The equation of the line that passes through the origin and is parallel to line ab is 5x + 3y = 0.
What is the general equation of a straight line?
The general equation of a straight line is-
y = mx + c
where, m is the gradient and is given by y₂ - y₁/x₂ - x₁
and, c is the value where the line cuts y-axis.
Now, line ab passes through a(-3, 0) and b(-6, 5)
Therefore, slope of ab = y₂ - y₁/x₂ - x₁
⇒ slope of ab = 5 - 0/-6 - (-3)
⇒ slope of ab = 5 /-6 + 3
⇒ slope of ab = -5/3
⇒ m = -5/3
Since,required line is parallel to ab.
Therefore,slope of the line is -5/3.
It is also given that line passes through origin,so (x₁,y₁) is (0,0).
So, equation of required is-
y-y₁ = m(x-x₁)
⇒ y-0 = -5/3(x-0)
⇒ y = -5/3 x
⇒ 3y = -5x
⇒ 5x + 3y = 0
Hence, the required equation of line is 5x + 3y = 0.
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