Question # 01:
Find equation of the line which is perpendicular to the line
and has y-intercept = 6.
Solution:
Y = 5x + 9
From the given equation slop of line is 5
Since slop of perpendicular line = m = - 1/5
And Y intercept is = 6
By slop intercept form:
Y = mx + b
Y = -1/5x + 6
For the equation of the line which is perpendicular to the line:
Y = 5x + 9 and has Y-intercept = 6
Y = 5x + 9 is derivative w.r.t x
Dy/dx = 5 + 0
M=5
Then slope of normal is m= -1/5
for normal equation is (Y-Y1)m = (X-X1)
(Y-6)-1/5 =( x - 0)
5y – 30 -1 = 5x
5y- 31 = 5x
5y = 5x+31
y = (5x+31)/5
y = x+(31/5)
This is normal equation
for perpendicular is
Y = mx + b
Y=x+(31/5)
Find equation of the line which is perpendicular to the line
and has y-intercept = 6.
Solution:
Y = 5x + 9
From the given equation slop of line is 5
Since slop of perpendicular line = m = - 1/5
And Y intercept is = 6
By slop intercept form:
Y = mx + b
Y = -1/5x + 6
For the equation of the line which is perpendicular to the line:
Y = 5x + 9 and has Y-intercept = 6
Y = 5x + 9 is derivative w.r.t x
Dy/dx = 5 + 0
M=5
Then slope of normal is m= -1/5
for normal equation is (Y-Y1)m = (X-X1)
(Y-6)-1/5 =( x - 0)
5y – 30 -1 = 5x
5y- 31 = 5x
5y = 5x+31
y = (5x+31)/5
y = x+(31/5)
This is normal equation
for perpendicular is
Y = mx + b
Y=x+(31/5)
Question # 02:
Find the vertex, x and y intercepts of the parabola given by the equation
Solution:
y = 3x2 - 2x + 1
In this problem: a = 3, b = -2 , and c = 1
Since "a" is positive we'll have a parabola that opens upward (is U shaped).
To find the x-intercepts we plug in 0 for y:
once it's found that the discriminant is 4-12=-8, the solutions must therefore be imaginary numbers (complex conjugates, that is), so there are no x-intercepts.
Join Our Groups:
Groups.Google.Com/group/vubest
facebook.com/groups/vubest
If you Like This. Please Like us onfacebook - Thumbs Up
Comments
Post a Comment