If the area of the triangle formed by the points $(- 2, 3), (4, - 5)$ and $(- 3, y)$ is 10 square units then $y =$

\frac{19}{3}

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\frac{21}{3}

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\frac{23}{3}

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\frac{25}{3}

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Solution

The correct option is C $\frac{23}{3}$
Area of triangle having vertices $(x_{1}, y_{1}), (x_{2}, y_{2})$ and $(x_{3}, y_{3})$ is given by
$= \frac{1}{2} \times [x_{1} (y_{2} - y_{3}) + x_{2} (y_{3} - y_{1}) + x_{3} (y_{1} - y_{2})]$
$Area = \frac{1}{2} [- 2 (- 5 - y) + 4 (y - 3) - 3 (3 + 5)]$
$= \frac{1}{2} (10 + 2 y + 4 y - 12 - 24)$
$= \frac{1}{2} (6 y - 26) = 3 y - 13$
$3 y - 13 = 10 s q . u n i t$
$y = \frac{23}{3}$

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Similar questions

If the area of the triangle formed by the points $(x, 2 x), (- 2, 6)$ and $(3, 1)$ is $5$ square units , then $x$
(a) $23$
(b) $35$
(c) $3$
(d) $5$

x^{2} - 3 x y + y^{2} = 0

x + y + 1 = 0

If the vertices of a triangle are A (-2, -3), B (3, 2) and C (-1, -8), then the area of triangle ADC, where D is the mid-point of side BC, is:

A B

A C

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x + y = 5

7 x - y = 3

B C

△ A B C = 5 {unit}^{2}

(- 2, 5), (x, - 3)

(3, 2)

14

x =

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