The figure below shows parallelograms abce, pqrs, and tnlm on a coordinate plane. points s, p, q, and r are midpoints of abce t, n, l, and m are midpoints of pqrs write the equation of line ml. show your work.

The equation of line ML is y = 3

Step-by-step explanation:

* Lets revise how to find the mid-point of a segment passes through

 two points (x1 , y1) , (x2 , y2)

- The rule of the mid-point is ((x1 + x2)/2 , (y1 + y2)/2)

- The equation of the line which passes through (x1 , y1) , (x2 , y2) is

 y = mx + c, where m is the slope of the line and c is the y-intercept

- The rule of the slope is m = (y2 - y1)/(x2 - x1)

- If the line is horizontal (the two point have the same y- coordinates),

 then its slope = 0 and its equation is y = c ⇒ the line intersect the

 y-axis at point (0 , c)

* Now lets solve the problem

- From the graph the vertices of ABCE are:

 A (-3 , -3) , B (5 , -3) , C (5 , 5) , E (-3 , 5)

∵ S is the mid-point of AE

∴ S = [(-3 + -3)/2 , (-3 + 5)/2] = [-6/2 , 2/2]

∴ S = (-3 , 1)

∵ R is the mid-point of EC

∴ R = [(5 + -3)/2 , (5 + 5)/2] = [2/2 , 10/2]

∴ R = (1 , 5)

∵ Q is the mid-point of BC

∴ Q = [(5 + 5)/2 , (-3 + 5)/2] = [10/2 , 2/2]

∴ Q = (5 , 1)

* Now lets find M and L to find the equation of ML

∵ M is the mid-point of SR

∵ S = (-3 , 1) , R = (1 , 5)

∴ M = [(-3 + 1)/2 , (1 + 5)/2] = [-2/2 , 6/2]

∴ M = (-1 , 3)

∵ L is the mid-point of RQ

∵ R = (1 , 5) , Q = (5 , 1)

∴ L = [(1 + 5)/2 , (5 + 1)/2] = [6/2 , 6/2]

∴ L = (3 , 3)

- The y-coordinates of point M and L are equal

∴ Segment ML is horizontal

∴ Its equation is y = c

∵ c = 3

∴ The equation of line ML is y = 3

y = 3

Step-by-step explanation:

< coa = 150 ( given)

as ba and bc are tangents

so oc and oa is perpendicular to cb and ab respectively because in circle normal passes through centre

so in quadrilateral oabc

< coa + < oab + < ocb + < abc = 360

150 +90+90 + < abc = 360

330 + < abc = 360

< abc = 360 -330 = 30

so < abc = 30

⭐dhruv⭐

468

step-by-step explanation: