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SSAT Upper Level Math : Coordinate Geometry

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Example Questions

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Given a line defined by the equation $y=\frac{9}{8}x+4$ , which of the following lines is perpendicular to ?

Possible Answers:

$y=-\frac{8}{9}x+4$

$y=\frac{9}{8}x+4$

$y=-\frac{9}{8}x+4$

$y=\frac{9}{8}x-4$

$y=4x+\frac{9}{8}$

Correct answer:

$y=-\frac{8}{9}x+4$

Explanation:

For a given line defined by the equation y=mx+b , any line perpendicular to must have a slope that is the negative reciprocal of 's slope , $-\frac{1}{m}$ .

In this instance, the slope of line is $m=\frac{9}{8}$ , so $-\frac{1}{m}=-\frac{1}{\frac{9}{8}}=-\frac{8}{9}$ . The only line provided with an equation that has this slope is $y=-\frac{8}{9}x+4$ .

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Example Question #2 : How To Find The Equation Of A Perpendicular Line

A given line is defined by the equation $y=\frac{5}{6}x-9$ . What is the slope of any line that is perpendicular to ?

Possible Answers:

Not enough information provided

$\frac{1}{9}$

$-\frac{6}{5}$

$-\frac{5}{6}$

Correct answer:

$-\frac{6}{5}$

Explanation:

For a given line defined by the equation y=mx+b , any line perpendicular to must have a slope that is the negative reciprocal of 's slope , $-\frac{1}{m}$ .

Since in this case $m=\frac{5}{6}$ ,

$-\frac{1}{m}=-\frac{1}{\frac{5}{6}}=\frac{6}{5}$ .

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Example Question #21 : How To Find The Equation Of A Perpendicular Line

Which of the following equations represents a line that goes through the point (-2, 6) and is perpendicular to the line 3x + 6y = 12 ?

Possible Answers:

y = 2x + 6

y = -2x + 10

y = 2x + 10

y = 2x + 2

y = -2x + 2

Correct answer:

y = 2x + 10

Explanation:

In order to solve this problem, we need first to transform the equation from standard form to slope-intercept form:

y = mx + b

Transform the original equation to find its slope.

3x + 6y = 12

First, subtract from both sides of the equation.

3x-3x + 6y = 12-3x

Simplify and rearrange.

6y = -3x + 12

Next, divide both sides of the equation by 6.

$\frac{6y}{6} = -\frac{3x}{6} + \frac{12}{6}$

$y = -\frac{1}{2}x+ 2$

The slope of our first line is equal to $-\frac{1}{2}$ . Perpendicular lines have slopes that are opposite reciprocals of each other; therefore, if the slope of one is x, then the slope of the other is equal to the following:

$-\frac{1}{x}$

Let's calculate the opposite reciprocal of our slope:

$-\frac{1}{2}\rightarrow -\left (-\frac{2}{1} \right )=2$

The slope of our line is equal to 2. We now have the following partial equation:

y = 2x + b

We are missing the y-intercept, . Substitute the x- and y-values in the given point (-2, 6) to solve for the missing y-intercept.

6 = 2(-2) + b

6 = -4 + b

Add 4 to both sides of the equation.

6+4 = -4+4 + b

10 = b

Substitute this value into our partial equation to construct the equation of our line:

y = 2x + 10

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

A line segment has the endpoints (-1, 3) and (4,-2) . What is the length of this line segment?

Possible Answers:

$5\sqrt2$

$\sqrt{30}$

Correct answer:

$5\sqrt2$

Explanation:

Use the following formula to find the distance between two points:

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Plug in the points that are given.

$\text{Distance}=\sqrt{(4-(-1))^2+(-2-3)^2}$

$\text{Distance}=\sqrt{(5)^2+(-5)^2}$

$\text{Distance}=\sqrt{25+25}=\sqrt{50}=\sqrt{5*5*2}=5\sqrt2$

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Example Question #2 : How To Find The Length Of A Line With Distance Formula

A line segment has end points at (8,1) and (-2,-3) . What is the length of this line segment?

Possible Answers:

7.46

10.77

9.88

6.32

Correct answer:

10.77

Explanation:

The distance between two points is given by the following equation:

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Now, using the two given points, plug them in to find the distance.

$\text{Distance}=\sqrt{(-2-8)^2+(-3-1)^2}=\sqrt{100+16}=\sqrt{116}=10.77$

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

Find the length of a line that has end points at $(8, -1)\text{ and }(-2, 1)$ .

Possible Answers:

$2\sqrt{26}$

126

$\sqrt{26}$

104

Correct answer:

$2\sqrt{26}$

Explanation:

Use the distance formula to find the length of this line segment:

$\text{Distance}=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Substitute in the values provided:

$\text{Distance}=\sqrt{(-2-8)^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}$

At this point, break down the square root:

$\sqrt{104}=\sqrt{2\cdot52}=\sqrt{2\cdot2\cdot26}=\sqrt{2\cdot2\cdot2\cdot13}$

You can remove two of the twos and put them on the outside of the square root symbol, and multiply the two and the thirteen that remain underneath the square root symbol:

$\sqrt{2\cdot2\cdot2\cdot13}=2\sqrt{2\cdot13}=2\sqrt{26}$

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Example Question #3 : How To Find The Length Of A Line With Distance Formula

A line segment has endpoints at $(-2, -1)\text{ and }(8, 1)$ . Find the length of this line.

Possible Answers:

$6\sqrt{3}$

$2\sqrt{26}$

Correct answer:

$2\sqrt{26}$

Explanation:

Use the distance formula to find the length of the line segment.

$D=\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}$

Where,

(x_1,y_1)=(-2,-1) and (x_2,y_2)=(8,1) .

$\text{Distance}=\sqrt{(8-(-2))^2+(1-(-1))^2}=\sqrt{100+4}=\sqrt{104}=2\sqrt{26}$

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Example Question #4 : How To Find The Length Of A Line With Distance Formula

Find the length of the line segment that has endpoints $(0, 4)\text{ and }(12, 6)$ .

Possible Answers:

10.659

12.412

8.451

12.166

Correct answer:

12.166

Explanation:

Use the distance formula to find the length of the line segment.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

(x_1,y_1)=(0,4) and (x_2,y_2)=(12,6) .

$\text{Distance}=\sqrt{(12-0)^2+(6-4)^2}=\sqrt{144+4}=\sqrt{148}=12.166$

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Example Question #151 : Geometry

Find the length of the line segment that has the endpoints $(2, 4)\text{ and }(7, 8)$ .

Possible Answers:

$\sqrt{29}$

$\sqrt{31}$

$\sqrt{41}$

Correct answer:

$\sqrt{41}$

Explanation:

Use the distance formula to find the length of the line.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

(x_1,y_1)=(2,4) and (x_2,y_2)=(7,8) .

$\text{Distance}=\sqrt{(7-2)^2+(8-4)^2}=\sqrt{25+16}=\sqrt{41}$

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Example Question #1 : How To Find The Length Of A Line With Distance Formula

One leg of a triangle has endpoints at the coordinates $(-1, -2)\text{ and }(-4, -1)$ . Find the length of this leg.

Possible Answers:

$\sqrt{10}$

$2\sqrt2$

Correct answer:

$\sqrt{10}$

Explanation:

Use the distance formula to find the length of the leg.

$D=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

Where,

(x_1,y_1)=(-1,-2) and (x_2,y_2)=(-4,-1) .

$\text{Length}=\sqrt{(-4-(-1))^2+(-1-(-2))^2}=\sqrt{9+1}=\sqrt{10}$

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SSAT Upper Level Math : Coordinate Geometry

Study concepts, example questions & explanations for SSAT Upper Level Math

All SSAT Upper Level Math Resources

Example Questions

Example Question #1 : How To Find The Equation Of A Perpendicular Line

Example Question #2 : How To Find The Equation Of A Perpendicular Line

Example Question #21 : How To Find The Equation Of A Perpendicular Line

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #2 : How To Find The Length Of A Line With Distance Formula

Example Question #1 : How To Find The Length Of A Line With Distance Formula

Example Question #3 : How To Find The Length Of A Line With Distance Formula

Example Question #4 : How To Find The Length Of A Line With Distance Formula

Example Question #151 : Geometry

Example Question #1 : How To Find The Length Of A Line With Distance Formula

All SSAT Upper Level Math Resources

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