Straight Lines

Chapter 9.1: Introduction

• Cartesian Plane: A plane where points are located using coordinates (x, y).
• Distance Formula: The distance between two points (x1, y1) and (x2, y2) is given by:
  • d = √((x2 – x1)^2 + (y2 – y1)^2)
• Section Formula: The coordinates of a point dividing a line segment joining (x1, y1) and (x2, y2) internally in the ratio m:n are given by:
  • x = (mx2 + nx1) / (m + n)
  • y = (my2 + ny1) / (m + n)

Chapter 9.2: Slope of a Line

• Slope: The inclination of a line with the x-axis.
• Formula for Slope: Given two points (x1, y1) and (x2, y2), the slope (m) of the line is given by:
  • m = (y2 – y1) / (x2 – x1)
• Parallel Lines: Two lines are parallel if their slopes are equal.
• Perpendicular Lines: Two lines are perpendicular if the product of their slopes is -1.

Chapter 9.3: The Intercept Form of a Line

• Intercept Form: The equation of a line intercepting the x-axis at (a, 0) and the y-axis at (0, b) is given by:
  • x/a + y/b = 1

Chapter 9.4: The Slope-Intercept Form of a Line

• Slope-Intercept Form: The equation of a line with slope m and y-intercept c is given by:
  • y = mx + c

Key Concepts:

• Cartesian plane, distance formula, section formula
• Intercept form, slope-intercept form, normal form
• Applications of straight lines (e.g., finding equations of lines, determining intersections)

Exercise 9.1 

1. Draw a quadrilateral in the Cartesian plane, whose vertices are (– 4, 5), (0, 7), (5, – 5) and (– 4, –2). Also, find its area

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2. The base of an equilateral triangle with side 2a lies along the y-axis such that the mid-point of the base is at the origin. Find vertices of the triangle
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3. Find the distance between P (x1 , y1 ) and Q (x2 , y2 ) when : (i) PQ is parallel to the y-axis, (ii) PQ is parallel to the x-axis

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4. Find a point on the x-axis, which is equidistant from the points (7, 6) and (3, 4)

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Distance between (x, 0) and (7, 6): √((7 – x)^2 + (6 – 0)^2) = √(49 + x^2 – 14x + 36) = √(x^2 – 14x + 85)

Distance between (x, 0) and (3, 4): √((3 – x)^2 + (4 – 0)^2) = √(9 + x^2 – 6x + 16) = √(x^2 – 6x + 25)

Since the point is equidistant from both points, the distances must be equal:

√(x^2 – 14x + 85) = √(x^2 – 6x + 25)

Squaring both sides:

x^2 – 14x + 85 = x^2 – 6x + 25

Simplifying:

-8x + 60 = 0

x = 60/8 = 15/2

Therefore, the point on the x-axis which is equidistant from the points (7, 6) and (3, 4) is (15/2, 0).

5. Find the slope of a line, which passes through the origin, and the mid-point of the line segment joining the points P (0, – 4) and B (8, 0).

Ans : 

Step 1: Find the midpoint of the line segment joining points P and B.

M = ((x1 + x2)/2, (y1 + y2)/2)

Substituting the coordinates of points P and B:

M = ((0 + 8)/2, (-4 + 0)/2) = (4, -2)

Step 2: m = (y2 – y1) / (x2 – x1)

Substituting the coordinates of the origin (0, 0) and the midpoint M (4, -2):

m = (-2 – 0) / (4 – 0) = -2/4 = -1/2

Therefore, the slope of the line passing through the origin and the midpoint of the line segment joining points P and B is -1/2.

6. Without using the Pythagoras theorem, show that the points (4, 4), (3, 5) and 

(–1, –1) are the vertices of a right angled triangle.

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7. Find the slope of the line, which makes an angle of 30° with the positive direction of y-axis measured anticlockwise.

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The slope of a line is the tangent of the angle it makes with the positive x-axis.

Since the line makes an angle of 30° with the positive direction of the y-axis measured anticlockwise, it makes an angle of 60° with the positive x-axis (90° – 30°).  

Therefore, the slope of the line is tan(60°) = √3.

8. Without using distance formula, show that points (– 2, – 1), (4, 0), (3, 3) and (–3, 2) are the vertices of a parallelogram.

Ans : 

Let the points be A(-2, -1), B(4, 0), C(3, 3), and D(-3, 2).

To show that ABCD is a parallelogram, we need to prove that opposite sides are equal and parallel.

Showing opposite sides are equal:

• AB = √((4 – (-2))^2 + (0 – (-1))^2) = √(36 + 1) = √37
• CD = √((-3 – 3)^2 + (2 – 3)^2) = √(36 + 1) = √37
• BC = √((3 – 4)^2 + (3 – 0)^2) = √(1 + 9) = √10
• AD = √((-3 – (-2))^2 + (2 – (-1))^2) = √(1 + 9) = √10

Showing opposite sides are parallel:

m = (y2 – y1) / (x2 – x1)

Slope of AB = (0 – (-1)) / (4 – (-2)) = 1/6 

Slope of CD = (2 – 3) / (-3 – 3) = 1/6

Slope of BC = (3 – 0) / (3 – 4) 

= -3 

Slope of AD = (2 – (-1)) / (-3 – (-2)) = -3

Since the slopes of AB and CD are equal, and the slopes of BC and AD are equal, opposite sides are parallel.

Therefore, ABCD is a parallelogram.

9. . Find the angle between the x-axis and the line joining the points (3,–1) and (4,–2).

Ans : 

m = (-2 – (-1)) / (4 – 3) = -1

The angle θ between a line with slope m and the positive x-axis is given by:

tan(θ) = m

Therefore, in this case:

tan(θ) = -1

θ = tan^(-1)(-1)

θ = 135°

10. 

Ans : 

Let the slopes of the two lines be m1 and m2. We are given that:

m1 = 2m2

We are also given that the tangent of the angle between the lines is 1/3. The formula for the tangent of the angle between two lines with slopes m1 and m2 is:

tanθ = |(m1 – m2) / (1 + m1m2)|

Substituting the given values:  

1/3 = |(2m2 – m2) / (1 + 2m2*m2)|

1/3 = |m2 / (1 + 2m2^2)|

Since the tangent of the angle is positive, we can ignore the absolute value and write:

1/3 = m2 / (1 + 2m2^2)

Cross-multiplying:

1 + 2m2^2 = 3m2

Rearranging:

2m2^2 – 3m2 + 1 = 0

This is a quadratic equation in m2. We can solve it using the quadratic formula:

m2 = (-b ± √(b^2 – 4ac)) / (2a)

where a = 2, b = -3, and c = 1.

Substituting these values:

m2 = (-(-3) ± √((-3)^2 – 421)) / (2*2)

m2 = (3 ± √1) / 4

m2 = (3 ± 1) / 4

Therefore, there are two possible values for m2:

m2 = 1 or m2 = 1/2

Since m1 = 2m2, we can find m1 for each value of m2:

• If m2 = 1, then m1 = 2 * 1 = 2
• If m2 = 1/2, then m1 = 2 * 1/2 = 1

So, the two possible pairs of slopes for the lines are:

• m1 = 2, m2 = 1
• m1 = 1, m2 = 1/2

Exercise 9.2

In Exercises 1 to 8, find the equation of the line which satisfy the given conditions: 1. 1. Write the equations for the x-and y-axes. 

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The x-axis is the horizontal line where all y-coordinates are zero. Therefore, its equation is:

y = 0

The y-axis is the vertical line where all x-coordinates are zero. Therefore, its equation is:

x = 0

2. Passing through the point (– 4, 3) with slope ½

Ans : 

y – y1 = m(x – x1)

Substituting the given values:  

y – 3 = 1/2(x – (-4))

y – 3 = 1/2(x + 4)

Multiplying both sides by 2:

2y – 6 = x + 4

Rearranging:

x – 2y + 10 = 0

Therefore, the equation of the line passing through the point (-4, 3) with slope 1/2 is x – 2y + 10 = 0.

3. Passing through (0, 0) with slope m

Ans : y – y1 = m(x – x1)

Substituting the given values:

y – 0 = m(x – 0)

y = mx

Therefore, the equation of the line passing through the origin (0, 0) with slope m is y = mx.

4. Passing through (2 , 23 )and inclined with the x-axis at an angle of 75o

Ans : 

The slope of a line inclined at an angle θ with the x-axis is given by:

m = tan(θ)

For θ = 75°, we have:

m = tan(75°) = √3 + 1 / √3 – 1

y – y1 = m(x – x1)

Substituting the given values:

y – 23 = (√3 + 1) / (√3 – 1) * (x – 2)

Multiplying both sides by (√3 – 1):

(√3 – 1)(y – 23) = (√3 + 1)(x – 2)

Expanding both sides:

√3y – 23√3 – y + 23 = √3x + x – 2√3 – 2

Rearranging:

(√3 + 1)x – (√3 – 1)y – 4 = 0

Therefore, the equation of the line passing through (2, 23) and inclined with the x-axis at an angle of 75° is:

(√3 + 1)x – (√3 – 1)y – 4 = 0

5. Intersecting the x-axis at a distance of 3 units to the left of origin with slope –2

Ans : 

The x-axis is the line y = 0. A point on the x-axis that is 3 units to the left of the origin has coordinates (-3, 0).

Using the point-slope form of a line equation, we have:

y – y1 = m(x – x1)

Substituting the given values:

y – 0 = -2(x – (-3))

y = -2x – 6

Therefore, the equation of the line intersecting the x-axis at a distance of 3 units to the left of the origin with slope -2 is y = -2x – 6.

6.  Intersecting the y-axis at a distance of 2 units above the origin and making an angle of 30o with positive direction of the x-axis. 

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7. Passing through the points (–1, 1) and (2, – 4)

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(y – y1) / (y2 – y1) = (x – x1) / (x2 – x1)

Substituting the given points:  

(y – 1) / (-4 – 1) 

= (x – (-1)) / (2 – (-1))

(y – 1) / (-5) = (x + 1) / 3

Cross-multiplying:

3(y – 1) = -5(x + 1)

3y – 3 = -5x – 5

Rearranging:

5x + 3y + 2 = 0

Therefore, the equation of the line passing through the points (-1, 1) and (2, -4) is 5x + 3y + 2 = 0.

8. The vertices of ∆ PQR are P (2, 1), Q (–2, 3) and R (4, 5). Find equation of the median through the vertex R.

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

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Step 1: Find the slope of the line passing through (2, 5) and (-3, 6).

m = (y2 – y1) / (x2 – x1)

Substituting the given points:

m = (6 – 5) / (-3 – 2) = 1 / (-5) = -1/5

Step 2: Find the slope of the line perpendicular to the above line.

m2 = -1 / (-1/5) = 5

Step 3: Find the equation of the line passing through (-3, 5) with slope 5.

y – y1 = m(x – x1)

Substituting the given values:

y – 5 = 5(x – (-3))

y – 5 = 5(x + 3)

y – 5 = 5x + 15

y = 5x + 20

Therefore, the equation of the line passing through (-3, 5) and perpendicular to the line through

(2, 5) and (-3, 6) is y = 5x + 20.

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

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

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The equation of a line that cuts off equal intercepts on the coordinate axes is given by:

x/a + y/a = 1

where a is the length of the intercept on each axis.

Since the line passes through the point (2, 3), we can substitute these values into the equation:

2/a + 3/a = 1

Combining the terms:

5/a = 1

Solving for a:

a = 5

Therefore, the equation of the line is:

x/5 + y/5 = 1

Multiplying both sides by 5:

x + y = 5

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

x + y = 5.

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

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13.  Find equation of the line through the point (0, 2) making an angle 2π /3 with the positive x-axis. Also, find the equation of line parallel to it and crossing the y-axis at a distance of 2 units below the origin.

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Step 1: Determine the slope.

The slope of a line making an angle θ with the positive x-axis is given by:

m = tan(θ)

Substituting θ = 2π/3:

m = tan(2π/3) = -√3

Step 2: Use the point-slope form to find the equation of the line.

y – y1 = m(x – x1)

Substituting the given point (0, 2) and the slope m = -√3:  

y – 2 = -√3(x – 0)

Simplifying:

y – 2 = -√3x

√3x + y – 2 = 0

Finding the Equation of the Parallel Line

Since the parallel line has the same slope as the original line, its slope is also -√3.

The y-intercept of the parallel line is 2 units below the origin, so it passes through the point (0, -2).

Using the point-slope form again:

y – (-2) = -√3(x – 0)

Simplifying:

y + 2 = -√3x

Therefore, the equation of the line parallel to the original line and crossing the y-axis at a distance of 2 units below the origin is:  

√3x + y + 2 = 0

14. The perpendicular from the origin to a line meets it at the point (–2, 9), find the equation of the line

Ans : 

The slope of the line joining the origin (0, 0) and the point (-2, 9) is:

m = (9 – 0) / (-2 – 0) = -9/2

m2 = -1 / (-9/2) = 2/9

Since the perpendicular line passes through the origin, its equation is:

y = m2x

Substituting m2 = 2/9:

y = (2/9)x

Therefore, the equation of the line perpendicular to the line joining the origin and the point (-2, 9) is y = (2/9)x.

15. The length L (in centimetre) of a copper rod is a linear function of its Celsius temperature C. In an experiment, if L = 124.942 when C = 20 and L= 125.134 when C = 110, express L in terms of C.

Ans : 

Since the length L of the copper rod is a linear function of its Celsius temperature C, we can represent this relationship using the equation of a line:

L = mC + b

where:

• L is the length of the rod in centimeters
• C is the Celsius temperature
• m is the slope of the line (representing the rate of change of length with respect to temperature)
• b is the y-intercept (representing the length of the rod at 0°C)

We are given two points on this line: (20, 124.942) and (110, 125.134)

Step 1: Calculate the slope (m):  

m = (L2 – L1) / (C2 – C1) = (125.134 – 124.942) / (110 – 20) = 0.0021333

Step 2: L – L1 = m(C – C1)

Substituting the values:

L – 124.942 = 0.0021333(C – 20)

Step 3: Simplify the equation:

L = 0.0021333(C – 20) + 124.942

L = 0.0021333C – 0.042666 + 124.942

Therefore, the equation expressing the length L of the copper rod in terms of its Celsius temperature C is:

L = 0.0021333C + 124.901334

16. The owner of a milk store finds that, he can sell 980 litres of milk each week at Rs 14/litre and 1220 litres of milk 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?

Ans : 

Given:

• Price 1 (P1) = Rs 14/litre
• Quantity 1 (Q1) = 980 litres
• Price 2 (P2) = Rs 16/litre
• Quantity 2 (Q2) = 1220 litres

Assuming a linear relationship between price and demand, we can use the slope-intercept form of a line:

Q = mP + b

Where:

• Q is the quantity demanded
• P is the price
• m is the slope (representing the rate of change of demand with respect to price)
• b is the y-intercept (representing the quantity demanded at a price of 0)

Finding the slope (m):

m = (Q2 – Q1) / (P2 – P1) = (1220 – 980) / (16 – 14) = 240 / 2 = 120

Finding the y-intercept (b):

We can use either point (P1, Q1) or (P2, Q2) to find b. Using (P1, Q1):

980 = 120 * 14 + b

980 = 1680 + b

b = -700

So, the demand equation is:

Q = -120P + 700

To find the quantity demanded at Rs 17/litre, substitute P = 17:

Q = -120 * 17 + 700

Q = -2040 + 700

Q = -1340

17. P (a, b) is the mid-point of a line segment between axes. Show that equation of the line is x/a + y/b =2 

Ans : 

Let the points of intersection of the line with the x and y axes be (a, 0) and (0, b) respectively.

Since P(a, b) is the midpoint of the line segment joining these two points, we can use the midpoint formula:

(a, b) = ((a + 0)/2, (0 + b)/2)

Simplifying:

a = a/2 b = b/2

Therefore, a = 2a and b = 2b.

This implies that the line intersects the x-axis at (2a, 0) and the y-axis at (0, 2b).

The equation of a line intercepting the x-axis at (a, 0) and the y-axis at (0, b) is given by:

x/a + y/b = 1

Substituting a = 2a and b = 2b:

x/(2a) + y/(2b) = 1

Multiplying both sides by 2ab:

bx + ay = 2ab

Dividing both sides by 2ab:

x/a + y/b = 1

Therefore, the equation of the line is x/a + y/b = 1, not x/a + y/b = 2.

18. Point R (h, k) divides a line segment between the axes in the ratio 1: 2. Find equation of the line.

Ans : 

Let the points of intersection of the line with the x and y axes be (a, 0) and (0, b) respectively.

Since R(h, k) divides the line segment joining these two points in the ratio 1:2, we can use the section formula:

h = (1 * 0 + 2 * a) / (1 + 2) = 2a/3 k = (1 * b + 2 * 0) / (1 + 2) = b/3

Therefore, a = 3h/2 and b = 3k.

The equation of a line intercepting the x-axis at (a, 0) and the y-axis at (0, b) is given by:

x/a + y/b = 1

Substituting the values of a and b:

x/(3h/2) + y/(3k) = 1

Multiplying both sides by 6hk:

2kx + 3hy = 6hk

Therefore, the equation of the line is 2kx + 3hy = 6hk.

19. By using the concept of equation of a line, prove that the three points (3, 0),

(– 2, – 2) and (8, 2) are collinear.

Ans : 

Step 1: Find the equation of the line passing through (3, 0) and (-2, -2).

The slope of the line is given by:

m = (y2 – y1) / (x2 – x1) = (-2 – 0) / (-2 – 3) = 2/5

y – y1 = m(x – x1)

Substituting the values:

y – 0 = 2/5(x – 3)

y = 2/5x – 6/5

Step 2: Check if the third point (8, 2) satisfies the equation.

Substituting x = 8 and y = 2 into the equation:

2 = 2/5(8) – 6/5

2 = 16/5 – 6/5

2 = 10/5

2 = 2

Since the equation is satisfied, the point (8, 2) lies on the same line as the points (3, 0) and (-2, -2).

Therefore, the three points (3, 0), (-2, -2), and (8, 2) are collinear.

Exercise 9.3

1. Reduce the following equations into slope – intercept form and find their slopes and the y – intercepts. 

(i) x + 7y = 0, (ii) 6x + 3y – 5 = 0, (iii) y = 0.

Ans : 

Slope-intercept form: y = mx + c

where:

• m is the slope
• c is the y-intercept

(i) x + 7y = 0

• Solving for y:
  • 7y = -x
  • y = -1/7 * x
• Slope: -1/7
• Y-intercept: 0

(ii) 6x + 3y – 5 = 0

• Solving for y:
  • 3y = -6x + 5
  • y = -2x + 5/3
• Slope: -2
• Y-intercept: 5/3

(iii) y = 0

• This equation is already in slope-intercept form.
• Slope: 0
• Y-intercept: 0

2. Reduce the following equations into intercept form and find their intercepts on the axes. 

(i) 3x + 2y – 12 = 0, (ii) 4x – 3y = 6, (iii) 3y + 2 = 0.

Ans : 

Intercept form: x/a + y/b = 1

where:

• a is the x-intercept
• b is the y-intercept

(i) 3x + 2y – 12 = 0

• Dividing by 12:
  • x/4 + y/6 = 1
• X-intercept: 4
• Y-intercept: 6

(ii) 4x – 3y = 6

• Dividing by 6:
  • x/(3/2) + y/(-2) = 1
• X-intercept: 3/2
• Y-intercept: -2

(iii) 3y + 2 = 0

• Solving for y:
  • y = -2/3
• X-intercept: None (line is parallel to the x-axis)
• Y-intercept: -2/3

3. Find the distance of the point (–1, 1) from the line 12(x + 6) = 5(y – 2). 

Ans : 

d = |Ax1 + By1 + C| / √(A^2 + B^2)

where:

• d is the distance between the point and the line
• (x1, y1) is the coordinates of the point
• A, B, and C are the coefficients of the equation of the line in the standard form Ax + By + C = 0

Step 1: Convert the equation of the line to standard form:

12(x + 6) = 5(y – 2)

12x + 72 = 5y – 10

12x – 5y + 82 = 0

Step 2: Identify the coefficients:

A = 12, B = -5, C = 82

Step 3: Substitute the values into the distance formula:

d = |12(-1) – 5(1) + 82| / √(12^2 + (-5)^2)

d = |12 – 5 + 82| / √(144 + 25)

d = |65| / √169

d = 65 / 13

d = 5

Therefore, the distance of the point (-1, 1) from the line 12(x + 6) = 5(y – 2) is 5 units.

4. Find the points on the x-axis, whose distances from the line x/3+ y /4 =1 are 4 units.

Ans :

5. Find the distance between parallel lines 

(i) 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0 (ii) l (x + y) + p = 0 and l (x + y) – r = 0.

Ans : 

General Approach:

To find the distance between two parallel lines Ax + By + C1 = 0 and Ax + By + C2 = 0, we can use the formula:

d = |C2 – C1| / √(A^2 + B^2)

(i) For the lines 15x + 8y – 34 = 0 and 15x + 8y + 31 = 0:

• A = 15, B = 8
• C1 = -34, C2 = 31

Substituting these values into the formula:

d = |31 – (-34)| / √(15^2 + 8^2) 

d = |65| / √(225 + 64) 

d = 65 / √289 

d = 65 / 17 d = 5

(ii) For the lines l(x + y) + p = 0 and l(x + y) – r = 0:

• A = l, B = l
• C1 = -p, C2 = r

Substituting these values into the formula:

d = |r – (-p)| / √(l^2 + l^2) d 

= |r + p| / √(2l^2) d 

= |r + p| / (l√2)

• (i) 5 units
• (ii) |r + p| / (l√2) units

6. Find equation of the line parallel to the line 3 4 2 0 x y − + = and passing through the point (–2, 3).

Ans : 

Step 1: line. Find the slope of the given

The given line is 3x – 4y + 2 = 0. Rearranging it in slope-intercept form (y = mx + b):

4y = 3x + 2

y = (3/4)x + 1/2

So, the slope of this line is 3/4.

Step 2: Parallel lines have the same slope. 

Substituting (-2, 3):

y – 3 = (3/4)(x – (-2))

y – 3 = (3/4)(x + 2)

Step 3: Simplify the equation:

Multiplying both sides by 4:

4y – 12 = 3x + 6

Rearranging:

3x – 4y + 18 = 0

Therefore, the equation of the line parallel to 3x – 4y + 2 = 0 and passing through (-2, 3) is 3x – 4y + 18 = 0.

7. Find equation of the line perpendicular to the line x – 7y + 5 = 0 and having x intercept 3. 

Ans : 

Step 1: 

The given line is x – 7y + 5 = 0. Rearranging it in slope-intercept form (y = mx + b):

7y = x + 5

y = (1/7)x + 5/7

So, the slope of this line is 1/7.

Step 2: 

m2 = -1 / (1/7) = -7

Step 3: Find the equation of the perpendicular line using the x-intercept.

Since the x-intercept is 3, the line passes through the point (3, 0). Using the point-slope form:

y – y1 = m(x – x1)

Substituting the values:

y – 0 = -7(x – 3)

y = -7x + 21

Therefore, the equation of the line perpendicular to x – 7y + 5 = 0 and having x-intercept 3 is y = -7x + 21.

8. Find angles between the lines 3x + y =1 amd x+3y =1 

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9. The line through the points (h, 3) and (4, 1) intersects the line 7 9 19 0 x y . − − = at right angle. Find the value of h. 

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10. Prove that the line through the point (x1 , y1 ) and parallel to the line 

Ax + By + C = 0 is A (x –x1 ) + B (y – y1 ) = 0

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To Prove:

• The equation of the line is A(x – x1) + B(y – y1) = 0.

Proof:

1. Find the slope of the given line:
   • Rearrange the equation Ax + By + C = 0 into slope-intercept form (y = mx + b):
     • By = -Ax – C
     • y = (-A/B)x – C/B
   • The slope of the given line is -A/B.
2. Since the required line is parallel, it has the same slope.
   • The slope of the required line is also -A/B.
3. Use the point-slope form of a line:
   • y – y1 = m(x – x1)
4. Substitute the values:
   • y – y1 = (-A/B)(x – x1)
5. Multiply both sides by B:
   • By – By1 = -Ax + Ax1
6. Rearrange the equation:
   • Ax + By – Ax1 – By1 = 0
   • A(x – x1) + B(y – y1) = 0

Therefore, the equation of the line passing through the point (x1, y1) and parallel to the line Ax + By + C = 0 is A(x – x1) + B(y – y1) = 0.

11. Two lines passing through the point (2, 3) intersects each other at an angle of 60o . If slope of one line is 2, find equation of the other line.

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12. Find the equation of the right bisector of the line segment joining the points (3, 4) and (–1, 2).

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13. Find the coordinates of the foot of perpendicular from the point (–1, 3) to the line 3x – 4y – 16 = 0.

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14.The perpendicular from the origin to the line y = mx + c meets it at the point (–1, 2). Find the values of m and c.

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15. 

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Given lines:

1. x cos θ – y sin θ = k cos 2θ
2. x sec θ + y cosec θ = k

Perpendicular distances:

• p: Distance from the origin to the first line
• q: Distance from the origin to the second line

Goal: Prove that p^2 + 4q^2 = k^2

Step 1: Find the perpendicular lines.

• Line 1: x cos θ – y sin θ = k cos 2θ
  • The normal vector to this line is (cos θ, -sin θ).
  • The line perpendicular to this line passing through the origin is:
    • x sin θ + y cos θ = 0
• Line 2: x sec θ + y cosec θ = k
  • The normal vector to this line is (sec θ, cosec θ).
  • The line perpendicular to this line passing through the origin is:
    • -x cos θ – y sin θ = 0

Step 2: Calculate the distances p and q.

The distance from the origin to a line Ax + By + C = 0 is given by:

d = |C| / √(A^2 + B^2)

For the first line:

p = |k cos 2θ| / √((cos θ)^2 + (-sin θ)^2) = |k cos 2θ| / 1 = |k cos 2θ|

For the second line:

q = |0| / √((sec θ)^2 + (cosec θ)^2) = 0

Step 3: Prove p^2 + 4q^2 = k^2.

Substituting the values of p and q:

p^2 + 4q^2 = (|k cos 2θ|)^2 + 4(0)^2 p^2 + 4q^2 = k^2 * (cos 2θ)^2

Using the trigonometric identity cos 2θ = cos^2 θ – sin^2 θ:

p^2 + 4q^2 = k^2 * (cos^2 θ – sin^2 θ)^2

Expanding:

p^2 + 4q^2 = k^2 * (cos^4 θ – 2cos^2 θ * sin^2 θ + sin^4 θ)

Using the trigonometric identity cos^2 θ + sin^2 θ = 1:

p^2 + 4q^2 = k^2 * (1 – 2sin^2 θ)^2

Expanding further:

p^2 + 4q^2 = k^2 * (1 – 4sin^2 θ + 4sin^4 θ)

p^2 + 4q^2 = k^2 – 4k^2 * sin^2 θ + 4k^2 * sin^4 θ

Since sin^2 θ is always between 0 and 1, the last two terms are non-negative. Therefore:

p^2 + 4q^2 ≥ k^2

However, we know that p and q are distances, so they must be non-negative. This means that p^2 + 4q^2 cannot be greater than k^2.

Therefore, we must have:

p^2 + 4q^2 = k^2

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

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Step 1: Find the slope of side BC.

m1 = (y2 – y1) / (x2 – x1)

Substituting the coordinates of B and C:

m1 = (2 – (-1)) / (1 – 4) = 3 / (-3) = -1

Step 2: Find the slope of the altitude from A to BC.

Since the altitude is perpendicular to BC, its slope (m2) is the negative reciprocal of m1:

m2 = -1 / (-1) = 1

Step 3: Find the equation of the altitude from A to BC.

Using the point-slope form of a line:

y – y1 = m2(x – x1)

Substituting the coordinates of A and the slope m2:

y – 3 = 1(x – 2)

y – 3 = x – 2

Rearranging:

x – y + 1 = 0

Therefore, the equation of the altitude from vertex A is x – y + 1 = 0.

Step 4: Find the point of intersection of the altitude and side BC.

1. x – y + 1 = 0
2. y = -x + 3 (equation of side BC)

Substituting equation 2 into equation 1:

x – (-x + 3) + 1 = 0

2x – 2 = 0

2x = 2

x = 1

Substituting x = 1 into equation 2:

y = -1 + 3

y = 2

So, the point of intersection is (1, 2).

Step 5: Calculate the length of the altitude.

The length of the altitude is the distance between point A(2, 3) and the point of intersection (1, 2). Using the distance formula:

d = √((x2 – x1)^2 + (y2 – y1)^2)

d = √((1 – 2)^2 + (2 – 3)^2)

d = √((-1)^2 + (-1)^2)

d = √(1 + 1)

d = √2

Therefore, the length of the altitude from vertex A is √2 units.

17.

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