Question

What is the equation of a line perpendicular to `y + 2x = 17` and goes through point `(-3/2, 6)`?

Answer 1

The equation of the line is `2x-4y= -27`

Explanation:

Slope of the line, `y+2x=17 or y= -2x +17; [y=mx+c]`

is `m_1= -2` [Compared with slope-intercept form of equation]

The product of slopes of the pependicular lines is `m_1*m_2=-1`

`:.m_2=(-1)/-2=1/2`. The equation of line passing through

`(x_1,y_1)` having slope of `m` is `y-y_1=m(x-x_1)`.

The equation of line passing through `(-3/2,6)` having slope of

`1/2` is `y-6=1/2(x+3/2) or 2y-12 =x+3/2`. or

`4y-24 =2x+3 or 2x-4y= -27`

The equation of the line is `2x-4y= -27` [Ans]

Answer 2

`y = 1/2x +6 3/4`

or

`2x -4y = -27`

Explanation:

The given line `y+2x =17` can be rewritten as `y = -2x +17`

The gradient : `m = -2`

If lines are perpendicular, their slopes are negative reciprocals of each other and their product is `-1`

`m_1 = -2" " rarr" " m_2 =1/2`

We have the slope and the point `(-3/2, 6)`

use the formula `" "y - y_1 = m(x-x_1)`

`y -6 = 1/2(x-(-3/2))`

`y -6 = 1/2(x+3/2)`

`y = 1/2x +3/4 +6`

`y = 1/2x +6 3/4`

You can also change this to standard form:

`xx 4`

`4y = 2x +27`

`2x -4y = -27`