What is the area of triangle whose vertices are (-4, -4), (-3, 2), (3, -16)?(a) 24 sq. units(b) 27 sq. units(c) 32 sq. units(d) 37 sq. unitsI have been asked this question during an interview for a job.Enquiry is from Slope of a Line in division Straight Lines of Mathematics – Class 11

What is the area of triangle whose vertices are (-4, -4), (-3, 2), (3,

1.	What is the area of triangle whose vertices are (-4, -4), (-3, 2), (3, -16)?(a) 24 sq. units(b) 27 sq. units(c) 32 sq. units(d) 37 sq. unitsI have been asked this question during an interview for a job.Enquiry is from Slope of a Line in division Straight Lines of Mathematics – Class 11
Answer» Right answer is (b) 27 sq. units To EXPLAIN I would say: We know, area of TRIANGLE joining vertices (x1, y1), (x2, y2) and (x3, y3) is (1/2)* determinant \(\begin{pmatrix}-4 & -4 &1 \\-3& 2 & 1\\ 3& -16 & 1\end{pmatrix}\) is \(\frac{1}{2}\){(-4)(2+16) – (-4)(-3-3) + (1)(48-6)} = \(\frac{1}{2}\)\|(-72)+(-24)+42\| = 27 square units.

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