In order to attack this question, we can plug in the value of x for each coordinate to see if the proper value of y matches. for 

(2, 7) if we plug in 2 for x, we can see that

which does in fact = 7. 

For (-5, -3.5), if we take the same step we can see that 

does give us a y value of -3.5.

For (14, 25), plugging in x = 14 gets us to

which simplifies to 25.

So, by default we can see that our correct answer must be 

(6, 12), since an input of 6 gives us 

which equals 13, not 12. So, (6, 12) is the only point *not* on the line. 

In order to find the area of the triangle, we can use relevant points such as the point of intersection and the x intercepts. In this case, the point of intersection is the point at which the equations of the lines are equal. So, if 

then x = 4, and if we plug 4 back into our equation we can see that y = 6. So, the point of intersection is (4, 6). 

If we set y = 0 for each equation, we can also find the x intercepts, in this case, (1, 0) and (16, 0).

The triangle will appear as follows

At this point, we can use the x intercepts to find the base, 16-1 or 15, and the point of intersection to find the height, in this case 6.

If the area of a triangle is 

we can fill in the unknowns with

to arrive at an area of 45.

We can quickly visualize this line to draw conclusions.

In order to do so, plug in 0 for x to find a point on the line:

3(0)+2y=6

y=3

Thus, (0,3) is a point on the line.

Plug in 0  for y to find a second point on the line:

3x+2(0)=6

x=2

(2,0) is another point on the line.

Now we know that the line passes through the points (2,0) and (0,3).  

A quick sketch of the two points reveals that the line passes through all but the third quadrant.

*note - a line with a positive slope will always pass through quadrants I and III, while a line with a negative slope will always pass through quadrants II and IV.*

As you unpack the given information, a few things should stand out:

1. Triangles ABC and ADE are similar. They each have a right angle and they each share the angle at point A, meaning that their lower-left-hand angles (at points B and D) will be the same also since all angles in a triangle must sum to 180.  

2. Side BC has to measure 6, as you’re given one side (AC = 8) and the hypotenuse (AB = 10) of a right triangle. You can use Pythagorean Theorem to solve, or you can recognize the 3-4-5 side ratio (which here amounts to a 6-8-10 triangle). 

3. This then allows you to use triangle similarity to determine the side lengths of the large triangle. Since the hypotenuse is 20 (segments AB and BD, each 10, combine to form a side of 20) and you know it’s a 3-4-5 just like the smaller triangle, you can fill in side DE as 12 (twice the length of BC) and segment CE as 8.

So you now know the dimensions of the parallelogram: BD is 10, BC is 6, CE is 8, and DE is 12. The sum of those four sides is 36.

Because the triangles are similar, you can tell that if the hypotenuse of the larger triangle is 15 and the hypotenuse of the smaller triangle is 10, then the sides have a ratio of 3:2 between the triangles.  This allows you to fill in the sides of XYZ: side XY is 6 (which is 2/3 of its counterpart side AB which is 9) and since YZ is 8 (which is 2/3 of its counterpart side, BC, which is 12).

Since the area of a triangle is  Base * Height, if you know that you have a base of 8 and a height of 6, that means that the area is .

One important concept to recognize in this problem is that the triangles ABD and ECD are similar. Each has a right angle and they share the same angle at point D, meaning that their third angles (BAD and CED, the angles at the upper left of each triangle) must also have the same measure.

With that knowledge, you can use the given side lengths to establish a ratio between the side lengths of the triangles. If BC is 2 and CD is 8, that means that the bottom side of the triangles are 10 for the large triangle and 8 for the smaller one, or a 5:4 ratio.

You’re then told the area of the larger triangle. Knowing that the area is 25 and that area =  Base x Height, you can plug in 10 as the base and determine that the height, side AB, must be 5.

Since you know that the smaller triangle’s height will be  the length of 5, you can then conclude that side EC measures 4, and that is your right answer.

In beginning this problem, it is important to note that the two triangles pictured, ABC and CED, are similar. They each have a right angle and they share the vertical angle at point C, meaning that the angles at A and D must also be congruent and therefore the triangles are similar.

This means that their side lengths will be proportional, allowing you to answer this question. You’re asked to match the ratio of AB to AC, which are the side across from angle C and the hypotenuse, respectively.  In triangle CED, those map to side ED and side CD, so the ratio you want is ED:CD.

If the perimeter of triangle ABC is twice as long as the perimeter of triangle DEF, and you know that the triangles are similar, that then means that each side length of ABC is twice as long as its corresponding side in triangle DEF. That also means that the heights have the same 2:1 ratio: the height of ABC is twice the length of the height of DEF.

Since the formula for area of a triangle is  Base x Height, you can express the area of triangle DEF as bh and the area of ABC as .  You’ll then see that the areas of ABC to DEF are  and bh, for a ratio of 4:1.

NOTE: It can seem surprising that the ratio isn’t 2:1 if each length of one triangle is twice its corresponding length in the other. But keep in mind that for an area you multiply two lengths together, and go from a unit like “inches” to a unit like “square inches.”  Because each length is multiplied by 2, the effect is exacerbated.

A key to solving this problem comes in recognizing that you’re dealing with similar triangles. Because lines BE, CF, and DG are all parallel, that means that the top triangle ABE is similar to two larger triangles, ACF and ADG. You know that because they all share the same angle A, and then if the horizontal lines are all parallel then the bottom two angles of each triangle will be congruent as well. You’ve established similarity through Angle-Angle-Angle.

This means that the side ratios will be the same for each triangle.  And for the top triangle, ABE, you know that the ratio of the left side (AB) to right side (AE) is 6 to 9, or a ratio of 2 to 3.  Note then that the remainder of the given information provides you the length of the entire right-hand side, line AG, of larger triangle ADG.  If AE is 9, EF is 10, and FG is 11, then side AG is 30.  And since you know that the left-hand side has a 2:3 ratio to the right, then line segment AD must be 20.

This problem tests the concept of similar triangles. First, you should recognize that triangle ACE and triangle BDE are similar. You know this because they each have the same angle measures: they share the angle created at point E and they each have a 90-degree angle, so angle CAE must match angle DBE (the top left angle in each triangle.

Because these triangles are similar, their dimensions will be proportional. Since sides, AC and BD - which are proportional sides since they are both across from the same angle, E - share a 3:2 ratio you know that each side of the smaller triangle (BDE) will be  as long as its counterpart in the larger triangle (ACE). Consequently, if the bottom side CE in the larger triangle measures 30, then the proportional side for the smaller triangle (side DE) will be  as long, measuring 20.

Since the question asks for the length of CD, you can take side CE (30) and subtract DE (20) to get the correct answer, 10.