# SAT Coordinate Geometry

To solve problems involving coordinate geometry on the SAT, you need to know:

- the distance formula

The distance, \(d\), between the points \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is

\[d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}.\]

- the midpoint formula

The midpoint of a line segment with endpoints \((x_{1}, y_{1})\) and \((x_{2}, y_{2})\) is

\[\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right).\]

- If \(a\) and \(b\) are on a number line, and \(a<b\), then the distance between \(a\) and \(b\) is \(b-a\).
- how to find the perimeter and area of simple geometric figures.

#### Contents

## General Examples

If \(2AB=CD\) in the figure above, what is the value of \(x\)?

(A) \(\ \ -6\)

(B) \(\ \ 3\)

(C) \(\ \ 4\)

(D) \(\ \ 5\)

(E) \(\ \ 6\)

Correct Answer: C

Solution:

To find the length of \(\overline{AB},\) we find the difference between the \(y\)-coordinates of its endpoints, \(A\) and \(B:\) \[AB=x-(-1)=x+1.\] Similarly, we find the length of \(\overline{CD}\) by finding the difference between the \(x\)-coordinates of its endpoints, \(C\) and \(D:\) \[CD=4-(-6)=4+6=10.\] Using \(2AB=CD,\) we find \(x:\) \[\begin{align} 2AB&=CD\\ 2(x+1)&=10\\ x+1&=5\\ x&=4. \end{align}\]

Incorrect Choices:

(A)

This wrong choice is the \(y\)-coordinate of points \(A\) and \(B.\) Note that the \(x\)-coordinate of any point in the fourth quadrant is positive. Therefore, \(x\) cannot equal -6.

(B)

Tip: If a diagram is drawn to scale, trust it.

This wrong choice is the \(x\)-coordinate of points \(C\) and \(D.\) Since points \(B\) and \(C\) have the same \(y\)-coordinate, if they also have the same \(x\)-coordinate, they would be the same point. But according to the graph, this is not the case.

(D)

If you solve for the length of \(\overline{AB}\), which equals \(x+1\), instead of for \(x\), you will get this wrong answer.

(E)

This wrong choice is just meant to confuse you.

In the figure above, the circle is tangent to the lines \(x=-9\) and \(y=2,\) and the circle's center has a \(y\)-coordinate equal to 5. Which of the following are the coordinates of point \(P?\)

(A) \(\ \ (-12, 2)\)

(B) \(\ \ (-9, 2)\)

(C) \(\ \ (-6, 2)\)

(D) \(\ \ (-6, 5)\)

(E) \(\ \ (2, -6)\)

Correct Answer: C

Solution 1:

A theorem from geometry states that if a circle is tangent to a line, then that line is perpendicular to the circle's radius at the point of tangency. We use this theorem to conclude that lines \(y=2\) and \(x=-9\) are perpendicular to the radius of the circle at points \(P\) and \(Q,\) respectively.

Let us call the center of the circle \(C,\) as shown above. Because \(\overline{CP}\) is perpendicular to the line \(y=2,\) and because \(C\) has a \(y\)-coordinate equal to 5, and \(P\) has a \(y\)-coordinate equal to 2, the radius of the circle is the distance from \(C\) to \(P\), or \(5-2=3.\) This means that \(CQ=3\) also. Since point \(P\) is 3 units away from the line \(x=-9,\) the \(x\)-coordinate of point \(P\) is \(-9+3=-6.\) Because point \(P\) lies on the line \(y=2,\) its \(y\)-coordinate equals 2. Therefore, \(P=(2, -6).\)

Solution 2:

Tip: Draw a picture.

Tip: If you can, verify your choice.

We could approximate where the points in the answer choices are located. Remember that unless stated otherwise, diagrams on the SAT are drawn to scale. As you read, refer to the graph above.

(A) \((-12, 2)\) is located to the left of the line \(x=-9\), but point \(P\) is located to its right. Therefore, we can eliminate choice (A).

(B) \((-9, 2)\) is the intersection of lines \(x=-9\) and \(y=2\), and it is located to the left of point \(P\). We eliminate choice (B).

(C) \((-6, 2)\) is on the line \(y=2,\) and it is located approximately where point \(P\) is located. It is likely the answer.

(D) \((-6, 5)\) has the same \(y\)-coordinate as the center of the circle, but point \(P\) is located below the center. We eliminate choice (D).

(E) \((2, -6)\) is located in the fourth quadrant, far away from point \(P\). We don't need to plot this point to know that it should be eliminated from the choices.We've eliminated all choices except for choice (C). It is the correct answer.

Incorrect Choices:

(A),(B),(D), and(E)

Solution 2 explains why these choices are wrong.

## Examples for Midpoint Formula

## Examples for Distance Formula

There are \(3\) meters from \(A\) to \(B\), \(22\) meters from \(B\) to \(C\), \(5\) meters from \(C\) to \(D\), and \(7\) meters from \(D\) to \(E\), as shown in the figure above. What is the distance from \(A\) to \(E\), in meters?

(A) \(\ \ 22\)

(B) \(\ \ 18\)

(C) \(\ \ 17\)

(D) \(\ \ 15\)

(E) \(\ \ 8\)

Correct Answer: C

Solution:

Tip: Draw a picture.

Tip: Distance formula: \(d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}.\)

To find the distance from \(A\) to \(E\), we superimpose the given graph onto the \(xy\)-plane so that \(A\) has coordinates \((0, 0),\) and \(\overline{AB}\) lies on the \(x\)-axis, as shown above. We calculate that \(E\) will have the coordinates \((8, 15)\). To find \(AE,\) we use the distance formula:

\[\begin{array}{r c l} AE=\sqrt{(15-0)^{2}+(8-0)^{2}}&=& \sqrt{15^{2}+8^{2}}\\ &=&\sqrt{225+64}\\ &=&\sqrt{289}\\ &=&17. \end{array}\]

Therefore, \(\overline{AE}\) is \(17\) meters long.

Incorrect Choices:

(A)

Tip: If a diagram is drawn to scale, trust it.

It should be obvious that the distance between \(A\) and \(E\) is smaller than the distance between \(B\) and \(C.\) Use the edge of a paper to convince yourself of this. Eliminate this choice.

(B)

In the case you decide to estimate the length of \(\overline{AE},\) and you have good reasons to eliminate all choices except for (B) and (C)... well then, you will still have to guess. At least you will have \(50\%\) chance of guessing correctly. Better to learn the distance formula.

(D)and(E)

Tip: Just because a number appears in the question doesnâ€™t mean it is the answer.

If point \(A\) lies on the origin and segment \(\overline{AB}\) lies on the \(x\)-axis, imagine drawing a vertical line from point \(E\) to the \(x\)-axis, and let that vertical line cross the \(x\)-axis at point \(P.\) Then \(\triangle AEP\) is a right triangle whose leg \(\overline{EP}\) is \(15\) meters long. \(\overline{AE},\) the hypotenuse in the right triangle, has to be longer than the leg, \(\overline{EP}.\) So, \(AE\) has to be greater than \(15\) meters. Eliminate both choices (D) and (E).

## Review

If you thought these examples difficult and you need to review the material, these links will help:

## SAT Tips for Coordinate Geometry

- If a diagram is drawn to scale, trust it.
- Draw a picture.
- Read the diagram carefully.
- Distance formula: \(d=\sqrt{(x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}}.\)
- Midpoint formula: \(\left(\frac{x_{1}+x_{2}}{2}, \frac{y_{1}+y_{2}}{2}\right).\)
- \(A_{\triangle}=\frac{1}{2}bh.\)
- The area of a parallelogram is \(A=bh.\)
- SAT General Tips

**Cite as:**SAT Coordinate Geometry.

*Brilliant.org*. Retrieved from https://brilliant.org/wiki/sat-coordinate-geometry/