If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form

Question

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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unraveled · Accepted Answer

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.

saik1993 · Answer

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

madgmat2019 · Answer

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

anvesh004 · Answer

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.

eakabuah · Answer

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.

exc4libur · Answer

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.

2y=x-3…y=x/2-3/2…slope: 1/2…negative reciprocal: -(1/1/2)=-2= 
  3y=-x+3…y=-x/3+1…slope: -1/3…negative reciprocal: -(1/-1/3)=3
  y=-2x+1…slope: -2…negative reciprocal: -(1/-2)=1/2=

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)

lacktutor · Answer

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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Ted2007 · Answer

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.

jcgomezlv · Answer

Can someone draw a graph? I still do not see it clearly

Abhishek009 · Answer

Capture.PNG 1. x-2y-3=0 Or, y = \frac{x}{2} - \frac{3}{2} 2. x+3y-3=0 Or, y = \frac{-x}{3} + 1 3. 2x+y-1=0 Or, y = -2x -1 Now, we find that lines (1) and (3) are perpendicular as product of their slopes is -1 Now, plug in the options given either in (1) or in (3) to mark the correct answer, hope this helps !!!

blonpina · Answer

I draw and it took me 04:24

If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form

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GMAT Problem Solving (PS) Questions

If the lines x - 2y - 3 = 0, x + 3y - 3 = 0 and 2x + y - 1 = 0 form

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