IPMAT Indore 2025 (MCQ) - The area of the triangle, formed by the straight lines y = 0, 12x - 5y = 0, and 3x + 4y = 7 is | PYQs + Solutions

IPMAT Indore 2025

Geometry

Straight Lines

Medium

The area of the triangle, formed by the straight lines $y = 0, 12x - 5y = 0,$ and $3x + 4y = 7$ is

\frac{35}{27}

\frac{14}{9}

\frac{28}{9}

\frac{35}{54}

✅ Correct Option: 2

- Line 1:

y = 0

(the x-axis)

- Line 2:

12x - 5y = 0

y = \dfrac{12x}{5}

- Line 3:

3x + 4y = 7

y = \dfrac{7-3x}{4}

Find intersection points:

Point A (lines 1 and 2):

- When

y = 0

12x = 0 \implies x = 0

So point A is

(0,0)

Point B (lines 1 and 3):

- When

y = 0

3x = 7 \implies x = \frac{7}{3}

So point B is

(\frac{7}{3},0)

- Point C (lines 2 and 3):

\frac{12x}{5} = \frac{7-3x}{4}

\frac{48x}{5} = 7-3x

48x + 15x = 35

63x = 35

x = \dfrac{35}{63} = \dfrac{5}{9}

Substitute into line 2:

y = \dfrac{12 \cdot \frac{5}{9}}{5} = \dfrac{4}{3}

So point C is

(\frac{5}{9}, \frac{4}{3})

Calculate the area using the formula for triangle area given vertices:

Area

=\dfrac{1}{2}|x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|

=\dfrac{1}{2}|0 \cdot (0-\dfrac{4}{3}) + \dfrac{7}{3} \cdot \dfrac{4}{3} + \dfrac{5}{9} \cdot 0|

=\dfrac{1}{2} \cdot \dfrac{7 \cdot 4}{3 \cdot 3}

=\dfrac{1}{2} \cdot \dfrac{28}{9}

=\dfrac{14}{9}

square units

Alternatively, using

\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}

Base along x-axis

=\frac{7}{3}

Height to point C

=\frac{4}{3}

Area

=\dfrac{1}{2} \cdot \dfrac{7}{3} \cdot \dfrac{4}{3} = \dfrac{14}{9}

square units