Question

the straight line 3 X + 4 Y = 5 and 4 X- 3 Y = 15 intersect at a point A on these lines the points B and C are chosen so that AB= AC then find the possible equation of the line BC passing through the point (1,2).

Answer 1

first solve the given equation and find the value of X AND Y then the given pt and u find that pt u have 

Answer 2

Answer:

Step-by-step explanation:

Given that the two lines intersect at a point it means they share a point.

We therefore need to get that common point.

We get the common point by solving the two equations simultaneously.

SOLUTION :

3X + 4Y = 5........1)

4X - 3Y = 15.........2)

We multiply equation 1 by 4 and equation 2 by 3 then subtract 1 from 2.

-25Y = 25

Y = - 1

Substitute in equation 1.

3X - 4 = 5

3X = 9

X = 3

The common point is :

(3, - 1)

The gradient of BC is :

(-1 - 2) / (3 - 1) = - 3/2 = - 1.5

Equation of BC :

(Y + 1 ) / (X - 3) = - 3/2

2(Y + 1) = - 3(x - 3)

2Y + 2 = - 3X + 9

2Y = - 3X + 9 - 2

2Y = - 3X + 7

Y = - 3/2X + 7/2