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The vertices of ∆ ABC 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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Let three vertices be A (−2, 1) and B (5, 4) and C(2, −3)

Area of the triangle having vertices (x1,y1), (x2,y2) and (x3,y3

\(\frac{1}2\) |x1(y- y3)+x2(y- y1)+x3(y- y2)| 

Area of ∆ABC 

\(\frac{1}2\) |-2(4 – (-3)) + 5(-3 -1) + 2(1 – 4)| 

\(\frac{1}2\) |-14 - 20 - 6| 

= 20 sq. units 

Now to find length of BC, 

By distance formula,

XY = \(\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\)

For BC, 

BC = \(\sqrt{(2-5)^2+(-3-4)^2}\)

\(\sqrt{9+49}\)

\(\sqrt{58}\) sq. units

Area of ∆ABC = \(\frac{1}2\) × Base × Altitude

∴ 20 = \(\frac{1}2\) × \(\sqrt{58}\) × Altitude

∴ Altitude = \(\frac{40}{\sqrt{58}}\) units 

Hence, 

the length of altitude through A is \(\frac{40}{\sqrt{58}}\) units.

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