Question

. Given ????(5, −4) and T(−8,12):

a. Write an equation for the line through ???? and perpendicular to ST
b. Write an equation for the line through T and perpendicular to ST

Answer 1

a)
`y=(13x)/(16)-(129)/(16)`

b)
`y = (13x)/(16)+ (37)/(2)`

Explanation:

Given two points:
`S(5,-4)` and
`T(-8,12)`

Since in both questions,a and b, we're asked to find lines that are perpendicular to ST. So, we'll do that first!

Perpendicular to ST:

the equation of any line is given by:
`y = mx + c` where, m is the slope(also known as gradient), and c is the y-intercept.

to find the perpendicular of ST we first need to find the gradient of ST, using the gradient formula.

`m = (y_2 - y_1)/(x_2 - x_1)`

the coordinates of S and T can be used here. (it doesn't matter if you choose them in any order: S can be either x_1 and y_1 or x_2 and y_2)

`m = (12 - (-4))/((-8) - 5)`

`m = (-16)/(13)`

to find the perpendicular of this gradient: we'll use:

`m_1m_2=-1`

both
`m_1`and
`m_2` denote slopes that are perpendicular to each other. So if
`m_1 = (12 - (-4))/((-8) - 5)`, then we can solve for
`m_2` for the slop of ther perpendicular!

`\left((-16)/(13)\right)m_2=-1`

`m_2=(13)/(16)`:: this is the slope of the perpendicular

a) Line through S and Perpendicular to ST

to find any equation of the line all we need is the slope
`m` and the points
`(x,y)`. And plug into the equation:
`(y - y_1) = m(x-x_1)`

side note: you can also use the
`y = mx + c` to find the equation of the line. both of these equations are the same. but I prefer (and also recommend) to use the former equation since the value of 'c' comes out on its own.

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

we have the slope of the perpendicular to ST i.e
`m=(13)/(16)`

and the line should pass throught S as well, i.e
`(5,-4)`. Plugging all these values in the equation we'll get.

`(y - (-4)) = (13)/(16)(x-5)`

`y +4 = (13x)/(16)-(65)/(16)`

`y = (13x)/(16)-(65)/(16)-4`

`y=(13x)/(16)-(129)/(16)`

this is the equation of the line that is perpendicular to ST and passes through S

a) Line through T and Perpendicular to ST

we'll do the same thing for
`T(-8,12)`

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

`(y -12) = (13)/(16)(x+8)`

`y = (13x)/(16)+ (104)/(16)+12`

`y = (13x)/(16)+ (37)/(2)`

this is the equation of the line that is perpendicular to ST and passes through T