Question

PLEASE!!!!!!!! Solve it fast...The straight lines y=m1x+c1 , y=m2x+c2 , and x=0 intersect in the three points P,Q, snd R.Find the area of the triangle PQR.What is the value of the area if c1=c2????Please solve it in detail.The answer is {(c1-c2)^2}/{2(m1-m2)}

Answer 1

hope it helps....... :)

Answer 2

Solution:

Given

y = m1x+c1 ---(1)

y = m2x+c2 ----(2)

x = 0 ------------(3)

substitute x = 0 in equations (1)&(2) , we get

case1:

if x = 0 => y = c1

Let P(x,y) = (0,c1) -----(3)

case 2:

if x = 0 => y = c2

Let Q(x,y) = (0,c2)----(4)

We know that ,

(1) = (2)

=> m1x+c1 = m2x+c2

=> m1x-m2x = c2 - c1

=> x(m1-m2) = c2-c1

=> x = (c2-c1)/(m1-m2) ---(5)

substitute (5) in equation (1), we get

y = m1[(c2-c1)/(m1-m2)]+c1

=> y = [m1(c2-c1)+c1(m1-m2)]/(m1-m2)

=> y = (m1c2-m1c1+c1m1-c1m2)/(m1-m2)

=> y = (m1c2-c1m2)/(m1-m2)

Let R(x,y) = [(c2-c1)/(m1-m2),(m1c2-m2c1)/(m1-m2)]---(6)

Now,

Area of ∆PQR

= 1/2|x1(y2-y3)+x2(y3-y1)+x3(y1-y2)|

= 1/2|0+0+[(c2-c1)/(m1-m2)](c1-c2)|

= 1/2[(c1-c2)²/(m1-m2)|

= (c1-c2)²/(m1-m2)

Hence , proved

If c1 = c2 then three points are

collinear.

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