1.

Find the area of a triangles whose vertices are(a, c + a), (a, c) and (-a, c – a)

Answer»

Let A = (x1,y1) = (a, c + a), B = (x2, y2) = (a, c) and C = (x3, y3) = (- a, c – a) be the given points

Then,

The area of ∆ABC is given by

= \(\frac{1}{2}\)[a ( – {c – a}) + a(c – a – (c + a)) +( – a)(c + a – a)]

= \(\frac{1}{2}\) [a(c – c + a) + a(c – a – c – a) – a(c + a – c)]

= \(\frac{1}{2}\)[a × a + ax( – 2a) – a × a]

= \(\frac{1}{2}\)[a– 2a– a2]

= \(\frac{1}{2}\)×(-2a)2

= – a2



Discussion

No Comment Found

Related InterviewSolutions