Find the single equation of a pair of lines which pass through the point (0, 0) perpendicular to the lines represented by the equation x ^ 2 - 7xy + 12y ^ 2 = 0

 Solutions 

The given equation can be factored as follows: ${𝑥}^2-7𝑥𝑦+12{𝑦}^2=(𝑥-3𝑦)(𝑥-4𝑦)=0$

This equation represents two lines:

1. Line 1: $𝑥-3𝑦=0$ or $𝑦=\frac{1}{3}𝑥$
2. Line 2: $𝑥-4𝑦=0$ or $𝑦=\frac{1}{4}𝑥$

Now, we need to find the equations of the lines passing through the point (0, 0) and perpendicular to these lines.

1. Line 1: Slope = $\frac{1}{3}$
2. Line 2: Slope = $\frac{1}{4}$

The slopes of perpendicular lines are negative reciprocals of these slopes:

1. Perpendicular to Line 1: Slope = $-\frac{1}{\frac{1}{3}}=-3$
2. Perpendicular to Line 2: Slope = $-\frac{1}{\frac{1}{4}}=-4$

Using the point-slope form of the equation of a line, the equations of the perpendicular lines passing through (0, 0) are:

1. Perpendicular to Line 1: $𝑦-0=-3(𝑥-0)$ which simplifies to $𝑦=-3𝑥$

1. Perpendicular to Line 2: $𝑦-0=-4(𝑥-0)$ which simplifies to $𝑦=-4𝑥$

Now, combining these equations into a single equation, we get: $𝑦=-3𝑥\pm -4𝑥$ $𝑦=-7𝑥$

Therefore, the single equation of a pair of lines passing through the point (0, 0) and perpendicular to the lines represented by ${𝑥}^2-7𝑥𝑦+12{𝑦}^2=0$ is $𝑦=-7𝑥$, as derived using class 10 coordinate geometry concepts.