Correct answer is (a) True
Explanation: A (x1, y1), B (x2, y2) and C (x3, y3) are the vertices of a triangle ABC.
Draw AL, CM, BN perpendicular to x – axis.
Then, ML = (x1 – x2), LN = (x3 – x1) and MN = (x3 – x2).
∴ area of ∆ABC = ar(trap.BMLA) + ar(trap.ALNC) + ar(trap.BMNC)
= {\(\frac {1}{2}\) (AL + BM) × ML} + {\(\frac {1}{2}\) (AL + CN) × LN} – {\(\frac {1}{2}\) (CN + BM) × MN}
= {\(\frac {1}{2}\) (y1 + y2 ) × (x1 – x2)} + {\(\frac {1}{2}\) (y1 + y3 ) × (x3 – x1)} – {\(\frac {1}{2}\) (y2 + y3 ) × (x3 – x2)}
= \(\frac {1}{2}\) {x1 (y1 + y2 – y1 – y3 ) + x2 (y2 + y3 – y1 – y2 ) – x3 (y1 + y3 – y2 – y3 ) }
= \(\frac {1}{2}\) {x1 (y2 – y3 ) + x2 (y3 – y1 ) + x3 (y1 – y2 ) }