a) B(7,12) ; D(2,-3)
The equation of BD is a linear one of the form y =mx+b, where m is the slope and b, the y-intercept.
1st) calculate the sole m =(y₂-y₁)/(x₂-x₁) = (-3-12)/(2-7) = (-15)/(-5) = +3
y(AB) = 3x + b. To calculate b, plug any coordinates values of B or D.
Let it be B
12 = 3(7)+b →12 = 21 +b and b= -9
So y(BD) = 3x - 9
b) x + 3y = 23 → 3y = 23-x → y = -(1/3)x + 23/3
So y(AC) = -(1/3)x + 23/3
To find the coordinates of the intersection points of BD and AC, let's make
y(BD) = y(AC)
3x-9 = (-1/3)x + 23/3. Solving this equation gives x = 5
Replace x = 12 in any of te 2 equations → 6
Then coordinate of the intersection, = E(5,6)
