Question

What is the equation of the line that is perpendicular to the line passing through (3,18) and (6,14) at midpoint of the two points?

Answer 1

$let \: l1 \: be \: the \: our \: required \: line \\ point \: passing \: through \: l1 \\ = p(x.y) = (\frac{3 + 6}{2} . \frac{18 + 14}{2} ) \\ p(x.y) = (\frac{9}{2} .16) \\ \: slope \: of \: line2 = \frac{14 - 18}{ 6 - 3} = - \frac{4}{ 3} \\ as \: line \: is \: perpendicular \\ so \: slope \: of \: line1 \: m= \frac{3}{4} \\ eq \: of \: line1 \: = \\ y = mx + c \\ y = \frac{3}{4} x + c \\ put \: x \: and \: y \\ 16 = \frac{9}{2} . \frac{3}{4} + c \\ c = \frac{64 - 27}{8} = \frac{37}{8} \\ our \: req \: eqn = y = \frac{3}{4} x + \frac{37}{8}$

Answer 2

Answer:

Hey there!

If need more explanation feel free to ask.

Step-by-step explanation:

Step 1 : Find out the slope of the perpendicular line (m1).

=> slope of the line passing through (3,18) and (6,14) = m2 = (Y2 - Y1)/(X2 - X1)

     m2 = -4/3

=> since the relation between the slope of perpendicular lines is,

M1 * M2 = -1, therefore, M1 = 3/4.

Step 2 : Equation of the line.

=> the coordinates of the mid point will be Xm = (3+6)/2 = 9/2 ,

    Ym = (18+14)/2=16

=> Applying Y -y1 = m (X-x1) for equation of the line.

=> we get, 8y = 6x + 101 ...............Answer.