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Find the single equation of pair of straight lines passing through origin and parallel to the lines a1x + b₁y+c₁ = 0 and a2x + b2y + c2 = 0.
ANSWER
[tex]a {x}^{2} + 2hxy + b {y}^{2} = 0 [/tex]

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Respuesta :

EvelynSwift

Answer:

ax^2+2hxy +by^2 = 0

Step-by-step explanation:

To find the equation follow these steps:

1. We have the equations of the given lines:

Line 1: [tex](a_1x + b_1y + c_1 = 0)[/tex]

Line 2: [tex](a_2x + b_2y + c_2 = 0)[/tex]

2. To find the equation of the pair of straight lines passing through the origin, we’ll use the concept of the joint equation of lines:


3. The joint equation of the given lines is given by: [tex][L_1 \cdot L_2 = 0][/tex]

4. Expanding the above equation, we get an equation of the form[tex][ax^2 + 2hxy + by^2 = 0][/tex]

5. If (a), (b), and (h) are not all zero, then this represents the general equation of a second-degree homogeneous equation in (x) and (y), which corresponds to a pair of straight lines passing through the origin.

6. Therefore, the required equation of the pair of straight lines is:

[tex][a{x^2} + 2hxy + b{y^2} = 0][/tex]

I hope this helps! If you have any further questions, feel free to ask.

msm555

Answer:

[tex]\sf x^2 + 2mxy + y^2 = 0[/tex]

Step-by-step explanation:

To find the equation of the pair of straight lines passing through the origin and parallel to the lines [tex]\sf a_1x + b_1y + c_1 = 0[/tex] and [tex]\sf a_2x + b_2y + c_2 = 0[/tex], let's first find the slopes of the given lines.

The slopes of the lines [tex]\sf a_1x + b_1y + c_1 = 0[/tex] and [tex]\sf a_2x + b_2y + c_2 = 0[/tex] can be found by rearranging them into the form [tex]\sf y = mx + b[/tex],

where

  • [tex]\sf m[/tex] is the slope.

The slopes are then [tex]\sf -\dfrac{a_1}{b_1}[/tex] and [tex]\sf -\dfrac{a_2}{b_2}[/tex], respectively.

Since the lines we want are parallel to these, their slopes will also be [tex]\sf -\dfrac{a_1}{b_1}[/tex] and [tex]\sf -\dfrac{a_2}{b_2}[/tex].

Now, let [tex]\sf m[/tex] be the slope of the desired lines. The equation of a line passing through the origin with slope [tex]\sf m[/tex] is [tex]\sf y = mx[/tex].

The equation of the pair of lines can then be expressed as:

[tex]\sf a(x^2) + 2hxy + b(y^2) = 0[/tex]

Here, [tex]\sf a = b = 1[/tex] (since the coefficient of [tex]\sf x^2[/tex] and [tex]\sf y^2[/tex] in [tex]\sf y = mx[/tex] is 1), and [tex]\sf h = m[/tex].

Substitute these values into the equation:

[tex]\sf x^2 + 2mxy + y^2 = 0[/tex]

Multiply the entire equation by [tex]\sf \dfrac{1}{b} = 1[/tex] to get:

[tex]\sf x^2 + 2mxy + y^2 = 0[/tex]

Therefore, the single equation of the pair of straight lines passing through the origin and parallel is:

[tex]\sf x^2 + 2mxy + y^2 = 0[/tex]

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