Question

Find the value of k so that the area of triangle abc with a(k+1,1) b(4,-3) and c(7,-k) is 6 sq units

Answer 1

solve this nd get your answer

Answer 2

answer : k = 3

area of triangle formed by points $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ is given by,

$\textbf{area of triangle}=\frac{1}{2}|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|$

here, area of triangle formed by a(k+1, 1), b(4, -3) and c(7,-k) is 6 sq units.

6 = 1/2 |(k+1)(-3+k)+4(-k-1)+7(1+3)|

or, 12 = |(k+1)(-3+k)-4k-4+28|

or, 12 = |k² -2k - 3 - 4k + 24|

or, 12 = |k² - 6k + 21|

or, 12 = k² - 6k + 21 or -12 = k² - 6k + 21

or, 12 = k² - 6k + 21

or, k² - 6k + 9 = 0 => k = 3

again, -12 = k² - 6k + 21

or, k² - 6k + 33 = 0

D = (6)² - 4 × 33 < 0 so, doesn't have real roots.

hence, only one value of k e.g., k = 3