KarunMendiratta wrote:
Up for GRE Quant Challenge!!
A line k has a positive slope m and it passes through the point (-4, 6). If the area of the triangle formed by line k, x-axis and y-axis is 54 square units. What is the possible value for \(m\)?
A. 0.25
B. 0.75
C. 1.25
D. 1.75
E. Cannot be determined from the given information
Explanation:Drop a perpendicular from D(-4, 6) to x-axis (as shown in the figure)
Attachment:
If the area of the triangle formed by line.png
AC = \(x\)
CO = 4
CD = OE = 6
BE = \(y\)
△ACD ~ △AOE
i.e. \(\frac{x}{(x+4)} = \frac{6}{(6+y)} = \sqrt{\frac{Ar. (△ACD)}{ Ar. (△AOE)}} = \sqrt{\frac{3x}{54}}\)
Solve, \(\frac{x}{(x+4)} = \sqrt{\frac{3x}{54}} = \sqrt{\frac{x}{18}}\)
Squaring both sides;
\(\frac{x^2}{(x+4)^2} = \frac{x}{18}\)
\(\frac{x}{(x+4)^2} = \frac{1}{18}\)
\(18x = x^2 + 8x + 16\)
\(x^2 - 10x + 16 = 0\)
\(x^2 - 8x - 2x + 16 = 0\)
\(x(x-8) - 2(x-8) = 0\)
\((x-2)(x-8) = 0\)
\(x = 2\) or \(8\)
Taking \(x=8\), and solve for \(y\);
\(\frac{8}{(8+4)} = \frac{6}{(6+y)}\)
\(48+8y = 72 + 12y\)
\(4y = 36\)
\(y=9\)
Therefore, A is (-12, 0) and B(0, 9)
Apply slope formulae;
\(m = \frac{(9-0)}{(0+12)} = \frac{9}{12} = \frac{3}{4}\)
Hence, option B
[NOTE: if you take \(x=2\), you get \(m=3\) - which is not in the option choices]but sir how should one come up solution on time or writing exam.....
any suggestion to develop skill to solve problem just like in given question....