Ques. -20

Geometry Level 4

Given the four lines with the equations ( x + 2 y 3 = 0 ) (x+2y-3=0) , ( 3 x + 4 y 7 = 0 ) (3x+4y-7=0) , ( 2 x + 3 y 4 = 0 ) (2x+3y-4=0) , ( 4 x + 5 y 6 = 0 ) (4x+5y-6=0) , then :


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They are all concurrent. They form a cyclic quadrilateral. They are the sides of a non-cyclic quadrilateral. None of the given is true.

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2 solutions

Thomas Lee
May 30, 2015

I personally like to rewrite the equations into y = m x + b y=mx+b and when you do that, you get

y = 1 / 2 x + 3 / 2 y=-1/2x+3/2

y = 3 / 4 + 7 / 4 y=-3/4+7/4

y = 2 / 3 x + 4 / 3 y=-2/3x+4/3

y = 4 / 5 x + 6 / 5 y=-4/5x+6/5

I did a little sketch of them and found that they do not form a quadrilateral nor do they meet at a single point; therefore, the answer is None of the given is true

PS: A picture to help visualize the problem .

Umm, i found out a point of intersection, and saw all of them didnt satisfied which eliminates that concurrent one. Then I saw that if opposite angles had slope -ve of each other as tan(180-A)=-tan A. It also didnt happened. Then i checked for their points of intersection, they didnt even made a quadrilateral( I should have checked this first, then i shouldnt have to check for cyclic quad) and so the rem. one is None!

Md Zuhair - 3 years, 2 months ago

All have -tive slope between -1/2 and -4/5. There is no pattern either in slope or y-intercept. Sketching them shows first three line intersecting near (-1,2) and last one
passing them. Since three of them seem to intersect at a point, or very near one another, none of the other options is possible.

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