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?

1 Answer

6x-8y+121=0

Explanation:

The slope m of line perpendicular to the line joining the points (x_1, y_1)\equiv(3, 18) & (x_2, y_2)\equiv(6, 14) is given as

m=-1/{\frac{y_2-y_1}{x_2-x_1}}

=-1/{\frac{14-18}{6-3}}

=3/4

Now, the mid-point of line joining the points (x_1, y_1)\equiv(3, 18) & (x_2, y_2)\equiv(6, 14) are given as

(\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2})

\equiv(\frac{3+6}{2}, \frac{18+14}{2})

\equiv(\frac{9}{2}, 16)

Now, the equation of line passing through the point (x_0, y_0)\equiv(9/2, 16) & having slope m=3/4 is given as

y-y_0=m(x-x_0)

y-16=3/4(x-9/2)

6x-8y+121=0