Question

Consider the line y=3x-2.
(a)
Find the equation of the line that is parallel to this line and passes through the point (4, 4).
(b)
Find the equation of the line that is perpendicular to this line and passes through the point (4, 4)

Answer 1

a) The equation of the line that is parallel to this line and passes through the point (4, 4) is y = 3x - 8

b) The equation of the line that is perpendicular to this line and passes through the point (4, 4) is $y = \frac{-x}{3} + \frac{16}{3}$

Solution:

Given that line is y = 3x - 2

Let us first slope of given line

The slope intercept form is given as:

y = mx + c  --- eqn 1

Where "m" is the slope of line and "c" is the y - intercept

On comparing the given equation y = 3x - 2 with slope intercept form, we get "m" = 3

Thus the slope of given line is 3

a) Find the equation of the line that is parallel to this line and passes through the point (4, 4)

Let us find the slope of line parallel to given line

We know that slopes of parallel lines are equal

Therefore slope of line parallel to given line is also 3

Now find the equation of line with slope 3 and passes through point (4, 4)

Substitute m = 3 and (x, y) = (4, 4) in eqn 1

4 = 3(4) + c

4 = 12 + c

c = -8

Thus the required equation of line parallel to given line is:

Substitute m = 3 and c = -8 in eqn 1,

y = 3x - 8

Thus the required equation of line parallel to given line is found

b)  Find the equation of the line that is perpendicular to this line and passes through the point (4, 4)

Let us find the slope of line perpendicular to given line

We know product of slopes of given line and slope of line perpendicular to given line is equal to -1

slope of given line x slope of line perpendicular to given line = -1

3 x slope of line perpendicular to given line = -1

$\text{ slope of line perpendicular to given line } = \frac{-1}{3}$

Now find the equation of line with slope $\frac{-1}{3}$ and passes through point (4, 4)

Substitute $m = \frac{-1}{3}$ and (x, y) = (4, 4) in eqn 1

$4 = \frac{-1}{3}(4) + c\\\\4 = \frac{-4 + 3c}{3}\\\\12 = -4 + 3c\\\\16 = 3c\\\\c = \frac{16}{3}$

Thus the required equation of line is:

Substitute $m = \frac{-1}{3}$ and $c = \frac{16}{3}$ in eqn 1

$y = \frac{-1}{3}x + \frac{16}{3}\\\\y = \frac{-x}{3} + \frac{16}{3}$

Thus the equation of line perpendicular to given line is found