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Mathematics 45 Online
OpenStudy (ayyookyndall):

Help!? What is the first step in constructing a perpendicular at a point on a line? a. Place the compass on the end of the line. b. Place the compass on the point. c. Draw an arc from each end of the line. d. Draw a straight line through the point. Find the slope of a line perpendicular to the line containing the points C(–12, 11) and D(3, 9). a. 15 over 2 b. –2 over 15 c. 2 over 15 d. –15 over 2 *Please explain

OpenStudy (ayyookyndall):

@DullJackel09 @arabpride @StudyGurl14 @mathmath333 @iGreen

OpenStudy (studygurl14):

c, for the first one

OpenStudy (studygurl14):

but technically, to do c, you have to do a...so yea, this is a stupid question

OpenStudy (ayyookyndall):

So, which one is it?

OpenStudy (studygurl14):

first, find the slope of the line going through the points by using the slope formula

OpenStudy (studygurl14):

(for the second one)

OpenStudy (ayyookyndall):

For #2 I got c.

OpenStudy (ayyookyndall):

@StudyGurl14

OpenStudy (studygurl14):

That is incorrect.

OpenStudy (ayyookyndall):

Okay. So how do I find the slopes? @StudyGurl14

OpenStudy (studygurl14):

Slope formula: Where \[(x_1,y_1)\]and \[(x_2,y_2)\]are points on the same line... \[slope=\frac{y_2-y_1}{x_2-x_1}\]

OpenStudy (studygurl14):

But remember, perpendicular lines have opposite reciprocal slopes

OpenStudy (ayyookyndall):

Okay. Before you go, can you help me with one more question?

OpenStudy (studygurl14):

Sure :)

OpenStudy (ayyookyndall):

An irregular quadrilateral has exterior angle measures of 93, 85, and 89. What is the measure of the fourth interior angle? 273 267 93 87 *What is a quadrilateral equal to?

OpenStudy (ayyookyndall):

@StudyGurl14

OpenStudy (studygurl14):

quadrilaterals interior angles add up to 360 degrees

OpenStudy (ayyookyndall):

So I got 87.. Am I correct? @StudyGurl14

OpenStudy (studygurl14):

Yep. great job!

OpenStudy (ayyookyndall):

Okay. Can you help me with other questions? I am so confused in Geometry. @StudyGurl14

OpenStudy (studygurl14):

post new question in new thread and link me :)

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