Test:26 QNo:2

VITEEE Preparation › Disucss Test Questions questions › Test:26 QNo:2

The two points of a triangle are (4,3) and (9,8). The third point lies on the line x-y+3=0. Find the area.

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5 Answers

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scintillating dev ·2010-09-30 00:16:10

plz help

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nihal raj ·2010-09-30 00:37:29

answer is 10 ? am i right???

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scintillating dev ·2010-09-30 00:42:09

no, its 20. but what's the idea needed???

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yagyadutt Mishra ·2010-09-30 00:53:34

It is easy but you have to brush up your mind for few second....

Now here we go :)

Point B = (4,3) Point C = ( 9,8) and Another point A is on x-y+3 = 0

Slope of line BC = (8-3)/(9-4) = 1

Slope of the line on which A is located = 1..

So the line BC is parallel to x-y+3 = 0

So distance between x-y+3 = 0 and the line BC will be the height of the triangle .
Base will be BC

So apply area = 1/2 * base * height

Now equation of BC = > x-y+ c = 0 => x-y-1 = 0

Distance between BC and Line = 4/root(2) = 2√2

Length of BC = √ ( 8-3)^2 + ( 9-4)^2 = 5√2

SO area = 1/2 * 5√2* 2√2 = 10 ans

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scintillating dev ·2010-09-30 01:21:59

thanx[1]

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Your Answer

Basic Integral Calculus Operators Symbols Matrix Cases

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