Given directed line segment ab find the coordinates of p such thats the ratio of ap to pd is 2:1

The point P on line AB such that the ratio of AP to PB is 2:1 is P = (0, 3)

Step-by-step explanation:

The dimension of the graph can be taken as 1 box = 1 unit

Therefore;

The coordinates of the point A = (-2, -7)

The coordinates of the point B = (1, 8)

For the a point P that gives the ratio AP to PB = 2:1 is given as follows;

The fraction of AP to AB = 2/(2 + 1)×AB = 2/3 × AB

Which gives;

The coordinate of the point P = (-2 + 2/3×(1 - (-2)), -7 + 2/3×(8 - (-7)))

P = (0, 3)

Therefore;

The coordinates of P such that the ratio of AP to PB is 2:1 is P = (0, 3).

∠1≅∠2 by the alternate exterior angles theorem.

step-by-step explanation:

given, a ∥ b and ∠1 ≅ ∠3 .we have to prove that e ∥ f

we know that ∠1≅∠3 and that a || b because they are given. we see that by the alternate exterior angles theorem. therefore, ∠2≅∠3 by the transitive property. so, we can conclude that e || f by the converse alternate exterior angles theorem.

we have to fill the missing statement.

transitivity property states that if a = b and b = c, then a = c.

now, given ∠1≅∠3 and by transitivity property ∠2≅∠3 .

hence, to apply transitivity property one angle must be common which is not in result after applying this property which is ∠1.

the only options in which ∠1 is present are ∠1 and ∠2, ∠1 and ∠4

∠1 and ∠4 is not possible ∵ after applying transitivity we didn't get ∠4.

hence, the missing statement is ∠1≅∠2.

so, ∠1≅∠2 by the alternate exterior angles theorem.

answer: 193

explanation:

200-7 = 193