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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.
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)`=1/2xxBCxx`ength of altitude though A
20`=1/2xxsqrt 58xx `lenght of altitude through A
Lenght of altitude through A`=40/sqrt58`
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