IPMAT Indore 2025 · Quantitative Ability · Question 30

Step 1 — find the three vertices by taking the lines two at a time.

y=0 with 12x-5y=0: putting y=0 gives 12x=0, so {V}_{1}=\left( 0,0 \right)

y=0 with 3x+4y=7: putting y=0 gives 3x=7, so {V}_{2}=\left( 7/3,0 \right)

12x-5y=0 with 3x+4y=7: from the first, y=12x/5. Substituting:

3x+4\cdot \frac{12x}{5}=7 \Rightarrow \frac{15x+48x}{5}=7 \Rightarrow 63x=35 \Rightarrow x=\frac{5}{9}, y=\frac{4}{3}

{V}_{3}=\left( 5/9, 4/3 \right)

Step 2 — use the easy base. Two vertices lie on y=0, so take that as the base:

\text{base}=\frac{7}{3}-0=\frac{7}{3}

The height is just the y-coordinate of the third vertex:

\text{height}=\frac{4}{3}

Step 3 — area.

\frac{1}{2}\times \frac{7}{3}\times \frac{4}{3}=\frac{28}{18}=\frac{14}{9}

Answer: Option 2 — \frac{14}{9}