+0

0
132
1

Draw a line perpendicular to the line that contains the points

(1, 6) and  (4, 4) and passes through the point  (−2, 6).

Can't seem to figure out how to do the numbers by drawing a perpendicular line

Oct 3, 2018

#1
+5225
+1

ok two steps to this

first we find the slope of the line passing through the two points

$$m = \dfrac{4-1}{4-6} = -\dfrac{3}{2}$$

now we want a line perpendicular to this so we have a slope of

$$m_{\perp} = -\dfrac{1}{m} = -\left(\dfrac{1}{-\frac 3 2}\right) = \dfrac 2 3$$

now using point slope formula with the point (-2,6) we have

$$(y-6) = \dfrac 2 3 (x - (-2)) \\ y = \dfrac 2 3 x + \dfrac{22}{3}$$

.
Oct 3, 2018

#1
+5225
+1

ok two steps to this

first we find the slope of the line passing through the two points

$$m = \dfrac{4-1}{4-6} = -\dfrac{3}{2}$$

now we want a line perpendicular to this so we have a slope of

$$m_{\perp} = -\dfrac{1}{m} = -\left(\dfrac{1}{-\frac 3 2}\right) = \dfrac 2 3$$

now using point slope formula with the point (-2,6) we have

$$(y-6) = \dfrac 2 3 (x - (-2)) \\ y = \dfrac 2 3 x + \dfrac{22}{3}$$

Rom Oct 3, 2018