Answer:
a)[tex]y=\dfrac{13x}{16}-\dfrac{129}{16}[/tex]
b)[tex]y = \dfrac{13x}{16}+ \dfrac{37}{2}[/tex]
Step-by-step explanation:
Given two points: [tex]S(5,-4)[/tex] and [tex]T(-8,12)[/tex]
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: [tex]y = mx + c[/tex] 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.
[tex]m = \dfrac{y_2 - y_1}{x_2 - x_1}[/tex]
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)
[tex]m = \dfrac{12 - (-4)}{(-8) - 5}[/tex]
[tex]m = \dfrac{-16}{13}[/tex]
to find the perpendicular of this gradient: we'll use:
[tex]m_1m_2=-1[/tex]
both [tex]m_1[/tex]and [tex]m_2[/tex] denote slopes that are perpendicular to each other. So if [tex]m_1 = \dfrac{12 - (-4)}{(-8) - 5}[/tex], then we can solve for [tex]m_2[/tex] for the slop of ther perpendicular!
[tex]\left(\dfrac{-16}{13}\right)m_2=-1[/tex]
[tex]m_2=\dfrac{13}{16}[/tex]:: 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 [tex]m[/tex] and the points [tex](x,y)[/tex]. And plug into the equation: [tex](y - y_1) = m(x-x_1)[/tex]
side note: you can also use the [tex]y = mx + c[/tex] 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.
[tex](y - y_1) = m(x-x_1)[/tex]
we have the slope of the perpendicular to ST i.e [tex]m=\dfrac{13}{16}[/tex]
and the line should pass throught S as well, i.e [tex](5,-4)[/tex]. Plugging all these values in the equation we'll get.
[tex](y - (-4)) = \dfrac{13}{16}(x-5)[/tex]
[tex]y +4 = \dfrac{13x}{16}-\dfrac{65}{16}[/tex]
[tex]y = \dfrac{13x}{16}-\dfrac{65}{16}-4[/tex]
[tex]y=\dfrac{13x}{16}-\dfrac{129}{16}[/tex]
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 [tex]T(-8,12)[/tex]
[tex](y - y_1) = m(x-x_1)[/tex]
[tex](y -12) = \dfrac{13}{16}(x+8)[/tex]
[tex]y = \dfrac{13x}{16}+ \dfrac{104}{16}+12[/tex]
[tex]y = \dfrac{13x}{16}+ \dfrac{37}{2}[/tex]
this is the equation of the line that is perpendicular to ST and passes through T
