Question attached write equation parallel to a given line write equation of a line perpendicular to

5 answers
Question:

Question attached
write equation parallel to a given line
write equation of a line perpendicular to a given line

[tex]Question attached write equation parallel to a given line write equation of a li[/tex]


Answers

36. y=x

37. x=4

38. y=3x-10

39. y=-25x+49

40. [tex]y=-\dfrac{1}{18}x-\dfrac{89}{18}[/tex]

41. y=-x+3

Step-by-step explanation:

36. Parallel lines have the same slope. The slope of the line [tex]y=x+42[/tex] is [tex]m=1,[/tex] so the equation of a parallel line is

[tex]y=x+b[/tex]

This line passes through the point (2,2), so its coordinates satisfy the equation:

[tex]2=2+b\\ \\b=0[/tex]

and the equation of the line is [tex]y=x[/tex]

37. The line [tex]x=03[/tex] is vertical ine, so parallel line is also vertical line with equation [tex]x=a.[/tex] Substitute the coordinates of the point (4,3):

[tex]4=a[/tex]

hence the equation is [tex]x=4[/tex]

38. Parallel lines have the same slope. The slope of the line [tex]y=3x+24[/tex] is [tex]m=3,[/tex] so the equation of a parallel line is

[tex]y=3x+b[/tex]

This line passes through the point (2,-4), so its coordinates satisfy the equation:

[tex]-4=3\cdot 2+b\\ \\b=-10[/tex]

and the equation of the line is [tex]y=3x-10[/tex]

39. Parallel lines have the same slope. The slope of the line [tex]y=-25x+35[/tex] is [tex]m=-25,[/tex] so the equation of a parallel line is

[tex]y=-25x+b[/tex]

This line passes through the point (2,-1), so its coordinates satisfy the equation:

[tex]-1=-25\cdot 2+b\\ \\b=49[/tex]

and the equation of the line is [tex]y=-25x+49[/tex]

40. Perpendicular lines have slopes satisfying [tex]m_1\cdot m_2=-1[/tex] Since the line [tex]y=18x+26[/tex] has the slope [tex]m_1=18,[/tex] perpendicular line has the slope [tex]m_2 =-\dfrac{1}{18}.[/tex]

The equation is

[tex]y=-\dfrac{1}{18}x+b[/tex]

This line passes through the point (1,-5), so its coordinates satisfy the equation:

[tex]-5=-\dfrac{1}{18}\cdot 1+b\\ \\b=-5+\dfrac{1}{18}=-\dfrac{89}{18}[/tex]

and the equation of the line is [tex]y=-\dfrac{1}{18}x-\dfrac{89}{18}[/tex]

41. Perpendicular lines have slopes satisfying [tex]m_1\cdot m_2=-1[/tex] Since the line [tex]y=x+2[/tex] has the slope [tex]m_1=1,[/tex] perpendicular line has the slope [tex]m_2 =-1.[/tex]

The equation is

[tex]y=-x+b[/tex]

This line passes through the point (4,-1), so its coordinates satisfy the equation:

[tex]-1=-4+b\\ \\b=3[/tex]

and the equation of the line is [tex]y=-x+3[/tex]

Please, post just one question at a time.  I am arbitrarily focusing on #24 this time thru. 
                                                                                         2x-3
2x - 4y = 3  can easily be solved for y:  2x-3=4y, or y =
                                                                                            4

The slope of this line is m = 2/4, or m = 1/2.

We need the equation of a new line parallel to this one.   Use the slope-intercept form of the equation of a straight line:  y = mx + b.

Let m = 1/2 and arbitrarily choose y-intercept b = 0.  Then we have

y = (1/2)x + 0, or y = (1/2)x.

Want the equation of a line perpendicular to 2x - 4y = 3?

We already know that the slope of this given line is 1/2.  The slope of a line perpendicular to this given line is the neg. reciprocal of 1/2, or m = -2.

Again, use the slope-intercept formula y = mx + b:

y = -2x + 0 (my choice of b = 0 is arbitrary; could have chosen some other value for b)

The format of a line equation is y = mx + b
When two lines are parallel, their 'm' variables are equal.
Knowing this, the unknown line equation, so far, would look like this:
y = 9x + b
Since we know that the line equation goes through the point (2, 7)
7 = 18 + b
b = -11
y = 9x - 11

X=-y+1
hope this helps

Y= 4x + 1 is the answer for this question, hope this helped u :)

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