Respuesta :
The pair of equations that has (4, 8) as its solution are the two equations represented by the lines which intersect at point (4, 8). The lines are line b and line d.
Therefore the equations are equation b and equation a.
Therefore the equations are equation b and equation a.
Answer with explanation:
Eqation of line joining two points (a,b) and (c,d) is given by
[tex]\rightarrow \frac{y-b}{x-a}=\frac{b-d}{a-c}[/tex]
Eqation of line a, which joins two points (-6,16) and (9,-4) is
[tex]\rightarrow \frac{y-16}{x+6}=\frac{16-(-4)}{-6-9}\\\\\rightarrow -15 y+240=20 x+120\\\\20x+15y=120\\\\4x+3y=24 [/tex]
⇒Putting, x=4 and , y=8 in above equation
4 × 4+3×8=16+24=40≠24
Point (4,8) does not lie on this line.
⇒⇒Eqation of line b, which joins two points (2,20) and (8,0) is
[tex]\rightarrow \frac{y-20}{x-2}=\frac{20-0}{2-8}\\\\\rightarrow -6 y+120=20 x-40\\\\20x+6y=160\\\\10x+3y=80 [/tex]
⇒Putting, x=4 and , y=8 in above equation
10 × 4+3×8=40+24=64≠80
Point (4,8) does not lie on this line.
⇒⇒Eqation of line c, which joins two points (7,-6) and (6,20) is
[tex]\rightarrow \frac{y-20}{x-6}=\frac{20-(-6)}{6-7}\\\\\rightarrow -y+20=26 x-156\\\\26x+6y=176\\\\13x+3y=88 [/tex]
⇒Putting, x=4 and , y=8 in above equation
13× 4+3×8=52+24=76≠88
Point (4,8) does not lie on this line.
⇒⇒Eqation of line d, which joins two points (7,20) and (0,-7) is
[tex]\rightarrow \frac{y-20}{x-7}=\frac{20-(-7)}{7-0}\\\\\rightarrow 7y-140=27x-189\\\\27x-7y=189-140\\\\27x-7y=49 [/tex]
⇒Putting, x=4 and , y=8 in above equation
27× 4-7×8=108-56=52≠24
Point (4,8) does not lie on this line.
⇒None of the two lines has point of Intersection at point (4,8).