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If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 00:16
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Competition Mode Question If the lines \(x2y3=0\), \(x+3y3=0\) and \(2x+y1=0\) form a right angled triangle, then the vertex containing the right angle is A. (3,0) B. (0,1) C. (1,1) D. (1,1) E. (1,1) Are You Up For the Challenge: 700 Level Questions
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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 00:41
For two lines to be at right angles, their slopes should be in inverse and sign should be opposite. (Product of slopes of perpendicular lines = 1) Line 1 and Line 3 are having slopes inverse and opposite. Substitute the options in line 1 and line 3



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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 02:09
to find angle b/w x−2y−3=0, x+3y−3=0and 2x+y−1=0 we have to find tanθ = (a1b2  b1a2)/(a1a2 + b1b2)
tan90=undefined.......for that value of (a1b2  b1a2)/(a1a2 + b1b2) have to be undefined
i.e (a1a2 + b1b2) value must be zero that value is zero for the lines x−2y−3=0and 2x+y−1=0.....POI b/w them is (1,1)
OA:C



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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 02:13
Ans is C.
Among 3 equations, find pair of equations that are perpendicular. i.e., product of their slopes should be 1. Line1 has slope 1/2 and Line3 has slope 2
Now just substitute each point into Eq1&3. Option that satisfies both Eq1 and Eq3 but not Eq2 is the correct answer.



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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 02:26
If the lines x−2y−3=0, x+3y−3=0 and 2x+y−1=0 form a right angled triangle, then the vertex containing the right angle is A. (3,0) B. (0,1) C. (1,1) D. (1,1) E. (1,1) Any of the two lines among the three an intersect each other perpendicularly. For that it is to be checked whether product of their slopes is 1. Now, in the form y = mx + c the lines can be written as \(y = \frac{x}{2}  \frac{3}{2}\) \(y = \frac{x}{3} + 1\) \(y = 2x + 1\) Clearly, the first and third lines intersect each other perpendicularly since they have slopes 1/2 and 2 respectively.So, wherever they intersect is the required point. Now solving first and third line equations gives x = 1 and y = 1. IMO Answer C.
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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 04:49
We are given the equation for three lines: x2y3=0, implying y=x/23 .............(1) x+3y3=0, implying y=x/3+1 ..........(2) 2x+y1=0, implying y=2x+1 ...........(3)
The slope(m1) of Line 1 =1/2 The slope(m2) of Line 2 =1/3 The slope(m3) of Line 3 =2
Two lines are perpendicular if the product of their slopes equals 1 In other words, m1*m2=1 or m1 is the negative reciprocal of m2. Looking at the slopes of three lines, we can tell that Line 1(m=1/2) and Line 3(m=2) are perpendicular to each other. In order to determine the points of intercept, we need to equate the two lines and solve for the x and y.
hence, 2x+1=1/2x3/2, x=1 y=1/23/2=1 Hence the vertex of the right angle is (1,1)
The answer is, therefore, option C.



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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 04:57
Quote: If the lines x−2y−3=0, x+3y−3=0 and 2x+y−1=0 form a right angled triangle, then the vertex containing the right angle is
A. (3,0) B. (0,1) C. (1,1) D. (1,1) E. (1,1) The vertex containing the rightangle is point where two lines meet at right angles, or the perpendicular bisector; Thus, we need to find a slope of a line that is the negative reciprocal of the slope of another line; Finally, we equate both lines to find the point they intersect. [1] 2y=x3…y=x/23/2…slope: 1/2…negative reciprocal: (1/1/2)=2=[line: 3] [2] 3y=x+3…y=x/3+1…slope: 1/3…negative reciprocal: (1/1/3)=3 [3] y=2x+1…slope: 2…negative reciprocal: (1/2)=1/2=[line: 1] [equate] y=x/23/2 and y=2x+1: (x3)/2=2x+1, x3=4x+2…5x=5…x=1 y=2x+1…y=2(1)+1…y=1 point(x,y)=(1,1) Ans (C)



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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 14:34
\(x —2y—3= 0 —> y= x/2–1.5\)
\(x+ 3y—3= 0 —> y= —x/3 —1\)
\(2x+ y —1= 0 —> y= —2x+ 1\)
—> the question asks the vertex containing the right angle.
—>In order to be perpendicular to each other, the product of the slopes of the lines must be equal to —1. y = x/2–1.5 y= —2x+1 —> these two lines are perpendicular to each other and they meet: —> x/2–1.5= —2x+1 x = 1 and y = —1 (1,—1) The answer is C
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Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 20:32
C, The first step is to find the pair of lines contains the right angle. To do so, we make use of a equation that line 1 and line 2 is perpendicular to each other if (slope of line 1)(slope of line 2) = 1 by converting the 3 equations to the form of y = mx + c where m is the slope. y = (1/2) x + 1/2 and y = 2x +1 since (1/2) * (2) = 1, solve the x  2y  3 = 0 and 2x + y  1 = 0 and that is the vertex of the right angle.
Hope this explanation is understandable.




Re: If the lines x  2y  3 = 0, x + 3y  3 = 0 and 2x + y  1 = 0 form
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21 Nov 2019, 20:32






