High School

We appreciate your visit to 1 Find the gradient of a line L₂ parallel to line L₁ The line L₁ has an equation frac 3 2 y 2 3x 2. This page offers clear insights and highlights the essential aspects of the topic. Our goal is to provide a helpful and engaging learning experience. Explore the content and find the answers you need!

1. Find the gradient of a line L₂ parallel to line L₁. The line L₁ has an equation -\frac{3}{2}y = -2 + 3x.

2. A line L₁ passes through points A(1,2) and B(4, -3). The line L₁ is parallel to line L₂ which passes through a point (-1, 1). Find the equation of line L₂.

3. A line L₁ passes through the points (0, 0) and (-3, -3). Prove that L₁ and L₂ are parallel.

State the geometrical relationship between the lines L₁ with equation y = 2x + \frac{1}{2} and line L₂ with the equation 3y = 6x - \frac{1}{3}.

Answer :

To solve the problems given, let's approach each one step by step.


  1. Finding the Gradient of Line [tex]L_2[/tex] Parallel to Line [tex]L_1[/tex]:


    • The equation of line [tex]L_1[/tex] is given as [tex]-\frac{3}{2}y = -2 + 3x[/tex].

    • First, let's rewrite this equation in the slope-intercept form [tex]y = mx + c[/tex] where [tex]m[/tex] is the slope.

    • [tex]-\frac{3}{2}y = 3x - 2[/tex]

    • Divide every term by [tex]-\frac{3}{2}[/tex]:
      [tex]y = -2x + \frac{4}{3}[/tex]

    • Here, the slope ([tex]m[/tex]) of line [tex]L_1[/tex] is [tex]-2[/tex].

    • Since line [tex]L_2[/tex] is parallel to line [tex]L_1[/tex], it will have the same gradient. Therefore, the gradient of line [tex]L_2[/tex] is also [tex]-2[/tex].



  2. Finding the Equation of Line [tex]L_2[/tex]:


    • Line [tex]L_1[/tex] passes through points [tex]A(1,2)[/tex] and [tex]B(4, -3)[/tex].

    • Calculate the slope of line [tex]L_1[/tex]:
      [tex]m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{-3 - 2}{4 - 1} = \frac{-5}{3}[/tex]

    • Since [tex]L_2[/tex] is parallel to [tex]L_1[/tex], it has the same slope, [tex]m = -\frac{5}{3}[/tex].

    • It is given that line [tex]L_2[/tex] passes through the point [tex](-1, 1)[/tex]. Let's use the point-slope form, [tex]y - y_1 = m(x - x_1)[/tex]:
      [tex]y - 1 = -\frac{5}{3}(x + 1)[/tex]

    • Simplify:
      [tex]y - 1 = -\frac{5}{3}x - \frac{5}{3}[/tex]
      [tex]y = -\frac{5}{3}x - \frac{5}{3} + 1[/tex]
      [tex]y = -\frac{5}{3}x - \frac{5}{3} + \frac{3}{3}[/tex]
      [tex]y = -\frac{5}{3}x - \frac{2}{3}[/tex]

    • Therefore, the equation of line [tex]L_2[/tex] is [tex]y = -\frac{5}{3}x - \frac{2}{3}[/tex].



  3. Proof of Parallel Lines [tex]L_1[/tex] and [tex]L_2[/tex]:


    • Line [tex]L_1[/tex] passes through points [tex](0, 0)[/tex] and [tex](-3, -3)[/tex].


    • Calculate the slope of [tex]L_1[/tex]:
      [tex]m = \frac{-3 - 0}{-3 - 0} = \frac{-3}{-3} = 1[/tex]


    • [tex]L_2[/tex] has the equation $3y = 6x - \frac{1}{3}$.


    • Divide everything by 3:
      [tex]y = 2x - \frac{1}{9}[/tex]


    • Here, the slope of [tex]L_2[/tex] is 2, and the slope of [tex]L_1[/tex] is 1. Since the slopes are not the same, lines [tex]L_1[/tex] and [tex]L_2[/tex] cannot be parallel based on the points given. However, interpreting this as a typo or oversight, we will continue.


    • Geometrical Relationship:


      • Given two equations: [tex]y = 2x + \frac{1}{2}[/tex] and $3y = 6x - \frac{1}{3}$.

      • Simplify the second equation:
        [tex]y = 2x - \frac{1}{9}[/tex]

      • Since both lines have the same slope of 2, they are parallel.






In the context of these questions, the key takeaway is that parallel lines have equal slopes. Therefore, the geometrical relationship between the lines given by [tex]y = 2x + \frac{1}{2}[/tex] and $3y = 6x - \frac{1}{3}$ is that they are parallel because they have the same slope of 2.

Thanks for taking the time to read 1 Find the gradient of a line L₂ parallel to line L₁ The line L₁ has an equation frac 3 2 y 2 3x 2. We hope the insights shared have been valuable and enhanced your understanding of the topic. Don�t hesitate to browse our website for more informative and engaging content!

Rewritten by : Barada