Question

Through: (2,5), perp. to y =- 2/3x +3

Answer 1

Find the equation of line passes through point (2, 5) and perpendicular to line having equation y =- 2/3x +3

Answer:

The equation of line passes through point (2, 5) and perpendicular to line having equation y =- 2/3x +3 in slope intercept form is
`y = (3)/(2)x + 2`

Solution:

Given that line passes through (2, 5) and perpendicular to line having equation
`y = (-2)/(3)x + 3`

Let us first find slope of original line

The slope intercept form of line is given as:

y = mx + c ------ eqn 1

Where "m" is the slope of line and "c" is the y - intercept

On comparing the slope intercept form y = mx + c and given equation of line
`y = (-2)/(3)x + 3` we get

`m = (-2)/(3)`

Thus slope of given line is
`m = (-2)/(3)`

We know that product of slopes of given line and slope of line perpendicular to given line is always -1

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

`\begin{array}{l}{(-2)/(3) * \text { slope of line perpendicular to it }=-1} \\\\ {\text { slope of line perpendicular to it }=(3)/(2)}\end{array}`

Let us find equation of line with slope
`m = (3)/(2)` and passes through (2, 5)

Substitute
`m = (3)/(2)` and (x, y) = (2, 5)

`5 = (3)/(2) * 2 + c\\\\5 = 3 + c\\\\c = 2`

Thus the required equation of line is:

Substitute
`m = (3)/(2)` and c = 2 in eqn 1

`y = (3)/(2)x + 2`

Thus the equation of line perpendicular to given line is found out