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# What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6,

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Joined: 02 Sep 2009
Posts: 55639
What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6,  [#permalink]

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19 Dec 2018, 01:23
00:00

Difficulty:

35% (medium)

Question Stats:

75% (01:34) correct 25% (02:12) wrong based on 108 sessions

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What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6, 6) in the coordinate plane?

A. 2
B. 3
C. 4
D. 5
E. 6

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Re: What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6,  [#permalink]

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19 Dec 2018, 22:03
Bunuel wrote:
What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6, 6) in the coordinate plane?

A. 2
B. 3
C. 4
D. 5
E. 6

Area of a triangle with its vertices $$(x_1,y_1), (x_2,y_2),$$ and $$(x_3,y_3)$$:-

A = $$\frac{{|x_{1}(y_{2}-y_{3})+x_{2}(y_{3}-y_{1})+x_{3}(y_{1}-y_{2})|}}{2}$$

Or, A=$$\frac{{|-2(4-6)+2(6-4)+6(4-4)|}}{2}$$
Or, A=$$\frac{{|-2(-2)+2(2)+6(0)|}}{2}$$
Or, A=4

Ans. (C)
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Re: What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6,  [#permalink]

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08 Jan 2019, 10:40
Top Contributor
Bunuel wrote:
What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6, 6) in the coordinate plane?

A. 2
B. 3
C. 4
D. 5
E. 6

When we sketch the points . . .

. . . we see that the triangle's base has length 4, and the triangle's height is 2

Area of triangle = (base)(height)/2
= (4)(2)/2
= 4

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Re: What is the area of a triangle with vertices (–2, 4), (2, 4) and (–6,   [#permalink] 08 Jan 2019, 10:40
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