What is the equation for 2 lines that are perpendicular to the line 3x - 4y + 5 = 0?
Answers
Answer:
Perpendicular lines have opposite-reciprocal slopes, so the slope of the line we want to find is 1/2. Plugging in the point given into the equation y = 1/2x + b and solving for b, we get b = 6. Thus, the equation of the line is y = ½x + 6. Rearranged, it is –x/2 + y = 6.
Answer:
Given point is : (2,3)
The given equation is : 3x + 4y - 5 = 0.
So, first we find the slope of the equation,
3x + 4y - 5 = 0
4y = 5 - 3x
So, y = 5/4 - 3x/4
The Slope is : -3/4
By the condition of perpendicularity,
M1 × M2 = -1
-3/4 × M2 = -1
M2 = 4/3
The Slope of perpendicular condition ( M2) is : 4/3.
Given point ( X1 , Y1 ) = (2,3).
On finding the equation ,
So, we know that : Y - Y1 = M2 ( X - X1 )
y - 3 = 4/3 ( x - 2 )
3y - 9 = 4x - 8
-9 +8 = 4x - 3y
4x - 3y = -1
4x - 3y + 1 = 0.
The resultant equation is : 4x - 3y + 1 =0.
What is the equation of a line through (1, 1) perpendicular to 3x+4y-5=0?
What is the equation of a line passing through (3, -4) and is perpendicular to the line 5x-2y+3?
How would one find the equation of a straight line passing through the point (6,-4) and perpendicular to the line 7x-6y+3=0?
What is the equation of a line through (5,1) perpendicular to 3x + 4y-5=0?
What is the equation of the line passing through (5,-3) and perpendicular to the line 2x -3y+14=0 is?
As it is perpendicular to 3x+4y-5=0, the product of slopes of both the lines should be -1.
Let m1 be the slope of the given line and m2 slope of the equation of the line to be found. Slope of the given line is -3/4. So, -3/4×m2=-1, m2= 4/3.
Now we know the slope of the line and one point is given, by applying the formula, y-y1=m(x-x1)
y−3=4/3(x−2)
3y−9=4x−8
4x−3y+1=0.
It is the required equation :)
The slope of the line 3x+4y-5=0
Or y =-(3/4)x+5/4
Is -3/4.
Let this be m1
To be perpendicular the product of two slopes should be -1.
m1*m2= -1
-3/4*(m2)=-1
m2=4/3.
Equation of line through (2,3) is
y =m x +c
3= (4/3)*2+c
3=(8/3)+c
c = 1/3
So the equation is
y = 4/3(x)+1/3
3y=4x+1
3x+4y-5=0
4y = -3x+5
y =( -3x/4)+ 5/4. [y= mx+c]
Therefore slope = (-3/4)
Slope of line perpendicular to the line
3x+4y-5=0 is (4/3)
Thus the equation of line which slope is m and passing through the point (x,y) is
y-y₁= m(x-x₁)
When m= 4/3
Points = (2,3)
y-3 =(4/3)(x-2)
4x-3y+1= 0
What is the equation of a line passing through (3, -4) and is perpendicular to the line 5x-2y+3?
Looks like you are doing your homework!
First of all, let me correct you! the equation you stated is not actually an equation, it is an expression.
So, i will answer this question with the equation : 5x-2y+3 = 0 (you have not written it's equivalent term there)
Now, the given line equation is: 5x-2y+3 =0
So, it's slope will be : -(5/-2) = 5/2
[i hope you know, how i stated this as the slope. If you don't, than let me know please]
Now, the slope of the required line will be :
-1/slope of given line (because lines are perpendicular to each other)
-1/ 5/2 = -2/5
Now, the equation of a line is :
(y-y’) = m(
What is the equation of a line through (5,1) perpendicular to 3x + 4y-5=0?
Slope of 3x+4y-5 = 0 is = -3/4.
Slope of a line perpendicular to 3x+4y-5=0
will be = 4/3.
eq. of required line is:-
y-1= 4/3.(x-5).
4x-20 = 3y-3.
4x-3y = 17 .Answer.
How do I find parametric equations of lines passing through the points (3,−1,−3) & perpendicular to line passing through point (3,−2,4) & (0,3,5) ?
⑴ Let A =(3，-2，4)
B=(0,3,5)
AB=(-3,5,1)
⑵ let (a,b,c)⊥AB
(a,b,c) .(-3,5,1)=0
3a+5b+c=0
let c=0,
b=-3a/5
let a=-5/3→
b=1
∴ a vector ⊥ AB is (-5/3,1,0)or (-5,3,0)
⑶ required equation is :
r=(3,-1,-3) +λ(-5,3,0)
What is the equation of a line through (1, 1) perpendicular to 3x+4y-5=0?
What is the equation of a line passing through (3, -4) and is perpendicular to the line 5x-2y+3?
How would one find the equation of a straight line passing through the point (6,-4) and perpendicular to the line 7x-6y+3=0?
What is the equation of a line through (5,1) perpendicular to 3x + 4y-5=0?
What is the equation of the line passing through (5,-3) and perpendicular to the line 2x -3y+14=0 is?
How do I find the equation of the line passing through the point (-5,7) and is perpendicular to the line given by y=-1/4*+3?
What is the equation of a line passing through the point (-2,3) and perpendicular to 7x+2y+3=0?
What is perpendicular to the line 2=10-x and passing through the point (-2,4)?
How do you find the equation of a line perpendicular to the line x-2y+3=0 and passing through the point (1, -2)?
What is the equation of the line passing through point (5,-4) and perpendicular to 3x-4y+7=0?
The line passing through (-3,4) and (0,3) is perpendicular through the line passing through (5,7) and (4,x) what is the value of x?
What is the equation of the line through the point (3,3) and perpendicular to the line 2y+3x=0?
What is the equation of a straight line passing through the point (2 , - 3) and perpendicular to the line 3x + 4y + 6 = 0?
What is the family of line perpendicular to 3x - 4y - 12=0?
What is the equation of a line perpendicular to 3x +4y=16 and passes through the point (-2, 7)?
Step-by-step explanation:
Hope that helps you a lot dear