Question

The triangle formed by the lines \(2x^2-3xy-2y^2 = 0\) and \(3x-y = 7\) is...

• A. Right angled but not isosceles
• B. isosceles with base angle \(30^{\circ}\)
• C. Right angled with one angle \(60^{\circ}\)
• D. Right angled isosceles

Hint:

Check the angle between the two lines and compare the two legs.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept:
We find the three vertices of the triangle, then check the side lengths and the angles.

Step 2: Split the pair of lines:
\(2x^2-3xy-2y^2=(2x+y)(x-2y)=0\). So the lines are \(y=-2x\) and \(y=\tfrac{x}{2}\). The product of the slopes is \(-2\times\tfrac12=-1\), so these two lines are perpendicular. The angle at the origin is \(90^\circ\).

Step 3: Vertices:
With \(3x-y=7\), i.e. \(y=3x-7\):
With \(y=-2x\): \(3x-7=-2x\), so \(x=\tfrac75\), \(y=-\tfrac{14}5\). Point \(P\left(\tfrac75,-\tfrac{14}5\right)\).
With \(y=\tfrac{x}{2}\): \(3x-7=\tfrac x2\), so \(x=\tfrac{14}5\), \(y=\tfrac75\). Point \(Q\left(\tfrac{14}5,\tfrac75\right)\).

Step 4: Lengths:
\(OP^2=\tfrac{49}{25}+\tfrac{196}{25}=\tfrac{245}{25}\), and \(OQ^2=\tfrac{196}{25}+\tfrac{49}{25}=\tfrac{245}{25}\). So \(OP=OQ\).

Step 5: Conclusion:
The triangle has a right angle at \(O\) and two equal legs, so it is right angled isosceles, option (D).

Final Answer:
The triangle is right angled and isosceles. \[ \boxed{\text{Right angled isosceles (D)}} \]