Answer:
Well we have to find eq of line passing through origin and perpendicular to y=5x-2
First you need slope
As slope of y=5x-2 line is 5
So, slope of perpendicular line is -1/5
Also it passes through origin
as we know eq of line is y-y1=m(x-x1)
substitute X1,y1 = (0,0)
it becomes y =mx
so, y = -x/5
5y +x = 0 is required equation
x + 5y = 0 is the required equation of the straight line here.
''A straight line is just a line with no curves. So, a line that extends to both sides till infinity and has no curves is called a straight line.''
Given, the equation of a straight line: y = 5x-2.
If we compare it with the general equation y = mx + c. Here, m = slope.
In the given equation, slope is = 5.
As the required equation is perpendicular to the given equation, then slope of the required equation m₁ = -(1/m) = -(1/5).
We need to find an equation of the line that passes through the origin(0,0).
The general equation that passes through the origin:
(y - y₁) = m₁(x - x₁)
⇒ y = (-1/5)x
⇒ x +5y = 0
Therefore, the required equation is x + 5y = 0.
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