Question

Find the equation of the line through (7,-3) and perpendicular to the line 2x-5y-8=0. Write your answer in general form.

Answer 1

Given:

The line passes through point (7,-3) and perpendicular to line 2x-5y-8=0.

Step-by-step explanation:

Simplify the equation 2x-5y-8=0 in slope intercept form.

`\begin{gathered} 2x-5y-8=0 \\ 5y=2x-8 \\ y=(2)/(5)x-(8)/(5) \end{gathered}`

So slope of line is 2/5.

Determine the slope of perpendicular line.

`\begin{gathered} m\cdot(2)/(5)=-1 \\ m=-(5)/(2) \end{gathered}`

The equation of line with slope m = -5/2 is,

`y=-(5)/(2)x+c`

Substitute 7 for x and -3 for y in the equation y = -5/2x + c to determine the value of c.

`\begin{gathered} -3=-(5)/(2)\cdot(7)+c \\ c=-3+(35)/(2) \\ =(-6+35)/(2) \\ =(29)/(2) \end{gathered}`

The value of c is 29/2.

The equation of line is,

`\begin{gathered} y=-(5)/(2)x+(29)/(2) \\ y=(-5x+29)/(2) \\ 2y+5x-29=0 \end{gathered}`

So answer is 2y + 5x -29 = 0