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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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Joined: 02 Sep 2009
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If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form  [#permalink]

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21 Nov 2019, 00:16
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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)

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Re: If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form  [#permalink]

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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  [#permalink]

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21 Nov 2019, 02:09
1
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  [#permalink]

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21 Nov 2019, 02:13
2
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  [#permalink]

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21 Nov 2019, 02:26
1
1
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.

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Re: If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form  [#permalink]

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21 Nov 2019, 04:49
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We are given the equation for three lines:
x-2y-3=0, implying y=x/2-3 .............(1)
x+3y-3=0, implying y=-x/3+1 ..........(2)
2x+y-1=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=1/2x-3/2, x=1
y=1/2-3/2=-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  [#permalink]

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21 Nov 2019, 04:57
1
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 right-angle 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=x-3…y=x/2-3/2…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/2-3/2 and y=-2x+1:
(x-3)/2=-2x+1, x-3=-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  [#permalink]

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21 Nov 2019, 14:34
1
$$x —2y—3= 0 —> y= x/2–1.5$$

$$x+ 3y—3= 0 —> y= —x/3 —1$$

$$2x+ y —1= 0 —> y= —2x+ 1$$

—> 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)

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Re: If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form  [#permalink]

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21 Nov 2019, 20:32
1
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   [#permalink] 21 Nov 2019, 20:32
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