Question

What is the equation of the line that passes through `(-2,7)` and is perpendicular to the line that passes through the following points: `(1,3),(-5,6)`?

Answer 1

`color (purple)(2x - y + 11 = 0`

Explanation:

`A(1,3), B(-5,6), C(-2,7)`

Slope of `vec(AB) = m = (y_b - y_a) / (x_b - x_a)`

`m = (6-3) / (-5 - 1) = -(1/2)`

Slope of perpendicular line `= -(1/m) = 2`

Equation of line passing through C and perpendicular to `vec(AB)` is

`y - y_c = -(1/m) * (x - x_c)`

`y - 7 = 2 * (x + 2)`

`y - 7 = 2x + 4`

`color (purple)(2x - y + 11 = 0`

Answer 2

`y=2x+11`

Explanation:

`"first find the slope m of the line passing through the"`
`"the 2 given points"`

`"to calculate the slope m use the "color(blue)"gradient formula"`

`•color(white)(x)m=(y_2-y_1)/(x_2-x_1)`

`"let "(x_1,y_1)=(1,3)" and "(x_2,y_2)=(-5,6)`

`m=(6-3)/(-5-1)=3/(-6)=-1/2`

`"given a line with slope m then the slope of a line"`
`"perpendicular to it is"`

`•color(white)(x)m_(color(red)"perpendicular")=-1/m`

`"hence "m_("perpendicular")=-1/(-1/2)=2`

`"the equation of a line in "color(blue)"slope-intercept form"` is.

`•color(white)(x)y=mx+b`

`"where m is the slope and b the y-intercept"`

`"here "m=2`

`y=2x+blarrcolor(blue)"is the partial equation"`

`"to find b substitute "(-2,7)" into the partial equation"`

`7=-4+brArrb=7+4=11`

`y=2x+11larrcolor(red)"is the required equation"`