Determine a ∈ ℝ so that A(1, 2), B(3, 4), and C(a, 5) are collinear.

AB = (2, 2), AC = (a - 1, 3). Collinear ⇒ (a - 1)/(2) = (3)/(2) ⇒ a - 1 = 3 ⇒ a = 4.

Daily Math Problems — June 10, 2026: Analytic Geometry, Trigonometry, Powers, Radicals, Logarithms

Problems & worked solutions

RO M1

Determine $a \in R$ so that $A (1, 2)$ , $B (3, 4)$ , and $C (a, 5)$ are collinear.

Show answer & worked solution

$\vec{A B} = (2, 2)$ , $\vec{A C} = (a - 1, 3)$ . Collinear ⇒ $\frac{a - 1}{2} = \frac{3}{2} \Rightarrow a - 1 = 3 \Rightarrow a = 4$ .

RO M1

The area of an equilateral triangle with side $a = 6$ equals:

Show answer & worked solution

$A = \frac{36 \sqrt{3}}{4} = 9 \sqrt{3}$ .

RO M1

The solution of $\log_{2} (2 x^{2} + x + 1) - \log_{2} (x^{2} - x + 2) = 1$ is:

Show answer & worked solution

$2 x^{2} + x + 1 = 2 (x^{2} - x + 2) \Rightarrow 2 x^{2} + x + 1 = 2 x^{2} - 2 x + 4 \Rightarrow 3 x = 3 \Rightarrow x = 1$ .

Verify the domain: at $x = 1$ , $2 x^{2} + x + 1 = 4 > 0$ and $x^{2} - x + 2 = 2 > 0$ , so both logs are defined. The solution set is ${1}$ .

Problems & worked solutions

RO M1

Determine $a \in R$ so that $A (1, 2)$ , $B (3, 4)$ , and $C (a, 5)$ are collinear.

Show answer & worked solution

$\vec{A B} = (2, 2)$ , $\vec{A C} = (a - 1, 3)$ . Collinear ⇒ $\frac{a - 1}{2} = \frac{3}{2} \Rightarrow a - 1 = 3 \Rightarrow a = 4$ .

RO M1

The area of an equilateral triangle with side $a = 6$ equals:

Show answer & worked solution

$A = \frac{36 \sqrt{3}}{4} = 9 \sqrt{3}$ .

RO M1

The solution of $\log_{2} (2 x^{2} + x + 1) - \log_{2} (x^{2} - x + 2) = 1$ is:

Show answer & worked solution

$2 x^{2} + x + 1 = 2 (x^{2} - x + 2) \Rightarrow 2 x^{2} + x + 1 = 2 x^{2} - 2 x + 4 \Rightarrow 3 x = 3 \Rightarrow x = 1$ .

Verify the domain: at $x = 1$ , $2 x^{2} + x + 1 = 4 > 0$ and $x^{2} - x + 2 = 2 > 0$ , so both logs are defined. The solution set is ${1}$ .