Geometry Problems: Lines and Angles

Question 1
Given that ABCD are collinear, implying they are on the same line, and
¿ x +14 , BC=x +11, CD=x +13 . Since these points are collinear, AD= AB+ BC+ CD
Therefore, AD= ( x +14 ) + ( x +11 )+ ( x +13 )=29
¿ 3 x+ 38=29
3 x=−9
x=−3
BC=x +11=−3+11=8
BC=8
Question 2
Given two endpoints with coordinates (x1 , y1 ¿ and (x2 , y2 ), we can obtain the coordinates of the
midpoint as ( x1 + x2
2 , y1+ y2
2 ). Hence, given a midpoint and one endpoint, we can obtain the other
endpoint.
End point1(6,-3) and midpoint(-2,-5)
Implying that 6+ x2
2 =−2 and −3+ y2
2 =−5.
Solving for these equations gives the following results:
x2= (−2∗2 )−6=−10 and y2= (−5∗2 ) +3=−7
Coordinates to the other endpoint are (−10 ,−7)
Question 3
First, plot the two points. To solve for the distance between the two points, we draw a right angled
triangle for ease of using the Pythagorean Theorem.
From the plot below, we obtain the values of the length of the base as 0.5 and the length of the height
as 2.5.
The Pythagorean rule states that a2+ b2=c2
Substituting the values gives 0.52 +2.52=6.5
Solving for c gives the distance between the two points as √6.5=2.5495

1.9 2 2.1 2.2 2.3 2.4 2.5 2.6
0
0.5
1
1.5
2
2.5
3
3.5
4
3.5
1
Y-Values
Question 4
The sum of angles on a straight line is equal to 1800
4 X +50+90=180
4 X +140=180
4 X=180−140=40
4 X
4 = 40
4
X =10 °
Question 5
The sum of angles at a point is equal to 360 °
3 X + ( X−69 ) +33=360
3 X + X−69+33=360
4 X−36=360
4 X=360+ 36
4 X=396
4 X
4 =396
4