Answer:
A) The equation to the line parallel and passes through (-5, -5) is [tex]y=-6x-35[/tex]
B) The equation of the line perpendicular and passes through (-5, -5) is [tex]y=\frac{1}{6}x-\frac{25}{6}[/tex]
Step-by-step explanation:
To recap, we must remember the equation of the straight line which is:
[tex]y=mx+c[/tex] where m is our gradient and c is our y-intercept
A) The equation which is parallel and passes through (-5, -5):
For a line to be parallel its gradient (m) must be the same however the y-intercept (c) must be different. We can write this so far as this:
[tex]y=-6x+c[/tex]
Therefore, for the line to pass through (-5, -5) we must substitute the x and y values in to make it equal to each other:
[tex]-5 = -6(-5)+c\\-5=30+c\\c=-35[/tex]
This gives us the equation:
[tex]y=-6x-35[/tex]
A) The equation which is perpendicular and passes through (-5, -5):
For a line to be perpendicular its gradient (m) must be the negative reciprocal to the original equation, while the y-intercept (c) can be any value. We can write this so far as this:
[tex]y=\frac{1}{6} x+c[/tex]
Therefore, for the line to pass through (-5, -5) we must substitute the x and y values in to make it equal to each other:
[tex]-5 = \frac{1}{6} (-5)+c\\-5=-\frac{5}{6} +c\\c=-\frac{25}{6}[/tex]
This gives us the equation:
[tex]y=\frac{1}{6}x-\frac{25}{6}[/tex]
Hope this helps!!!
Answer:
For a parallel line:
[tex] y = -6x - 35 [/tex]
For the perpendicular line:
[tex] y = \dfrac{1}{6}x - \dfrac{25}{6} [/tex]
Step-by-step explanation:
To find the equation of a line parallel to [tex] y = -6x + 4 [/tex], we note that parallel lines have the same slope. The slope of [tex] y = -6x + 4 [/tex] is [tex] -6 [/tex]. Therefore, any line parallel to it will also have a slope of [tex] -6 [/tex].
We use the point-slope form of the equation of a line:
[tex] y - y_1 = m(x - x_1) [/tex]
where [tex] m [/tex] is the slope, and [tex] (x_1, y_1) [/tex] is a point on the line.
Given the point [tex] (-5, -5) [/tex] and the slope [tex] m = -6 [/tex], we plug these values into the point-slope form to find the equation of the parallel line:
[tex] y - (-5) = -6(x - (-5)) [/tex]
[tex] y + 5 = -6(x + 5) [/tex]
[tex] y + 5 = -6x - 30 [/tex]
[tex] y = -6x - 30 - 5 [/tex]
[tex] y = -6x - 35 [/tex]
Now, for the line perpendicular to [tex] y = -6x + 4 [/tex], we need to find the negative reciprocal of the slope [tex] -6 [/tex]. The negative reciprocal of [tex] -6 [/tex] is [tex] \dfrac{1}{6} [/tex].
Again, using the point-slope form and the point [tex] (-5, -5) [/tex], we find the equation of the perpendicular line:
[tex] y - (-5) = \dfrac{1}{6}(x - (-5)) [/tex]
[tex] y + 5 = \dfrac{1}{6}(x + 5) [/tex]
[tex] y + 5 = \dfrac{1}{6}x + \dfrac{5}{6} [/tex]
[tex] y = \dfrac{1}{6}x + \dfrac{5}{6} - 5 [/tex]
[tex] y = \dfrac{1}{6}x - \dfrac{25}{6} [/tex]
Therefore, the equations of the lines are:
For a parallel line:
[tex] y = -6x - 35 [/tex]
For the perpendicular line:
[tex] y = \dfrac{1}{6}x - \dfrac{25}{6} [/tex]
