Question

FIND THE EQUATION OF LINE PASSING THROUGH THE POINT OF INTERSECTION OF THE LINE X-Y-1=0 AND 2X-3Y+1=0AND PARALLEL TO THE LINE 3X+4Y-13=0?




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Answer 1

Hi !

Solution is in this given attachment !

• Step 1:- Find the point of intersection.
• step :-2 put the value of X and y in the given parallel equation.
• step :- 3 ,let the value of constant term as k
• step :- 4 find value of k putting x,and y in the parallel given equation .
• step :- 5 then put the value of k in parallel given equation.

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Hope it's helpful

Answer 2

Answer:

y=(-3/4)(x)+5/2

Step-by-step explanation:

solve  

x-y=1 and 2x-3y=-1

multiply x-y=1 by 3

3x-3y=3----(1)

2x-3y=1----(2)

subtract (1) and (2) to get

x=4

substitute x=1 in x-y=1 to get  

y=3

point of intersection (x=4,y=3)

Two lines are parallel if their slopes are equal.

Slope of 3x+4y-13=0  

4y = -3x+13

y=(-3/4)x + 13/4  (y=mx+c)

Slope of line -3/4

Hence equation  of line will

y-y1 = m(x-x1) --- substitute point of intersection (x=4,y=3 m=(-3/4))

y-3= (-3/4)(x-4)

y=(-3/4)(x)+(-3/4)(-4)+3

y=(-3/4)(x)+6