Question

Prove that the lines 4x 3y - 8 = 0 and
3x + 4y + 5 = 0 are perpendicular and the lines
2x + 3y - 7 = 0 and 12x + 18y - 8 = 0 are
parallel.
​

Answer 1

≛ Solution

In the parametric equations , If two slopes(m) are equal , we say it as these lines are parallel to each other.

|| GIVEN ||

2x + 3y - 7 = 0 -------- ⓵

12x + 18y - 8 = 0 --------- ➁

Now we must write it in the format of || y = mx + c ||

where ,

⟿ x and y are the variables of the parametric equation.

⟿ m refers the slope of the equation.

So , Now the equation will be

2x + 3y = 7

↦ 3y = -2x + 7

$y = \dfrac{ - 2}{3}x + \dfrac{7}{3}$

Here the slope ,m =(-2/ 3) -------- ⓵

Now let us find for the other equation.

12x + 18y = 8

↦ 18y = - 12x + 8

$y = \dfrac{ - 12}{18}x + \dfrac{8}{18}$

Here the slope, m = (-12/18) = (-2/3)--------➁

Since both the slopes are equal , we say that these two parametric equations or lines are parallel.

_____________________ .

|| GIVEN ||

4x 3y - 8 = 0 -------- ⓵

3x + 4y + 5 = 0 --------- ➁

The same as before , we need to calculate slope from both the equatiions but the rule alone gets changed ,

For a perpendicular , when we multiply both the slopes , the product results as -1.

we must write it in the format of || y = mx + c ||

4x - 3y = 8

↦ 3y = -4x + 8

$y = \dfrac{ - 4}{3} x + \dfrac{8}{3}$

Here the slope,m1 = (-4/3) -------- ⓵

Now let us find for the other equation.

3x + 4y + 5 = 0

↦ 4y = -3x -5

$y = \dfrac{ - 3}{4} x + \dfrac{ - 5}{4}$

Here the slope,m2 = ( -3/4) --------- ➁

When we multiply m1 and m2 ,

$\dfrac{ - 4}{3} \times \dfrac{ - 3}{4} = - 1$

Hence these two lines are perpendicular in nature.

____________________.