Let 𝐴 ≡ (2,3) and 𝐵 ≡ (4,4). Find the equation of the straight line 𝑙 passing through the
points 𝐴 and 𝐵. Also find the equations of the straight lines 𝑙1 and 𝑙2 which passes through
point 𝐴 each by making an angle of 𝜋/4
with the straight line 𝑙. Here 𝑙1 is the straight line with
the higher gradient. Show that any co-ordinate on the straight line 𝑙 can be expressed in the
form of (2𝑡 + 2,𝑡 + 3). Here 𝑡 ∈ ℝ .

Let's break down the problem step by step.

### 1. Finding the Equation of the Straight Line \( l \) Passing Through Points \( A(2, 3) \) and \( B(4, 4) \):

First, we need to calculate the slope (gradient) \( m \) of the line \( l \) passing through points \( A(2, 3) \) and \( B(4, 4) \).

\[
m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{4 - 3}{4 - 2} = \frac{1}{2}
\]

Now, using the point-slope form of the equation of a line, \( y - y_1 = m(x - x_1) \), we substitute \( m = \frac{1}{2} \) and \( (x_1, y_1) = (2, 3) \):

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

Simplifying the equation:

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

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

Thus, the equation of the line \( l \) is:

\[
\boxed{y = \frac{1}{2}x + 2}
\]

### 2. Finding the Equations of Lines \( l_1 \) and \( l_2 \):

The lines \( l_1 \) and \( l_2 \) pass through point \( A(2, 3) \) and make an angle of \( \frac{\pi}{4} \) with the line \( l \).

If the slope of \( l \) is \( m \), the slopes of the lines making angles of \( \frac{\pi}{4} \) with \( l \) can be calculated using the formula:

\[
\text{slope of } l_1 = m_1 = \frac{m + \tan\frac{\pi}{4}}{1 - m\tan\frac{\pi}{4}} = \frac{\frac{1}{2} + 1}{1 - \frac{1}{2} \times 1} = \frac{\frac{3}{2}}{\frac{1}{2}} = 3
\]

\[
\text{slope of } l_2 = m_2 = \frac{m - \tan\frac{\pi}{4}}{1 + m\tan\frac{\pi}{4}} = \frac{\frac{1}{2} - 1}{1 + \frac{1}{2} \times 1} = \frac{-\frac{1}{2}}{\frac{3}{2}} = -\frac{1}{3}
\]

Thus, the equations of \( l_1 \) and \( l_2 \) using the point-slope form \( y - y_1 = m(x - x_1) \):

For \( l_1 \) (higher gradient \( 3 \)):

\[
y - 3 = 3(x - 2)
\]

\[
y = 3x - 6 + 3 = 3x - 3
\]

\[
\boxed{y = 3x - 3}
\]

For \( l_2 \) (lower gradient \( -\frac{1}{3} \)):

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

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

\[
y = -\frac{1}{3}x + \frac{2}{3} + 3 = -\frac{1}{3}x + \frac{11}{3}
\]

\[
\boxed{y = -\frac{1}{3}x + \frac{11}{3}}
\]

### 3. Showing that Any Point on the Line \( l \) Can Be Expressed as \( (2t + 2, t + 3) \):

Given the line \( l \) has the equation \( y = \frac{1}{2}x + 2 \), let's express \( x \) and \( y \) in terms of a parameter \( t \).

If \( x = 2t + 2 \), then substituting this into the equation of \( l \):

\[
y = \frac{1}{2}(2t + 2) + 2 = t + 1 + 2 = t + 3
\]

So, any point on the line \( l \) can be expressed as \( (x, y) = (2t + 2, t + 3) \), where \( t \in \mathbb{R} \).

### Final Results:

1. **Equation of line \( l \)**: \( \boxed{y = \frac{1}{2}x + 2} \)
2. **Equation of line \( l_1 \)** (higher gradient): \( \boxed{y = 3x - 3} \)
3. **Equation of line \( l_2 \)** (lower gradient): \( \boxed{y = -\frac{1}{3}x + \frac{11}{3}} \)
4. **Parametric form of the line \( l \)**: Any point on \( l \) can be expressed as \( (2t + 2, t + 3) \) where \( t \in \mathbb{R} \).

Would you like more details or have any further questions?

Here are 5 related questions:
1. How do you find the slope of a line given two points?
2. What are the trigonometric identities used in calculating the angles between two lines?
3. How do you derive the equation of a line using the point-slope form?
4. What is the significance of parametric equations in representing a line?
5. How do you determine whether two lines are perpendicular?

**Tip:** Always remember that the slope of a line is a measure of its steepness, and it is crucial in determining the angle between two lines.

Learn how to find the equation of a straight line passing through points A(2, 3) and B(4, 4), and the equations of lines making angles of π/4 with line l. Includes step-by-step solutions and parametric representation. Suitable for Grades 9-12.

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