Find the equation of the median through the vertex $R$ of triangle $PQR$ with vertices $P(2, 1)$, $Q(-2, 3)$, and $R(4, 5)$.

Correct Option: A. Solution: The midpoint of $PQ$ is $\left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = (0, 2)$. The slope of the median $R(4, 5)$ to $(0, 2)$ is $\frac{2 - 5}{0 - 4} = \frac{3}{4}$. Using point-slope form, $y - 5 = \frac{3}{4}(x - 4)$, which simplifies to $y = \frac{3}{4}x + 2$.

The owner of a milk store finds that he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres each week at Rs 16/litre. Assuming a linear relationship between selling price and demand, how many litres could he sell weekly at Rs 17/litre?

Correct Option: C. Solution: The slope $m$ is $\frac{1220 - 980}{16 - 14} = 120$. Using the equation $d = 1220 + 120(p - 16)$, at $p = 17$, $d = 740$.

Straight Lines

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Summary

Chapter 9: Straight Lines

Summary

Introduction to two-dimensional coordinate geometry.
Analytical geometry combines algebra and geometry, introduced by René Descartes.
Key concepts include coordinate axes, plotting points, distance between points, and section formulae.
Slope of a line is crucial for representing lines algebraically.
Distance formulas and equations for lines are essential for solving geometric problems.

Key Formulas/Definitions

Distance between two points:

$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Slope (m) of a line through points (x₁, y₁) and (x₂, y₂):

$m = \frac{y_2 - y_1}{x_2 - x_1}$
Equation of a line:

$y - y_1 = m(x - x_1)$
Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):

$\text{Area} = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$
Collinearity condition: Points A, B, and C are collinear if the slopes of AB and BC are equal.

Learning Objectives

Understand the basics of coordinate geometry.
Calculate the distance between points in a plane.
Determine the slope of a line and its significance.
Formulate the equations of lines in various forms.
Solve problems involving the area of triangles and collinearity.

Common Mistakes/Exam Tips

Mistake: Confusing the slope of a vertical line as a number; it is undefined.
Tip: Always check if points are collinear by comparing slopes.
Mistake: Forgetting to apply the absolute value when calculating area.
Tip: Use the correct formula for distance based on the context (between points or from a point to a line).

Important Diagrams

Fig 9.1: Shows points (6, -4) and (3, 0) on the XY-plane with distances marked.
Fig 9.12: Illustrates the concept of y-intercept and slope of a line.
Fig 9.17: Depicts a coordinate plane with labeled axes and lines intersecting at specific points.

Learning Objectives

Understand the concept of straight lines in coordinate geometry.
Recall the basics of coordinate geometry including plotting points and distance between points.
Identify and apply the slope of a line in various contexts.
Derive the equation of a line using different forms such as slope-intercept and intercept forms.
Calculate the distance from a point to a line.
Solve problems involving the equations of lines, including parallel and perpendicular lines.
Analyze the relationship between the slopes of two lines and their geometric implications (parallelism and perpendicularity).
Explore the concept of collinearity among points in a plane.

Detailed Notes

Chapter 9: Straight Lines

9.1 Introduction

Coordinate Geometry: A combination of algebra and geometry, first systematically studied by René Descartes in 1637.
Key Concepts: Coordinate axes, coordinate plane, plotting points, distance between points, section formulae.
Example: Points (6, -4) and (3, 0) in the XY-plane.

9.2 Slope of a Line

Definition: A line in a coordinate plane forms two supplementary angles with the x-axis.
Slope Formula: The slope (m) of a line passing through points (x₁, y₁) and (x₂, y₂) is given by:

$m = \frac{y_2 - y_1}{x_2 - x_1}$

9.3 Important Formulae

Distance between two points:
- Distance between points P (x₁, y₁) and Q (x₂, y₂):
$d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$
Coordinates of a point dividing a line segment:
- If point P divides the line segment joining A (x₁, y₁) and B (x₂, y₂) in the ratio m:n, then:
$P = \left( \frac{mx_2 + nx_1}{m+n}, \frac{my_2 + ny_1}{m+n} \right)$
Mid-point of a line segment:
- If m = n, the mid-point is:
$P = \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right)$
Area of Triangle:
- Area of triangle with vertices (x₁, y₁), (x₂, y₂), (x₃, y₃):
$Area = \frac{1}{2} \left| x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2) \right|$

9.4 Distance of a Point From a Line

Definition: The distance of a point from a line is the length of the perpendicular drawn from the point to the line.
Distance Formula: For line L: Ax + By + C = 0, the distance d from point P (x₁, y₁) is:

$d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$

9.5 Exercises

Example Problems:
1. Find the distance of the point (3, 5) from the line 3x - 4y - 26 = 0.
2. Find the equation of the line passing through (-3, 5) and perpendicular to the line through points (2, 5) and (-3, 6).
3. Find the area of the triangle formed by the lines y = x, x + y = 0, and x - k = 0.

Conclusion

The study of straight lines is fundamental in geometry and has numerous applications in various fields.

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Misunderstanding the Slope: Students often confuse the slope of a line with its angle. Remember that the slope is given by the formula $m = \frac{y_2 - y_1}{x_2 - x_1}$ .
Incorrectly Identifying Intercepts: When finding the intercepts of a line, ensure you set the appropriate variable to zero. For example, to find the y-intercept, set $x = 0$ .
Forgetting to Check for Parallel and Perpendicular Lines: Students may forget that parallel lines have equal slopes and that the product of the slopes of perpendicular lines is -1.
Misapplying Distance Formulas: Ensure you use the correct formula for the distance from a point to a line: $d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}}$ .

Tips for Success

Practice with Different Forms of Line Equations: Familiarize yourself with slope-intercept, point-slope, and intercept forms of line equations.
Visualize Problems: Draw diagrams whenever possible to better understand the relationships between points, lines, and angles.
Double-Check Calculations: Always review your calculations, especially when determining slopes and distances.
Understand the Concepts: Rather than memorizing formulas, focus on understanding the underlying concepts of coordinate geometry.

Practice & Assessment

Multiple Choice Questions

y = \frac{1}{2}x + 3

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

y = \frac{1}{2}x - 3

y = -\frac{1}{2}x - 3

Correct Answer: A

Solution:

The midpoint of

PQ

\left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = (0, 2)

. The slope of the median

R(4, 5)

(0, 2)

\frac{2 - 5}{0 - 4} = \frac{3}{4}

. Using point-slope form,

y - 5 = \frac{3}{4}(x - 4)

, which simplifies to

y = \frac{3}{4}x + 2

y - 5 = -\frac{1}{5}(x + 3)

y - 5 = 5(x + 3)

y - 5 = \frac{1}{5}(x + 3)

y - 5 = -5(x + 3)

Correct Answer: A

Solution:

The slope of the line through (2, 5) and (-3, 6) is

-\frac{1}{5}

. The perpendicular slope is 5. Therefore, the equation is

y - 5 = -\frac{1}{5}(x + 3)

x + y = 5

x - y = -1

x + y = 4

x - y = 1

Correct Answer: A

Solution:

If a line cuts off equal intercepts on the axes, its equation is of the form

x + y = c

. Substituting the point (2, 3) into the equation gives

2 + 3 = c

, so

c = 5

. Therefore, the equation is

x + y = 5

x + y = 9

x + y = 6

x + y = 5

2x + 2y = 9

Correct Answer: A

Solution:

The intercept form of the line is

\frac{x}{a} + \frac{y}{b} = 1

, where

a + b = 9

. Substituting the point (2, 2) into the equation, we get

\frac{2}{a} + \frac{2}{b} = 1

. Solving these equations simultaneously gives

a = 6

b = 3

. Thus, the equation is

x + y = 9

x + y = 5

x - y = 1

x + y = 6

x - y = 0

Correct Answer: C

Solution:

For a line with equal intercepts

a

on both axes, the equation is

x + y = a

. Substituting the point

(2, 3)

, we get

2 + 3 = a

, so

a = 5

. Therefore, the equation is

x + y = 6

3x - 4y - 12 = 0

4x + 3y - 9 = 0

3x + 4y - 12 = 0

3x - 4y + 12 = 0

Correct Answer: A

Solution:

The slope of the line

3x - 4y + 5 = 0

\frac{3}{4}

. A line parallel to this will have the same slope. Using the point-slope form,

y - 0 = \frac{3}{4}(x - 3)

, which simplifies to

3x - 4y - 12 = 0

(1, 2)

(2, 1)

(3, 0)

(0, 3)

Correct Answer: A

Solution:

The equation of the line perpendicular to

3x - 4y - 16 = 0

through the point (-1, 3) is

4x + 3y = c

. Substituting (-1, 3) into the equation gives

c = 5

. Solving the system

3x - 4y = 16

and

4x + 3y = 5

gives the intersection point as (1, 2).

y - 2 = \sqrt{3}x

y - 2 = x

y - 2 = \frac{1}{\sqrt{3}}x

y - 2 = -\sqrt{3}x

Correct Answer: A

Solution:

The slope of a line making an angle of 60° with the x-axis is

\tan(60°) = \sqrt{3}

. Using the point-slope form, the equation is:

y - 2 = \sqrt{3}x

-\frac{4}{3}

\frac{4}{3}

-\frac{3}{4}

\frac{3}{4}

Correct Answer: A

Solution:

The slope of a line passing through two points

(x_1, y_1)

and

(x_2, y_2)

is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting the given points

(3, 0)

and

(6, -4)

, we get

m = \frac{-4 - 0}{6 - 3} = -\frac{4}{3}

y = -3x + 3

y = 3x - 3

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

y = \frac{1}{3}x - 1

Correct Answer: A

Solution:

The slope of the line segment is

m = \frac{3 - 0}{2 - 1} = 3

. The perpendicular slope is

-\frac{1}{3}

. Using the point-slope form, the equation is

y = -3x + 3

L = 0.0021C + 124.9

L = 0.0019C + 124.942

L = 0.002C + 124.942

L = 0.0021C + 124.942

Correct Answer: A

Solution:

The slope

m

\frac{125.134 - 124.942}{110 - 20} = 0.0021

. Using the point-slope form with point

(20, 124.942)

L - 124.942 = 0.0021(C - 20)

, simplifying to

L = 0.0021C + 124.9

y = \sqrt{3}x + 2

y = \frac{1}{\sqrt{3}}x + 2

y = -\sqrt{3}x + 2

y = \frac{1}{2}x + 2

Correct Answer: A

Solution:

The slope of the line is

\tan(\frac{\pi}{3}) = \sqrt{3}

. Using the point-slope form, the equation is

y = \sqrt{3}x + 2

\frac{x}{6} + \frac{y}{3} = 1

\frac{x}{3} + \frac{y}{6} = 1

\frac{x}{4.5} + \frac{y}{4.5} = 1

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

Correct Answer: A

Solution:

Let the intercepts be

a

and

b

. Then

a + b = 9

. The equation of the line is

\frac{x}{a} + \frac{y}{b} = 1

. Substituting

(2, 2)

gives

\frac{2}{a} + \frac{2}{b} = 1

. Solving

a + b = 9

and

\frac{2}{a} + \frac{2}{b} = 1

gives

a = 6

b = 3

. Thus, the equation is

\frac{x}{6} + \frac{y}{3} = 1

4.5

3.6

2.8

5.2

Correct Answer: B

Solution:

First, find the slope of BC:

m_{BC} = \frac{2 + 1}{1 - 4} = -1

. The slope of the altitude from A is the negative reciprocal,

m = 1

. Using point A (2, 3), the equation of the altitude is

y - 3 = 1(x - 2)

, or

y = x + 1

. The length of the altitude is the perpendicular distance from A to BC. The equation of BC is

y + 1 = -1(x - 4)

, or

y = -x + 3

. The distance from A to BC is

\frac{|2(-1) + 3 - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{2}{\sqrt{2}} = \sqrt{2} \approx 1.41

. The correct length is 3.6.

y = \frac{1}{2}x + 3

y = -\frac{1}{2}x + 5

y = \frac{1}{2}x + 2

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

Correct Answer: A

Solution:

The midpoint of PQ is

\left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = (0, 2)

. The slope of the median from R to this midpoint is

\frac{2 - 5}{0 - 4} = \frac{3}{4}

. Using the point-slope form, the equation is

y - 5 = \frac{1}{2}(x - 4)

, which simplifies to

y = \frac{1}{2}x + 3

4 units

6 units

3 units

5 units

Correct Answer: A

Solution:

The vertical line segment extends from (6, -4) to (6, 0). The length is the difference in the Y-coordinates: |0 - (-4)| = 4 units.

y - 3 = -\frac{1}{2}(x - 2)

y - 3 = 2(x - 2)

y - 3 = \frac{1}{2}(x - 2)

y - 3 = -2(x - 2)

Correct Answer: A

Solution:

The slope of a line perpendicular to another line with slope

m

-\frac{1}{m}

. Given the slope is 2, the perpendicular slope is

-\frac{1}{2}

. Using the point-slope form

y - y_1 = m(x - x_1)

, we substitute

(x_1, y_1) = (2, 3)

and

m = -\frac{1}{2}

to get

y - 3 = -\frac{1}{2}(x - 2)

y = -5x - 10

y = x + 8

y = -x + 2

y = 5x + 20

Correct Answer: B

Solution:

The slope of the line through (2, 5) and (-3, 6) is 5\frac{6 - 5}{-3 - 2} = -\frac{1}{5}5. A line perpendicular to it has a slope of 5. Using the point (-3, 5), the equation is 5y - 5 = 5(x + 3)5, simplifying to 5y = 5x + 205.

y = 3x - 3

y = 3x + 3

y = -3x + 3

y = -3x - 3

Correct Answer: A

Solution:

The line dividing the segment in the ratio 1:n has a slope of 3. Thus, the equation is

y = 3x - 3

x - 2y = -5

x + 2y = 5

2x - y = 5

x - 2y = 5

Correct Answer: A

Solution:

The midpoint of the segment is

\left( \frac{3 + (-1)}{2}, \frac{4 + 2}{2} \right) = (1, 3)

. The slope of the line segment is

\frac{2 - 4}{-1 - 3} = \frac{-2}{-4} = \frac{1}{2}

. The slope of the perpendicular bisector is the negative reciprocal,

-2

. The equation of the bisector is

y - 3 = -2(x - 1)

, which simplifies to

x - 2y = -5

y - 3 = (2 - \sqrt{3})(x - 2)

y - 3 = (2 + \sqrt{3})(x - 2)

y - 3 = (-2 + \sqrt{3})(x - 2)

y - 3 = (-2 - \sqrt{3})(x - 2)

Correct Answer: A

Solution:

If the slope of one line is 2, the angle between the lines is 60°, and the tangent of the angle between two lines m1 and m2 is

\left|\frac{m_1 - m_2}{1 + m_1 m_2}\right| = \sqrt{3}

. Solving for m2 gives m2 = 2 - \sqrt{3}. Thus, the equation is y - 3 = (2 - \sqrt{3})(x - 2).

y = 3x - 1

y = -3x + 7

y = x - 1

y = -x + 4

Correct Answer: D

Solution:

The slope of the line through (1, 0) and (2, 3) is 5\frac{3 - 0}{2 - 1} = 35. A line perpendicular to it has a slope of 5-\frac{1}{3}5. Using the midpoint of the segment 5\left(\frac{1 + 2}{2}, \frac{0 + 3}{2}\right) = \left(\frac{3}{2}, \frac{3}{2}\right)5, the equation is 5y - \frac{3}{2} = -\frac{1}{3}(x - \frac{3}{2})5, simplifying to 5y = -x + 45.

y = -2x + 5

y = 2x + 1

y = -\frac{1}{2}x + 5

y = \frac{1}{2}x + 3

Correct Answer: D

Solution:

The midpoint of the segment is (1, 3). The slope of the segment is

\frac{1}{2}

, so the perpendicular slope is -2. Using the point-slope form, the equation is

y = \frac{1}{2}x + 3

y = \frac{9}{2}x + 9

y = -\frac{9}{2}x + 9

y = \frac{2}{9}x + 9

y = -\frac{2}{9}x + 9

Correct Answer: B

Solution:

The line through (-2, 9) and perpendicular to the line from the origin has a slope of

-\frac{x_1}{y_1} = -\frac{-2}{9} = \frac{9}{2}

. The equation is

y = \frac{9}{2}x + 9

2x + 9y = 85

2x - 9y = 85

9x + 2y = 85

9x - 2y = 85

Correct Answer: A

Solution:

The line through (-2, 9) and perpendicular to the origin has a slope of

m = -\frac{2}{9}

. Using the point-slope form, the equation is

y - 9 = -\frac{2}{9}(x + 2)

, which simplifies to

2x + 9y = 85

y - 3 = (x - 2)

y - 3 = \sqrt{3}(x - 2)

y - 3 = -\sqrt{3}(x - 2)

y - 3 = -1(x - 2)

Correct Answer: A

Solution:

The slope

m

of the line making an angle

\theta = 45^\circ

with the x-axis is

m = \tan(45^\circ) = 1

. Using the point-slope form

y - y_1 = m(x - x_1)

with the point (2, 3), the equation is

y - 3 = 1(x - 2)

, which simplifies to

y - 3 = x - 2

y - 3 = -5(x - 2)

y - 3 = \frac{1}{5}(x - 2)

y - 3 = 5(x - 2)

y - 3 = -\frac{1}{5}(x - 2)

Correct Answer: B

Solution:

First, find the slope of the line passing through (2, 5) and (-3, 6):

m = \frac{6 - 5}{-3 - 2} = -\frac{1}{5}

. The slope of the line perpendicular to it is the negative reciprocal, which is 5. Therefore, the equation of the line passing through (2, 3) is

y - 3 = 5(x - 2)

y = -\sqrt{3}x + 2

y = \sqrt{3}x + 2

y = -x + 2

y = x + 2

Correct Answer: A

Solution:

The slope of the line is

\tan\left(\frac{2\pi}{3}\right) = -\sqrt{3}

. Using point (0, 2), the equation is y = -\sqrt{3}x + 2.

y - 3 = 2(x - 2)

y - 3 = \sqrt{3}(x - 2)

y - 3 = \frac{1}{2}(x - 2)

y - 3 = -\sqrt{3}(x - 2)

Correct Answer: A

Solution:

The slope of the second line is given as 2. Using the point-slope form, the equation is:

y - 3 = 2(x - 2)

y - 3 = -1/2(x - 2)

y - 3 = 2(x - 2)

y - 3 = 1/2(x - 2)

y - 3 = -2(x - 2)

Correct Answer: A

Solution:

The slope of the line perpendicular to a line with slope 2 is -1/2. Using the point-slope form, the equation is y - 3 = -1/2(x - 2).

860

780

740

700

Correct Answer: C

Solution:

The slope

m

\frac{1220 - 980}{16 - 14} = 120

. Using the equation

d = 1220 + 120(p - 16)

, at

p = 17

d = 740

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

y = 3x - 2

y = -3x + 3

y = \frac{1}{3}x + \frac{1}{3}

Correct Answer: B

Solution:

The slope of the line joining (1, 0) and (2, 3) is 3. A line perpendicular to it has a slope of -

\frac{1}{3}

. Using the midpoint formula and the point-slope form, the equation is

y = 3x - 2

y = -\frac{5}{3}x - \frac{2}{3}

y = \frac{5}{3}x + \frac{2}{3}

y = \frac{3}{5}x - \frac{2}{3}

y = -\frac{3}{5}x + \frac{2}{3}

Correct Answer: A

Solution:

The slope of the line is

m = \frac{-4 - 1}{2 + 1} = -\frac{5}{3}

. Using the point-slope form

y - y_1 = m(x - x_1)

with point

(-1, 1)

, the equation is

y - 1 = -\frac{5}{3}(x + 1)

, which simplifies to

y = -\frac{5}{3}x - \frac{2}{3}

y = -\frac{5}{3}x - \frac{2}{3}

y = \frac{5}{3}x - \frac{2}{3}

y = -\frac{3}{5}x + \frac{7}{5}

y = \frac{3}{5}x + \frac{7}{5}

Correct Answer: A

Solution:

The slope of the line is

m = \frac{-4 - 1}{2 - (-1)} = -\frac{5}{3}

. Using the point-slope form, the equation is

y - 1 = -\frac{5}{3}(x + 1)

, which simplifies to

y = -\frac{5}{3}x - \frac{2}{3}

x + 2y = 8

2x - y = 1

x - 2y = 4

2x + y = 7

Correct Answer: B

Solution:

The slope of BC is (2 + 1)/(1 - 4) = -1. The altitude from A is perpendicular to BC, so its slope is 1. Using point A (2, 3), the equation is y - 3 = 1(x - 2), or 2x - y = 1.

1340 litres

1100 litres

980 litres

1400 litres

Correct Answer: A

Solution:

Let the price be

p

and the demand be

D

. The relationship is linear:

D = mp + c

. Using the points (14, 980) and (16, 1220), the slope

m = \frac{1220 - 980}{16 - 14} = 120

. The equation is

D = 120p + c

. Using

p = 14

980 = 120(14) + c

c = 300

. At

p = 17

D = 120(17) + 300 = 1340

litres.

x + y = 5

x - y = 1

x + y = 6

x - y = 2

Correct Answer: C

Solution:

For a line with equal intercepts, the equation is of the form

x + y = c

. Substituting (2, 3) gives

2 + 3 = c

, so

c = 5

. Therefore, the equation is

x + y = 5

4x + 3y = 5

4x - 3y = 5

3x + 4y = 5

3x - 4y = 5

Correct Answer: A

Solution:

The slope of the line 3x - 4y - 16 = 0 is

\frac{3}{4}

. The perpendicular slope is

-\frac{4}{3}

. Using the point-slope form with point (-1, 3), the equation is

y - 3 = -\frac{4}{3}(x + 1)

, simplifying to

4x + 3y = 5

x + 2y = 0

2x + y = 0

x - 2y = 0

2x - y = 0

Correct Answer: B

Solution:

Using the section formula, the coordinates of

R

are

\left( \frac{2x_1 + x_2}{3}, \frac{2y_1 + y_2}{3} \right)

. Solving for the line passing through

(0, 0)

and

(x_1, y_1)

gives

2x + y = 0

y = 2x - 3

y = x + 1

y = -x + 6

y = 3x - 7

Correct Answer: A

Solution:

The median through vertex R (4, 5) passes through the midpoint of side PQ. The midpoint of PQ is 5\left(\frac{2 + (-2)}{2}, \frac{1 + 3}{2}\right) = (0, 2)5. The slope of the median is 5\frac{5 - 2}{4 - 0} = \frac{3}{4}5. The equation of the median is 5y - 5 = \frac{3}{4}(x - 4)5, simplifying to 5y = 2x - 35.

y - 2 = \sqrt{3}(x - 0)

y - 2 = \frac{1}{\sqrt{3}}(x - 0)

y - 2 = \frac{1}{2}(x - 0)

y - 2 = 2(x - 0)

Correct Answer: A

Solution:

The slope of the line making an angle

\frac{\pi}{3}

with the x-axis is

\tan(\frac{\pi}{3}) = \sqrt{3}

. Thus, the equation is y - 2 = \sqrt{3}(x - 0).

y = \sqrt{3}x - 2

y = \sqrt{3}x - 4

y = \sqrt{3}x + 2

y = \frac{1}{\sqrt{3}}x - 2

Correct Answer: B

Solution:

The slope of the line making a 60° angle with the x-axis is

\tan(60°) = \sqrt{3}

. A parallel line has the same slope. Since it crosses the y-axis 2 units below the origin, the y-intercept is -2. The equation is

y = \sqrt{3}x - 4

-\frac{4}{3}

\frac{4}{3}

-\frac{3}{4}

\frac{3}{4}

Correct Answer: A

Solution:

The slope of a line passing through two points (x_1, y_1) and (x_2, y_2) is given by

m = \frac{y_2 - y_1}{x_2 - x_1}

. Substituting the given points A(3, 0) and B(6, -4), we get

m = \frac{-4 - 0}{6 - 3} = -\frac{4}{3}

L = 0.192C + 121.002

L = 0.002C + 124.942

L = 0.00214C + 124.57

L = 0.000192C + 124.942

Correct Answer: C

Solution:

The change in length per degree Celsius is 5\frac{125.134 - 124.942}{110 - 20} = 0.002145. Thus, the equation is 5L = 0.00214C + b5. Using 5L = 124.9425 when 5C = 205, 5124.942 = 0.00214(20) + b5, solving gives 5b = 124.575. Therefore, 5L = 0.00214C + 124.575.

(\frac{20}{25}, \frac{12}{25})

(\frac{20}{25}, \frac{16}{25})

(\frac{16}{25}, \frac{12}{25})

(\frac{16}{25}, \frac{20}{25})

Correct Answer: C

Solution:

Using the formula for the foot of the perpendicular, the coordinates are

\left(\frac{16}{25}, \frac{12}{25}\right)

-1

-5/3

5/3

Correct Answer: D

Solution:

Solution:

The excerpt states that the length

L

of a copper rod is a linear function of its Celsius temperature

C

Correct Answer: True

Solution:

The excerpt provides that

L = 124.942

Solution:

The excerpt clearly states that the QR code does not contain any scientific labels, formulas, or structures.

Correct Answer: True

Solution:

The excerpt mentions that the diagram labeled as 'Fig 9.7' has a point labeled 8 on the horizontal axis.

Straight Lines

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Summary

Chapter 9: Straight Lines

Summary

Key Formulas/Definitions

Learning Objectives

Common Mistakes/Exam Tips

Important Diagrams

Learning Objectives

Learning Objectives

Detailed Notes

Chapter 9: Straight Lines

9.1 Introduction

9.2 Slope of a Line

9.3 Important Formulae

9.4 Distance of a Point From a Line

9.5 Exercises

Conclusion

Exam Tips & Common Mistakes

Common Mistakes and Exam Tips

Common Pitfalls

Tips for Success

Practice & Assessment

Multiple Choice Questions

Q1.Find the equation of the median through the vertex RRR of triangle PQRPQRPQR with vertices P(2,1)P(2, 1)P(2,1), Q(−2,3)Q(-2, 3)Q(−2,3), and R(4,5)R(4, 5)R(4,5).

Correct Answer: A

Q2.What is the equation of the line passing through the point (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6)?

Correct Answer: A

Q3.Find the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2, 3).

Correct Answer: A

Q4.Find the equation of the line that passes through the point (2, 2) and cuts off intercepts on the axes whose sum is 9.

Correct Answer: A

Q5.What is the equation of the line that cuts off equal intercepts on the coordinate axes and passes through the point (2,3)(2, 3)(2,3)?

Correct Answer: C

Q6.A line passes through the point (3, 0) and is parallel to the line given by the equation 3x−4y+5=03x - 4y + 5 = 03x−4y+5=0. What is the equation of the line?

Correct Answer: A

Q7.Find the coordinates of the foot of the perpendicular from the point (-1, 3) to the line 3x−4y−16=03x - 4y - 16 = 03x−4y−16=0.

Correct Answer: A

Q8.Find the equation of the line that passes through the point (0, 2) and makes an angle of 60° with the positive x-axis.

Correct Answer: A

Q9.Consider a line passing through the points (3, 0) and (6, -4). What is the slope of this line?

Correct Answer: A

Q10.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. What is the equation of the line?

Correct Answer: A

Q11.The length LLL (in centimeters) of a copper rod is a linear function of its Celsius temperature CCC. If L=124.942L = 124.942L=124.942 when C=20C = 20C=20 and L=125.134L = 125.134L=125.134 when C=110C = 110C=110, express LLL in terms of CCC.

Correct Answer: A

Q12.What is the equation of the line passing through the point (0, 2) and making an angle of π3\frac{\pi}{3}3π​ with the positive x-axis?

Correct Answer: A

Q13.Find the equation of the line passing through the point (2,2)(2, 2)(2,2) and cutting off intercepts on the axes whose sum is 9.

Correct Answer: A

Q14.In the triangle ABC with vertices A (2, 3), B (4, -1), and C (1, 2), find the length of the altitude from vertex A.

Correct Answer: B

Q15.The vertices of triangle PQR are P (2, 1), Q (-2, 3), and R (4, 5). What is the equation of the median through vertex R?

Correct Answer: A

Q16.In a Cartesian coordinate system, the points (3, 0) and (6, -4) are given. A line is drawn from (6, -4) vertically upwards to the X-axis. What is the length of this vertical line segment?

Correct Answer: A

Q17.What is the equation of the line that passes through the point (2, 3) and is perpendicular to the line with slope 2?

Correct Answer: A

Q18.Find the equation of the line passing through (-3, 5) and perpendicular to the line through the points (2, 5) and (-3, 6).

Correct Answer: B

Q19.Find the equation of the line that divides the segment joining the points (1, 0) and (2, 3) in the ratio 1:n.

Correct Answer: A

Q20.A right bisector of a line segment joins the points (3, 4) and (-1, 2). What is the equation of this bisector?

Correct Answer: A

Q21.Two lines pass through the point (2, 3) and intersect each other at an angle of 60°. If the slope of one line is 2, what is the equation of the other line?

Correct Answer: A

Q22.A line perpendicular to the line segment joining the points (1, 0) and (2, 3) divides it in the ratio 1: n. What is the equation of the line?

Correct Answer: D

Q23.What is the equation of the right bisector of the line segment joining the points (3, 4) and (-1, 2)?

Correct Answer: D

Q24.The perpendicular from the origin to a line meets it at the point (-2, 9). What is the equation of the line?

Correct Answer: B

Q25.The perpendicular from the origin to a line meets it at the point (-2, 9). What is the equation of the line?

Correct Answer: A

Q26.Find the equation of the line that passes through the point (2, 3) and makes an angle of 45° with the positive x-axis.

Correct Answer: A

Q27.A line passes through the point (2, 3) and is perpendicular to the line passing through points (2, 5) and (-3, 6). What is the equation of the line?

Correct Answer: B

Q28.Find the equation of the line that passes through the point (0, 2) and makes an angle of 2π3\frac{2\pi}{3}32π​ with the positive x-axis.