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!
Asked
Active
Viewed 30 times
0
-
What makes you think it is a line ? – Shailesh Nov 04 '15 at 09:29
-
@ shailesh sir we have x-2y square under root and general expression of a line is ax+by+c . So basically it is a line. – Archis Welankar Nov 04 '15 at 09:36
-
@Claude can you explain it bit further. – Archis Welankar Nov 04 '15 at 09:38
1 Answers
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 ?.
