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The two perpendicular lines L and R intersect at the point [#permalink]

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13 Nov 2013, 01:20

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The two perpendicular lines L and R intersect at the point (3, 4) on the coordinate plane to form a triangle whose other two vertices rest on the x-axis. If one of the other two vertices is located at the origin, what is the x-coordinate of the other vertex?

The two perpendicular lines L and R intersect at the point (3, 4) on the coordinate plane to form a triangle whose other two vertices rest on the x-axis. If one of the other two vertices is located at the origin, what is the x-coordinate of the other vertex?

A. 11/8 B. 33/8 C. 5 D. 20/3 E. 25/3

Look at the diagram below:

Attachment:

Untitled.png [ 6.93 KiB | Viewed 3011 times ]

For a line that crosses two points \((x_1,y_1)\) and \((x_2,y_2)\), slope \(m=\frac{y_2-y_1}{x_2-x_1}\)

Thus the slope of line L passing (0, 0) and (3, 4) is \(m=\frac{y_2-y_1}{x_2-x_1}=\frac{4-0}{3-0}=\frac{4}{3}\).

The slope of line R, which is perpendicular to L, is negative reciprocal of the slope of L, hence its \(-\frac{3}{4}\).

Now, we need to find equation of R.

The equation of a straight line that passes through a point \(P_1(x_1, y_1)\) with a slope m is: \(y-y_1=m(x-x_1)\). Therefore, the equation of R is \(y-4=-\frac{3}{4}(x-3)\) --> \(y=-\frac{3}{4}x+\frac{25}{4}\). The x-intercept of this line is 25/3.

Re: The two perpendicular lines L and R intersect at the point [#permalink]

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16 Nov 2013, 12:55

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Hi honchos...you could even do this :- as bunuel has calculated, the slope of the line will be -3/4. Now suppose the co-ordinates of the point needed are (a,0), just equate the slopes. You will get the equation -3/4 = 4/(3-a) solving this you get the OA 25/3 i.e. E. Kudos me!
_________________

Re: The two perpendicular lines L and R intersect at the point [#permalink]

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22 Aug 2015, 07:07

honchos wrote:

The two perpendicular lines L and R intersect at the point (3, 4) on the coordinate plane to form a triangle whose other two vertices rest on the x-axis. If one of the other two vertices is located at the origin, what is the x-coordinate of the other vertex?

A. 11/8 B. 33/8 C. 5 D. 20/3 E. 25/3

I think this can be solved using similar triangle properties (right angled) Draw a perpendicular line from the top vertex to x-axis which intersects the axis at (3, 0) You have one right angled triangle with sides 3, 4, 5 Use the properties to get the other sides.

Re: The two perpendicular lines L and R intersect at the point [#permalink]

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21 Oct 2016, 09:00

Or you can use the distances: To find the l side you can simply find the distance from (3,4) to (0,0) which is sqr[ (3-0)^2 +(4-0)^2 ] = 5. The R side is the distance from (3,4) to (x,0) which is : sqr[ (x-3)^2 + (4-0)^2 ] Finally the other side is sqr[ (x-0)^2 +(0-0)^2 ]=x Using the property of pythagorean theorem (the triagle is 90 degrees) we find that : x^2=5^2+ {sqr[(x-3)^2 +16]}^2=> 6x=50=>x=25/3. Answer is E

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

Re: The two perpendicular lines L and R intersect at the point [#permalink]

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23 Oct 2017, 23:39

honchos wrote:

The two perpendicular lines L and R intersect at the point (3, 4) on the coordinate plane to form a triangle whose other two vertices rest on the x-axis. If one of the other two vertices is located at the origin, what is the x-coordinate of the other vertex?

A. 11/8 B. 33/8 C. 5 D. 20/3 E. 25/3

Another approach to this problem is By property of right angled triangle BD^2 = AD * DC 4^2 = 3 * DC DC = 16/3

Concentration: General Management, Entrepreneurship

GPA: 3.8

WE: Engineering (Energy and Utilities)

Re: The two perpendicular lines L and R intersect at the point [#permalink]

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23 Oct 2017, 23:43

honchos wrote:

The two perpendicular lines L and R intersect at the point (3, 4) on the coordinate plane to form a triangle whose other two vertices rest on the x-axis. If one of the other two vertices is located at the origin, what is the x-coordinate of the other vertex?

A. 11/8 B. 33/8 C. 5 D. 20/3 E. 25/3

It can be solved by using slope of a line and slope of perpendicular lines approach: Slope of AB = 4/3 Slope of AC = -3/4 = (4-0)/(3-x) -9+3x = 16 x = 25/3