Mathematical Equations and Solutions

Ans1. (a)
The equation of the straight line passing through (x1,y1) and (x2,y2) is
y-y1=((y2-y1)/(x2-x1))(x-x1)
Now the equation of the straight line passing through (-4, 20) and ( -2.5 , 15.5 ) is:
y-20=((15.5-20)/(-2.5-(-4)))(x-(-4))
Or, y-20= ((-4.5)/(1.5))(x+4)
Or, y-20=-3x-12
Or, 3x+y-8=0
Now taking the first point i.e. x=-2 and y=14, we get,
L.H.S.= 3*(-2)+14-8=0=R.H.S.
Taking x=-1.5 and y=12.5, we get,
L.H.S.= 3*(-1.5)+12.5-8=0=R.H.S.
Taking x=-1 and y=11, we get,
L.H.S.=3*(-1)+11-8=0=R.H.S.
Hence we can see that the points, (-2,14), (-1.5,12.5) and (-1,11) are collinear.
(b)
Here, L=2x-5y-9=0 -> eqn.1
And L’ = 3x+y-8=0 -> eqn.2
To find the intersection,
3*eqn.1→6x-15y-27=0→eqn.3
2*eqn.2→6x+2y-16=0→eqn.4
Now, eqn4-eqn.3→17y-11=0
or, y=11/17
Therefore, x= (9+5y)/2=104/17
So, they intersect at (104/17, 11/17)
Ans2.

(a) Plotting of 4*x+ y= 5
Plotting of 14x+11y=25

Plotting of 2*x+3*y=5
(b) Let us solve 4x+y=5 and 2x+3y=5
4x+y=5→eqn1
2x+3y=5→eqn2
2*Eqn2→4X+6y=10→eqn3
Eqn3 – Eqn1 →
5y=5
Or, y=1
Therefore x=(5-3*1)/2=1 (From Eqn2)
Putting x=1 and y=1 in 14x+11y-25=0
we get,
L.H.S.= 14*1+11*1-25=0=R.H.S.
There is a unique solution to the system because the three straight lines intersect at a unique point.
Ans3.
The minimum fare is $6.20 and so it is the y-intercept in the graph where y-axes represents total fare
and x-intercept represents distance (in km.). In other words, we have $6.20 is the fare just to sit in
the vehicle. So we can say that the straight line intersects the y-axis at the point (0,6.20).
Again, the fare per km. is $5.25 per km.
Therefore, the equation of the straight line is