Question

Consider the line y=7x-1. Find the equation of a line that is perpendicular and parallel to this line and passes through the point (2,-5)

Answer 1

Answer:

Equation of perpendicular line:

$y = - \frac{1}{7} x - 4 \frac{5}{7}$

Equation of parallel line:

y= 7x -19

Step-by-step explanation:

When an equation is written in the form of y= mx +c, we can easily identify its slope and y- intercept from its coefficient of x (value of m) and the value of c respectively.

y= 7x -1

Slope= 7

Y- intercept= -1

The product of the gradients of perpendicular lines is -1. Let the gradient of the perpendicular line be m.

m(7)= -1

$m = - \frac{1}{7}$

Substitute the value of m into the equation:

$y = - \frac{1}{7} x + c$

To find the value of c, substitute a pair of coordinates.

When x= 2, y= -5,

$- 5 = - \frac{1}{7} (2) + c$

$c = - 5 + \frac{2}{7}$

$c = - 4 \frac{5}{7}$

Thus the equation of the perpendicular line is $y = - \frac{ 1}{7} x - 4 \frac{5}{7}$.

______

Parallel lines have the same slope. Thus, m= 7.

y= 7x +c

To find the value of the y- intercept, substitute a pair of coordinates into the equation.

When x= 2, y= -5,

-5= 7(2) +c

-5= 14 +c

c= -5 -14

c= -19

Hence, the equation of the parallel line is y= 7x -19.