Question -
Answer -
We know that the area ofa triangle┬аwhose vertices are (x1,┬аy1), (x2,┬аy2), and
(x3,┬аy3)┬аis the absolutevalue of the determinant (╬Ф), where
It is given that thearea of triangle is 4 square units.
тИ┤╬Ф = ┬▒ 4.
(i) The area of thetriangle with vertices (k, 0), (4, 0), (0, 2) is given by the relation,
╬Ф =
тИ┤тИТk┬а+ 4 = ┬▒ 4
When тИТk┬а+ 4= тИТ 4,┬аk┬а= 8.
When тИТk┬а+ 4= 4,┬аk┬а= 0.
Hence,┬аk┬а=0, 8.
(ii) The area of thetriangle with vertices (тИТ2, 0), (0, 4), (0,┬аk) is given by therelation,
╬Ф =
тИ┤k┬атИТ 4 = ┬▒ 4
When┬аk┬атИТ4 = тИТ 4,┬аk┬а= 0.
When┬аk┬атИТ4 = 4,┬аk┬а= 8.
Hence,┬аk┬а=0, 8.