MHTCET 15 Apr 2026 Shift 1 Paper

Concept:A homogeneous quadratic equation represents two straight lines passing through the origin, and factorising it gives their individual equations.Explanation:Factorise the given pair of lines.$$2x^{2}-3xy-2y^{2}=0 \Rightarrow (2x+y)(x-2y)=0$$Hence, the two lines are $2x+y=0$ and $x-2y=0$.So, $y=-2x$ and $y=\frac{x}{2}$.Their slopes are $-2$ and $\frac{1}{2}$.Since $(-2)\times \frac{1}{2} = -1$, the two lines are perpendicular.Therefore, the triangle has a right angle at the origin.Now check whether the two perpendicular sides are equal.The third line is $3x-y=7$.Intersection with $y=-2x$:$$3x-(-2x)=7 \Rightarrow 5x=7 \Rightarrow x=\frac{7}{5}, \quad y=-\frac{14}{5}$$Distance from origin:$$\sqrt{\left(\frac{7}{5}\right)^{2}+\left(-\frac{14}{5}\right)^{2}}=\sqrt{\frac{49+196}{25}}=\frac{7\sqrt{5}}{5}$$Intersection with $y=\frac{x}{2}$:$$3x-\frac{x}{2}=7 \Rightarrow \frac{5x}{2}=7 \Rightarrow x=\frac{14}{5}, \quad y=\frac{7}{5}$$Distance from origin:$$\sqrt{\left(\frac{14}{5}\right)^{2}+\left(\frac{7}{5}\right)^{2}}=\sqrt{\frac{196+49}{25}}=\frac{7\sqrt{5}}{5}$$Both the perpendicular sides are equal in length.Thus, the triangle is right angled isosceles.Answer:Option D: Right angled isosceles.