problem related to the slope of a line.

Question

What is the slope of the line given by $\sqrt{x^2+4y^2-4xy+4} + x-2y=1$ . Not getting any start . Only observed we have $(x-2y)^2$ under the root . NOTE: root gets over after 4 so please dont misinterpret. How to proceed any clue would do. Thanks!

Answer 1

$$\sqrt{x^2 + 4y^2 - 4xy + 4} = 1 - (x - 2y)$$ $$\sqrt{(x - 2y)^2 + 4} = 1 - (x - 2y)$$

Squaring both sides:

$$(x - 2y)^2 + 4 = 1 - 2(x - 2y) + (x - 2y)^2$$

Collecting like terms and simplifying, we obtain:

$$2(x - 2y) = -3$$ $$2x - 4y = -3$$

Writing the equation of the line in slope-intercept form, we have:

$$y = \frac{1}{2}x + \frac{3}{4}.$$

Hence the slope of the line is ?.