Question

17. Write the equation of a line that is perpendicular to the given line and that passes through the

given point.
» _ 3 = 8(4 + 2); (-2, 3)
(1 point)
Oy +3 = -2(x - 2)
Oy -3 ==« +2)
Oy-2 =(x+3)

Answer 1

Option 2

The equation of a line that is perpendicular to the given line
`y - 3 = (8)/(3)(x + 2)` and that passes through the given point (-2, 3) is
`y-3=(-3)/(8)(x+2)`

Solution:

Given that line that is perpendicular to the given line
`y - 3 = (8)/(3)(x + 2)` and that passes through the given point (-2, 3)

We have to find the equation of line

The point slope form is given as:

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

Where "m" is the slope of line

Comparing the given equation
`y - 3 = (8)/(3)(x + 2)` with point slope form
`y - y_1 = m(x - x_1)`

`m = (8)/(3)`

Thus slope of line is
`m = (8)/(3)`

We know that product of slopes of perpendicular lines are always equal to -1

Slope of given line x slope of line perpendicular to given line = -1

`(8)/(3) * \text { slope of line perpendicular to given line }=-1`

`\begin{array}{l}{\text { slope of line perpendicular to given line }=-1 * (3)/(8)} \\\\ {\text { slope of line perpendicular to given line }=(-3)/(8)}\end{array}`

Now we have to find the equation of line having slope
`m = (-3)/(8)` and passes through point (-2, 3)

`\text {substitute } m=(-3)/(8) \text { and }\left(x_(1), y_(1)\right)=(-2,3) \text { in point slope form }`

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

`\begin{array}{l}{y-(3)=(-3)/(8)(x-(-2))} \\\\ {y-3=(-3)/(8)(x+2)}\end{array}`

Thus the required equation is found and option 2 is correct