Question

Slope of a line perpendicular to 7x - 5y = 15 is --------

Answer 1

To find:

Slope of a line perpendicular to 7x - 5y = 15.

Calculation:

Let the lines be L1 and L2 such that $L1\perp L2$

Equation of L1 is :

$7x - 5y = 15$

$= > 5y = 7x - 15$

$= > y = (\dfrac{7}{5}) x - \dfrac{15}{5}$

$= > y = (\dfrac{7}{5}) x - 3$

Slope of line L1 be m1 ;

$\therefore \: m1 = \dfrac{7}{5}$

Now , let slope of line L2 be m2 ;

We know that product of slopes of two perpendicular lines is -1.

$\therefore \: m1 \times m2 = - 1$

$= > \: \dfrac{7}{5} \times m2 = - 1$

$= > \: m2 = \dfrac{ - 5}{7}$

So, slope of line L2 is (-5/7).

Answer 2

Question:-

Slope of a line perpendicular to 7x - 5y = 15 is

Answer:-

For first line :-

7x - 5y = 15

→ 5y = 7x - 15

→ y = (7x - 15)/5

→ y = (7/5)x - 3

→ y = (7/5)x + (- 3)

We know, Equation of straight line is y = mx + c

By comparing,

m of first line = 7/5

→ m₁ = 7/5

We know that when two lines are perpendicular, then the product of their slopes is -1

Let the slope of second line be m₂

So,

m₁ * m₂ = -1

→7/5 * m₂ = -1

→ m₂ = -1 * 5/7

→ m₂ = -5/7

Ans. m₂ = -5/7