Question

find the value of K if area of triangle is 4unit and vertices (k,0),(4,0)(0,2)​

Answer 1

Answer:

(k,0) = (x1,y1)

(4,0) = (x2,y2)

(0,2) = (x3,y3)

area of triangle =

$\frac{1}{2} | x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|$

4 = 1/2 | k(0-2) + 4(2-0) + 0(0-0) |

4 = 1/2 | -2k + 8 |

4 ×2 = -2k + 8

8 = -2k + 8

0 = -2k

k = 0

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