Please answer and send graph along with it

Dear student,
The question requires knowledge of class 10 co-ordinate geometry where area of triangle is discussed. Still providing the solution with detailed explanation.
Put y=0 in both eqn.
  eqn.1: 5x-3*0=-10     => 5x=-10     =>x=-2      co-ordinate: (-2,0)
  eqn. 2:5x+2*0=15     =>5x=15     =>x=3      co-ordinate: (3,0)
for intersection point:
eqn.1 - eqn.2
=>5x-3y+10-(5x+2y-15)=0
=> -3y+10-2y+15=0    =>5y=25     =>y=5
put y=5 in eqn. 1:
5x-15+10=0                =>x=1
co-ordinate: (1,5)

Area of triangle=
 
$$\begin{gathered} x_1=-2; y_1=0 \\ {x}_2=3; y_2=0 \\ {x}_3=1; y_3=5 \\ =\frac{1}{2}\left|0-0+15-0+0-(-10)\right| \\ =\frac{1}{2}*25=12.5 sq unit \end{gathered}$$
Regards