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saw5 [17]
3 years ago
13

. Given ????(5, −4) and T(−8,12):

Mathematics
1 answer:
damaskus [11]3 years ago
7 0

Answer:

a)y=\dfrac{13x}{16}-\dfrac{129}{16}

b)y = \dfrac{13x}{16}+ \dfrac{37}{2}

Step-by-step 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 <u>we first need to find the gradient of ST, using the gradient formula.</u>

m = \dfrac{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 = \dfrac{12 - (-4)}{(-8) - 5}

m = \dfrac{-16}{13}

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

m_1m_2=-1

both m_1and m_2 denote slopes that are perpendicular to each other. So if m_1 = \dfrac{12 - (-4)}{(-8) - 5}, then we can solve for m_2 for the slop of ther perpendicular!

\left(\dfrac{-16}{13}\right)m_2=-1

m_2=\dfrac{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=\dfrac{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)) = \dfrac{13}{16}(x-5)

y +4 = \dfrac{13x}{16}-\dfrac{65}{16}

y = \dfrac{13x}{16}-\dfrac{65}{16}-4

y=\dfrac{13x}{16}-\dfrac{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) = \dfrac{13}{16}(x+8)

y = \dfrac{13x}{16}+ \dfrac{104}{16}+12

y = \dfrac{13x}{16}+ \dfrac{37}{2}

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

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