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Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A line k has a positive slope m and it passes through the point -4, 6
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06 Jun 2021, 11:32
06 Jun 2021, 11:32
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Question Stats:
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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
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
Retired Moderator
Joined: 16 Apr 2020
Status:Founder & Quant Trainer
Affiliations: Prepster Education
Posts: 1546
Given Kudos: 172
Location: India
WE:Education (Education)
A line k has a positive slope m and it passes through the point -4, 6
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07 Jun 2021, 06:45
07 Jun 2021, 06:45
1
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
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 [ 24.24 KiB | Viewed 1909 times ]
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]
General Discussion
Re: A line k has a positive slope m and it passes through the point -4, 6
[#permalink]
07 Jun 2021, 11:59
07 Jun 2021, 11:59
KarunMendiratta wrote:
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
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]
thank you for explanation,
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....
