Question

Hi. The problem is: A line that pass through point A(2,-3) and is perpendicular to line 2x + 5y = 4. I think the answer is y=4/5x-23/2. Im not sure

Answer 1

Given:

The point A = (2,-3)

The equation given is 2x +5y =4

We will obtain the slope (m1) from the equation of the line since the line with point (2,-3) is perpendicular to the equation.

The rule for perpendicularism is,

`m_1m_2=-1`

Isolate y from the equation

`\begin{gathered} 2x+5y=4 \\ 5y=4-2x \\ (5y)/(5)=-(2x)/(5)+(4)/(5) \\ y=-(2)/(5)x+(4)/(5) \end{gathered}`

The equation of a line is written in the form

`y=mx+b`

Therefore,

`m_2=-(2)/(5)`

Let us now obtain m1

`\begin{gathered} m_1(-(2)/(5))=-1 \\ -(2)/(5)m_1=-1 \\ -2m_1=-1*5 \\ -2m_1=-5 \\ m_1=(-5)/(-2)=(5)/(2) \\ \therefore m_1=(5)/(2) \end{gathered}`

The formula for the equation of a line given a point is,

`y-y_1=m(x-x_1)`

Where,

`(x_1,y_1)=(2,-3)`

Therefore,

`\begin{gathered} y-(-3)=(5)/(2)(x-2) \\ y+3=(5)/(2)* x-(5)/(2)*2 \\ y+3=(5)/(2)x-5 \\ y=(5)/(2)x-5-3 \\ \therefore y=(5)/(2)x-8 \end{gathered}`

Hence, the equation of the line is,

`y=(5)/(2)x-8`