Question 31


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Jambmaths question: 

The locus of a point equidistant from the intersection of lines $3x-7y+7=0$and $4x-6y+1=0$ is a

Option A: 

line parallel to $7x+13y+8=0$

Option B: 


Option C: 


Option D: 

bisector of the line  $7x+13y+8=0$

Jamb Maths Solution: 

You don’t need to crack your head, to solve the question.

The correct option is option B (Circle)

Now let me explain why

Where the two lines 3x –7y +7 = 0 and 4x –6y +1 = 0 intersect will be the centre of the circle which has radius r. The radius always equidistant to the circumference of the circle. I believe this explanation will drive home the point I am making.

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