What is the equation of the line that is perpendicular to the line y = 2x + 5 and passes through the point (-4, 2)?
Answers
Answer => Option - (A) :-
Given :-
- The equation line = y = 2x + 5
- Passing through the point = (-4 , 2)
To Find :-
- Write the equation of line that is perpendicular to line y = 2x + 5 and passes through (-4, 2).
Solution :-
To calculate this problem at first we have to focus , at first we have to use formula for sloping m as we know y = mx + b here y is horizontal axis and m is sloping line and b is constant. In Point (-4 , 2) x = - 4 which vertical axis and y = 2 which horizontal axis means base of the graph equation.
⇢ y = 2x + b
- As comparing we sloping (m) = 2
- so, its reciprocal sloping = -1/2
⇢ y = mx + b
⇢ y = -1/2 × X + b
- x = -4 , y = 2. b = 0
⇢ 2 = -1/2 × (-4) + b
⇢ 2 = 1/2 × 4 + b
⇢ 2 = 2 + b
⇢ b = 2 - 2
⇢ b = 0
⇢ y = mx + b
- substituting value we get :-
⇢ y = -1/2x + 0
⇢ y = -1/2x
Hence ,
- The required equation => y = -1/2x
Given :-
y = 2x + 5
Point = (-4,2)
To Find :-
Equation
Solution :-
We know that
Slope = m = y₂ - y₁/x₂ - x₁
Slope = 2/-4
Slope = 1/-2
Slope = -1/2
Now
We know that
y = mx + b
We need to find the value of b
y - mx = b
2 - (-1/2)(-4) = b
2 - (4/2) = b
2 - 2 = b
0 = b
Value of b is 0
y = mx
y = -1/2 × x
y = -1/2 x