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The vertices of ΔABC are (–2, 1), (5, 4) and (2, –3) respectively. Find the area of the triangle and the length of the altitude through A.

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Question

The vertices of ΔABC are (–2, 1), (5, 4)  and (2, –3) respectively. Find the area of the triangle and the length of the altitude through A.

Sum
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Solution

GIVEN: The vertices of triangle ABC are A (−2, 1) and B (5, 4) and C (2, −3)

TO FIND: The area of triangle ABC and length if the altitude through A

PROOF: We know area of triangle formed by three points (x1, y1), (x2, y2) and (x3, y3)is given 

By Δ `=1/2[x_1(y_1-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)  ` 

NOW AREA OF ΔABC 

Taking three point A(-2, 1) and B(5, 4) and C(2, -3)  

Area (ΔABC) `=1/2[{-8-15+2}-{5+8+6}]` 

`=1/2[{-21}-{19}]` 

`=1/2[{-40}]` 

`=1/2(40)` 

`=20` 

WE HAVE 

`BC=sqrt((5-2)^2+(4+3)^2)`  

`BC=sqrt((3)^2+(7)^2)`

`BC =sqrt(9+49 )` 

`BC=sqrt58 `

NOW  

 Area  (ΔABC) `=1/2 xx BC xx` length of altitude through A

20 `=1/2 xx sqrt 58 xx` length of altitude through A 

 Length of altitude through A `=40/sqrt58`

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Chapter 6: Co-ordinate Geometry - EXERCISE 6.5 [Page 6.41]

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R.D. Sharma Mathematics [English] Class 10
Chapter 6 Co-ordinate Geometry
EXERCISE 6.5 | Q 3. | Page 6.41
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