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(y2 - y3)+x2(y3 - y1)+x3(y1 - 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.