggests that the lines are perpendicular, thus forming an angle of $90^\circ$.

Key Concept: Slope Definition

c) $\frac{1}{\sqrt{3}}$

[Solution Description] The slope $m$ of a line inclined at an angle $\theta$ with the x-axis is given by $$m = \tan \theta$$. For $$\theta = 30^\circ$$, we have:

$$m = \tan 30^\circ = \frac{1}{\sqrt{3}}$$

Therefore, the slope of the line is $$\frac{1}{\sqrt{3}}$$.

Key Concept: Slope and Triangle Area, Slope and Line Intersection

c) 3

[Solution Description] To find the area of the triangle formed by points $A(1, 3)$, $B(-2, -1)$, and $C(4, 5)$, we use the formula for the area of a triangle given its vertices coordinates:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$$

Substituting the coordinates of the points:

$$\text{Area} = \frac{1}{2} \left| 1(-1-5) + (-2)(5-3) + 4(3+1) \right|$$

Simplifying the expression inside the absolute value:

$$\text{Area} = \frac{1}{2} \left| 1(-6) – 2(2) + 4(4) \right|$$

$$\text{Area} = \frac{1}{2} \left| -6 – 4 + 16 \right|$$

$$\text{Area} = \frac{1}{2} \times 6 = 3$$

Thus, the area of the triangle is 3 square units.

Key Concept: Slope Calculation

c) 2

[Solution Description] To calculate the slope of the line passing through two points, we use the formula $$m = \frac{y_2 – y_1}{x_2 – x_1}$$. With the given points $$(x_1, y_1) = (1, 2)$$ and $$(x_2, y_2) = (4, 8)$$, substitute these values into the formula:

$$m = \frac{8 – 2}{4 – 1} = \frac{6}{3} = 2$$

Hence, the slope of the line is 2.

Key Concept: Slope Calculation

b) $\frac{2}{5}$

[Solution Description] Convert the equation to slope-intercept form, $$y = mx + c$$. Start with:

$$\frac{x}{5} – \frac{y}{2} = 1$$

Multiply through by 10 to clear fractions:

$$2x – 5y = 10$$

Solve for $y$:

$$-5y = -2x + 10 \quad \Rightarrow \quad y = \frac{2}{5}x – 2$$

The slope $(m)$ of the line is $\frac{2}{5}$.

Key Concept: Parallel Lines

d) $m_1 = -3, m_2 = -3$

[Solution Description] Two lines are parallel if their slopes are equal. Consider each pair of slopes:

– Pair A: $$m_1 = 2, m_2 = 2$$

– Pair B: $$m_1 = -1, m_2 = 1$$

– Pair C: $$m_1 = 0, m_2 = 1$$

– Pair D: $$m_1 = -3, m_2 = -3$$

Only Pair A and Pair D have equal slopes.

Key Concept: Simple Calculation

a) $2$

[Solution Description] Use the formula $$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$$, where $A = 5$, $B = 12$, $C = -20$, and the point is $(2, 3)$:

$$d = \frac{|5(2) + 12(3) – 20|}{\sqrt{5^2 + 12^2}} = \frac{|10 + 36 – 20|}{\sqrt{25 + 144}} = \frac{|26|}{\sqrt{169}} = \frac{26}{13} = 2.$$

The distance is $2$.

Key Concept: Conversion Between Forms

a) $y = \frac{2}{3}x + 2$

[Solution Description] To convert the given line from general form $$Ax + By + C = 0$$ to slope-intercept form $$y = mx + c$$, solve for $y$.

Given:

$$2x – 3y + 6 = 0$$

Rearrange to isolate $y$:

$$-3y = -2x – 6$$

Divide each term by $-3$:

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

Hence, the equation in slope-intercept form is $$y = \frac{2}{3}x + 2$$.

Key Concept: Basic Distance

b) 5

[Solution Description] To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the formula:

$$PQ = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}.$$

Here, $x_1 = 2$, $y_1 = 3$, $x_2 = 5$, and $y_2 = 7$.

Step 1: Calculate $(x_2 – x_1)^2$:

$$(5 – 2)^2 = 3^2 = 9.$$

Step 2: Calculate $(y_2 – y_1)^2$:

$$(7 – 3)^2 = 4^2 = 16.$$

Step 3: Add the results from step 1 and 2:

$$9 + 16 = 25.$$

Step 4: Take the square root of the result from step 3:

$$\sqrt{25} = 5.$$

Therefore, the distance between these two points is 5.

Key Concept: Identify Vertices

a) (0,1), (0,5), (2,3)

[Solution Description] To find the vertices, solve for intersection points:

1. Intersection of $y = x + 1$ with $x = 0$: $(0, 1)$.

2. Intersection of $y = -x + 5$ with $x = 0$: $(0, 5)$.

3. Solve $y = x + 1$ and $y = -x + 5$ for the third vertex:

$$x + 1 = -x + 5$$

$$2x = 4$$

$$x = 2$$

$$y = x + 1 = 2 + 1 = 3$$

$$\text{Third vertex: } (2, 3)$$

Hence, the vertices are $(0, 1)$, $(0, 5)$, and $(2, 3)$.

Key Concept: Formula Recall

c) $d = \frac{|c_2 – c_1|}{\sqrt{1 + m^2}}$

[Solution Description] The formula for the distance between two parallel lines $$y = mx + c_1$$ and $$y = mx + c_2$$ is given by:

$$d = \frac{|c_2 – c_1|}{\sqrt{1 + m^2}}$$

Key Concept: Basic Concept

c) Introduction of analytical geometry

[Solution Description] René Descartes introduced analytical geometry, which combines algebra and geometry, allowing for a mathematical representation of geometric shapes.

Key Concept: Slope Zero or Undefined

d) Undefined

[Solution Description] For the points (5, 9) and (5, 15), since the $x$-coordinates are the same, the line is vertical.

The slope for a vertical line is undefined because it would require dividing by zero:

$$m = \frac{15 – 9}{5 – 5} = \frac{6}{0}$$

Since division by zero is not possible, the slope is undefined.

Key Concept: Intercept Form

b) $5x – 4y = 20$

[Solution Description] The equation of the line in intercept form is $$\frac{x}{a} + \frac{y}{b} = 1$$. Here, $a = 4$ and $b = -5$. Substituting these values, we get:

$$\frac{x}{4} + \frac{y}{-5} = 1$$

Multiplying through by 20 to clear fractions gives:

$$5x – 4y = 20$$

Therefore, the equation of the line is $5x – 4y = 20$.

Key Concept: Perpendicular Lines Identification

b) Yes

[Solution Description] Two lines are perpendicular if the product of their slopes is $-1$. For the given slopes $3$ and $-\frac{1}{3}$, we check:

$$3 \cdot \left(-\frac{1}{3}\right) = -1$$

Since the product is indeed $-1$, the lines are perpendicular.

Key Concept: Parallel Lines

a) $m_1 = 4, m_2 = 4$

[Solution Description] For two lines to be parallel, their slopes must be equal $m_1 = m_2$. Comparing each option:

– A) Slopes 4 and 4 are equal.

– B) Slopes 3 and -3 are not equal.

– C) Slopes $\frac{2}{3}$ and $-\frac{2}{3}$
are not equal.

– D) Slopes 5 and $-\frac{5}{5}$ are not equal.

Thus, option A is correct.

Key Concept: Perpendicular Line Equation

c) $y = -3x – 1$

[Solution Description] The original line has a slope $$m = \frac{1}{3}$$. A line perpendicular to this will have a slope that is the negative reciprocal, $m_p = -3$. Using the point-slope form of a line $y – y_1 = m(x – x_1)$ with point $(0, -1)$ and slope $m_p = -3$, we get:

$$y – (-1) = -3(x – 0)$$

Simplifying this equation:

$$y + 1 = -3x

y = -3x – 1$$

So, the equation of the line is $y = -3x – 1$.

Key Concept: Collinearity of Three Points

a) Yes

[Solution Description] To check if the points are collinear, calculate the slopes between each pair of points and compare them. The slope between $(1, 3)$ and $(4, 11)$ is $$m_1 = \frac{11 – 3}{4 – 1} = \frac{8}{3}$$. The slope between $(4, 11)$ and $(7, 19)$ is $$m_2 = \frac{19 – 11}{7 – 4} = \frac{8}{3}$$. Since $m_1 = m_2$, the points are collinear.

Key Concept: Slope Identification

b) 2

[Solution Description] To find the slope $m$ of a line passing through two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the formula:

$$m = \frac{y_2 – y_1}{x_2 – x_1}$$

Substituting the given points $(2, 3)$ and $(5, 9)$ into this formula gives:

$$m = \frac{9 – 3}{5 – 2} = \frac{6}{3} = 2$$

Therefore, the slope of the line is 2.

Key Concept: Midpoint Calculation

a) (9, 10)

[Solution Description] The midpoint of the line segment joining two points $(x_1, y_1)$ and $(x_2, y_2)$ is given by:

$$\left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)$$

Substitute $(x_1, y_1) = (5, 8)$ and $(x_2, y_2) = (13, 12)$:

$$\left(\frac{5 + 13}{2}, \frac{8 + 12}{2}\right) = \left(\frac{18}{2}, \frac{20}{2}\right) = (9, 10)$$

So, the midpoint is $(9, 10)$.

Key Concept: Slope Identification

b) 2

[Solution Description] To find the slope $m$ of a line that passes through the points $(x_1, y_1)$ and $(x_2, y_2)$, use the formula:

$$m = \frac{y_2 – y_1}{x_2 – x_1}$$

Substituting the given points $(x_1, y_1) = (4, 5)$ and $(x_2, y_2) = (9, 15)$ into the formula gives:

$$m = \frac{15 – 5}{9 – 4} = \frac{10}{5} = 2$$

Therefore, the slope of the line is $2$.

Key Concept: Path Optimization

a) $y = -\frac{1}{3}x + \frac{7}{3}$

[Solution Description] The shortest path from a point to a line is along the perpendicular. The line perpendicular to $y = 3x + 4$ will have a slope that is the negative reciprocal of 3, thus $-\frac{1}{3}$. Using the point-slope form with point $(1, 2)$:

$$y – 2 = -\frac{1}{3}(x – 1)$$

This simplifies to:

$$y = -\frac{1}{3}x + \frac{7}{3}$$

Hence, the equation of the path is $$y = -\frac{1}{3}x + \frac{7}{3}$$.

Key Concept: Slope Equality

a) $y = x + 2$ and $y = x – 4$

[Solution Description] Lines are parallel if their slopes are equal. For the given pairs:

– Pair (A): The slope of both lines is 1, so they are parallel.

– Pair (B): The slopes are different $(-2$ and $3)$, so not parallel.

– Pair (C): The slopes are different $(2$ and $-2)$, so not parallel.

– Pair (D): The slopes are different $(4$ and $-4)$, so not parallel.

Therefore, only pair (A) has parallel lines.

Key Concept: Slope from Inclination

a) $\sqrt{3}$

[Solution Description] The relationship between the inclination angle $\alpha$ and the slope $m$ is given by:

$$m = \tan \alpha$$

For an inclination of $$60^\circ$$:

$$m = \tan 60^\circ = \sqrt{3}$$

Therefore, the slope of the line is $\sqrt{3}$.

Key Concept: Conceptual Understanding

b) The points are collinear

[Solution Description] According to the concept of collinearity, if the area of a triangle is zero, then the points are collinear. We calculate the area using the formula:

$$\text{Area} = \frac{1}{2} \left| x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2) \right|$$

Substituting the coordinates $(1, 1)$, $(2, 2)$, and $(3, 3)$:

$$\text{Area} = \frac{1}{2} \left| 1(2-3) + 2(3-1) + 3(1-2) \right|$$

$$= \frac{1}{2} \left| 1(-1) + 2(2) + 3(-1) \right|$$

$$= \frac{1}{2} \left| -1 + 4 – 3 \right|$$

$$= \frac{1}{2} \times 0 = 0$$

Since the area is $0$, the points are collinear.

Key Concept: Inclination Identification

a) $0^\circ$

[Solution Description] A line parallel to the x-axis has no vertical change and thus has an inclination of $0^\circ$ since it aligns perfectly without any angle. Therefore, the inclination is $0^\circ$.

Key Concept: Distance Formula

b) 5

[Solution Description] To find the distance between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the distance formula $$d = \sqrt{(x_2 – x_1)^2 + (y_2 – y_1)^2}$$. For this problem:

Given points: $(1, 2) = (x_1, y_1)$ and $(4, 6) = (x_2, y_2)$.

Substitute into the formula:

$$d = \sqrt{(4 – 1)^2 + (6 – 2)^2}$$

$$= \sqrt{3^2 + 4^2}$$

$$= \sqrt{9 + 16}$$

$$= \sqrt{25}$$

$$= 5$$

Thus, the distance between the points is $5$.

Key Concept: Intercept Calculation

c) $(2, 0)$ and $(0, -6)$

[Solution Description] To find the x-intercept, set $y = 0$ in the equation and solve for $x$:

$$6x – 2(0) = 12 \implies 6x = 12 \implies x = \frac{12}{6} = 2$$

So, the x-intercept is $(2, 0)$.

To find the y-intercept, set $x = 0$ in the equation and solve for $y$:

$$6(0) – 2y = 12 \implies -2y = 12 \implies y = \frac{12}{-2} = -6$$

So, the y-intercept is $(0, -6)$.

Therefore, the intercepts are $(2, 0)$ and $(0, -6)$.

Key Concept: Equation Application

c) $x = -5$

[Solution Description] The given point is (-5, 4). A vertical line passes through the point with constant x-coordinate. Therefore, the equation of the line will be $x = -5$.

Key Concept: Undefined Slope

b) Undefined

[Solution Description] The slope $m$ of a line through points $$(x_1, y_1)$$ and $$(x_2, y_2)$$ is calculated as:

$$m = \frac{y_2 – y_1}{x_2 – x_1}$$

Given points $$(x_1, y_1) = (2, 5)$$ and $$(x_2, y_2) = (2, -3)$$, we have:

$$m = \frac{-3 – 5}{2 – 2} = \frac{-8}{0}$$

Since division by zero is undefined, the slope of the line is undefined.

Key Concept: Inclination Identification

c) $90^\circ$

[Solution Description] A line parallel to the y-axis is vertical. According to the syllabus, the inclination of a vertical line is $90^\circ$.

Key Concept: Basic Slope Formula

b) 2

[Solution Description] To find the slope $m$, use the formula $$m = \frac{y_2 – y_1}{x_2 – x_1}$$. Substitute the given points: $$x_1 = 1, y_1 = 2, x_2 = 3, y_2 = 6$$.

$$m = \frac{6 – 2}{3 – 1} = \frac{4}{2} = 2$$

Therefore, the slope of the line is 2.

Key Concept: Distance Formula

b) 5

[Solution Description] To find the distance $d$ between two points $(x_1, y_1)$ and $(x_2, y_2)$, we use the distance formula:

$$d = \sqrt{(x_2-x_1)^2 + (y_2-y_1)^2}$$

Substituting the given points $(1, 2)$ and $(4, 6)$, we have:

$$d = \sqrt{(4-1)^2 + (6-2)^2}$$

$$d = \sqrt{3^2 + 4^2}$$

$$d = \sqrt{9 + 16}$$

$$d = \sqrt{25}$$

$$d = 5$$

Thus, the distance between the points is 5 units.