Question

8) Solve graphically the pair of linear equations:
3x + 5y - 12 = 0 and 3x - 5y + 18 = 0, calculate the area of
triangle formed with the x-axis.​

Answer 1

NOTE - I USED GeoGebra TO DRAW THE LINES.

$\mathcal{\green{\underline{\blue{GIVEN:}}}}$

• We are given two equations : 3x + 5y - 12 = 0  and  3x - 5y + 18 = 0.

$\mathcal{\green{\underline{\blue{TO \:\: FIND:}}}}$

• The common solution or roots of both equations graphically.

• The area of the triangle formed by these two lines and the x - axis.

$\mathcal{\green{\underline{\blue{SOLUTION:}}}}$

Lets first draw the graphs.

The cyan coloured line is 3x + 5y - 12 = 0 and the blue coloured line is 3x - 5y + 18 = 0. Both of them intersect at (5,-0.6)

∴Roots are x=5, y= -0.6.

And the altitude/height of the triangle is AB, which is 5 cm. (see the first graph)

And the point C = (0,-3.6) and D = (0,2.4).

To find the distance, we use the distance formula:

$\boxed{\bold{\large{d= \sqrt{(x_1-x_2)^2+(y_1-y_2)^2} }}}$

$\bf \to d = \sqrt {(0 - 0)^2 + (2.4 - (-3.6))^2}$

$\bf \to d = \sqrt {(0)^2 + (6)^2}$

$\bf \to d=6cm.$

Thus, height = 6cm, base = 5cm.

$\boxed{\bold{\large{Area = \frac{1}{2} \times base \times height}}}$

$\bf Area=\frac{1}{2} \times 5 \times 6$

$\bf \to Area = 15 cm^2.$

$\huge\star\:\:{\orange{\underline{\pink{\mathbf{HOPE \:\: THIS \:\: HELPS \:\: :D}}}}}$