Question

Write the point-stope form of the linear equation that passes through the point (2, - 3) and is perpendicular to 3x - y = 4 .

Answer 1

Equation of a line that is perpendicular to y = 3x-4 and that passes through the point (2 -3) is
`y=-(1)/(3) x-(7)/(3)`

Solution:

The slope - intercept form equation of line is given as

y = mx+c ----- (1)

Where m is the slope of the line. The coefficient of “x” is the value of slope of the line.

Given that

3x -y = 4

Converting above equation in slope intercept form,

y = 3x - 4 ---- (2)

On comparing equation (1) and (2) we get slope of equation (2) is m=3

Consider equation of the line which is perpendicular to equation (2) is
`y=m_(1) x+b` --- eqn 3

If two lines having slope m1 and m1 are perpendicular then relation between their slope is
`m_(1) m_(2)=-1`

That is if slope of the line (2) is 3 then slope of equation (3) is
`m_(1)=-(1)/(3)`

On substituting value of m1 in equation (3), we get

`y=-(1)/(3) x+b` --- eqn 4

Given that equation (4) passes through (2, -3) that is x = 2 and y = -3. So on substituting value of x and y in equation (4),

`-3=-(2)/(3)+b`

On simplifying above equation,

`\begin{aligned} b &=(2)/(3)-3 \\ b &=(2-9)/(3) \\ b &=-(7)/(3) \end{aligned}`

On substituting value of b in equation (4),

`y=-(1)/(3) x-(7)/(3)`

Hence equation of a line that is perpendicular to y=3x-4 and that passes through the point (2 -3) is
`y=-(1)/(3) x-(7)/(3)`