If two lines are perpendicular n if one line form angle α with positive x-axis n or line form angle 90°+ α.

If inclination is α slope is given by tan α. Given that slope of line is negative which means tan α is negative. We know, tan α is negative in 2^nd quadrant i.e. α should be obtuse angle.

We know, area of triangle joining vertices (x₁, y₁), (x₂, y₂) and (x₃, y₃) is (1/2)* determinant \(\begin{pmatrix}-4 & -4 &1 \\-3& 2 & 1\\ 3& -16 & 1\end{pmatrix}\) is \(\frac{1}{2}\){(-4)(2+16) – (-4)(-3-3) + (1)(48-6)} = \(\frac{1}{2}\)|(-72)+(-24)+42| = 27 square units.

We know, slope of line joining two points (x₁, y₁) and (x₂, y₂) is given by(y₂-y₁)/(x₂-x₁).
Lines are perpendicular means m₁*m₂ = -1
=> \((\frac{x-4}{11-3})(\frac{6-2}{7-1})\) = -1
=> (x-4)(4) = (-1)(8)(6)
=> x-4 = -12 => x= -16.

Let general equation of line be y=m*x + c.
Given m=3 => y=3x+c
Substituting point (1, 2) in above equation we get 2=3*1 + c => c = -1
So, equation of line will be y = 3x-1.

Midpoint of QR line is \((\frac{3+7}{2}, \frac{5+9}{2})\) = (5, 7).
Equation of line joining (1, 2) and (5, 7) is \(\frac{y-2}{7-2} = \frac{x-1}{5-1}\)
=>\(\frac{y-2}{5} = \frac{x-1}{4}\) => 4y-8 = 5x-5 => 5x-4y+3 =0.

If x-intercept of a line is a and y-intercept of line is b so, equation of line is \(\frac{x}{a} + \frac{y}{b}\) = 1.
Equation of line is \(\frac{x}{4} + \frac{y}{2}\) = 1 => x+2y-4=0.

If a line is parallel to y-axis n angle formed by it with x-axis is 90°. So, its inclination is 90°.

We know, distance between two points (x₁, y₁) and (x₂, y₂) is \(\sqrt{(x_1-x_2)^2+(y_1-y_2)^2}\).
So, distance between (5, 12) from origin (0, 0) is \(\sqrt{(5-0)^2+(12-0)^2} = \sqrt{(5)^2+(12)^2}\) = 13 units.

If two lines are perpendicular n if one line form angle α with positive x-axis n or line form angle 90° + α.
If m₁ = tan α n m₂ will be tan (90°+ α) = – cot α = -1/tan α
=> m₁ * m₂ = – 1.