The line x=(c/2) divides the triangle with vertices (0,0),(1,1) and (9,1) into two regions. If the area of the two regions is the same, then c is equal to

Question

The line x=(c/2) divides the triangle with vertices (0,0),(1,1) and (9,1) into two regions. If the area of the two regions is the same, then c is equal to (A) 30 (B) 6 (C) (7/2) (D) 15

Tardigrade Admin · Accepted Answer

2 ⋅ (1/2) ⋅ | 9 1 1 (c/2) 1 1 (c/2) (c/18) 1 | = (1/2) | 9 1 1 1 1 1 0 0 1 | = (1/2) ⋅ 8 = 4 ⇒ (c/2)=3 or (c/2)=15 (not possible) ⇒ c=6

Q. The line $x = \frac{c}{2}$ divides the triangle with vertices $(0, 0), (1, 1)$ and $(9, 1)$ into two regions. If the area of the two regions is the same, then $c$ is equal to

$30$

$6$

$\frac{7}{2}$

$15$

$2 \cdot \frac{1}{2} \cdot ∣ ∣ 9 \frac{c}{2} \frac{c}{2} 11 \frac{c}{18} 111 ∣ ∣ = \frac{1}{2} ∣ ∣ 910110111 ∣ ∣ = \frac{1}{2} \cdot 8 = 4$
$\Rightarrow \frac{c}{2} = 3$ or $\frac{c}{2} = 15$ (not possible)
$\Rightarrow c = 6$

Q. The line x=2c​ divides the triangle with vertices (0,0),(1,1) and (9,1) into two regions. If the area of the two regions is the same, then c is equal to

30

6

27​

15

2⋅21​⋅∣∣​92c​2c​​1118c​​111​∣∣​=21​∣∣​910​110​111​∣∣​=21​⋅8=4 ⇒2c​=3 or 2c​=15 (not possible) ⇒c=6

Q. The line $x = \frac{c}{2}$ divides the triangle with vertices $(0, 0), (1, 1)$ and $(9, 1)$ into two regions. If the area of the two regions is the same, then $c$ is equal to

$30$

$6$

$\frac{7}{2}$

$15$

$2 \cdot \frac{1}{2} \cdot ∣ ∣ 9 \frac{c}{2} \frac{c}{2} 11 \frac{c}{18} 111 ∣ ∣ = \frac{1}{2} ∣ ∣ 910110111 ∣ ∣ = \frac{1}{2} \cdot 8 = 4$
$\Rightarrow \frac{c}{2} = 3$ or $\frac{c}{2} = 15$ (not possible)
$\Rightarrow c = 6$