Question

find the point of intersection of the lines 2x+y-3=0 , 3x-y-2=0 anyone knoe this question plz ans fast plzzzz

Answer 1

point the values of x And y by different equations on cartesion plane. Where the lines will interesect. that will be the point

Answer 2

Answer:

(1,1) is the required  intersection point of 2x +y -3 = 0 and 3x -y -2 = 0 .

Step-by-step explanation:

Explanation:

Given , two line  2x+ y -3 = 0   and  3x -y -2 = 0 .

Intersection point - In a plane, intersecting lines are any two or more lines that cross one another. The point of intersection, which can be found on all intersecting lines, is where the intersecting lines share a common point.

Formula to find the intersecting points  ;

x = $\frac{b_1 c_2 -b_2c_1}{a_1b_2 - a_2b_1}$  and y =$\frac{a_2 c_1 -a_1c_2}{a_1b_2 - a_2b_1}$

Step 1:

We have  2x + y -3 = 0 ........(i) and

3x -y -2 = 0 .........(ii)

$a_1 = 2 , a_2 = 3 , b_1 = 1, b_2 = -1 ,c_1 = -3 and c_2 = -2$

Therefore , x = $\frac{b_1 c_2 -b_2c_1}{a_1b_2 - a_2b_1}$ and y = $\frac{a_2 c_1 -a_1c_2}{a_1b_2 - a_2b_1}$ .

⇒ x = $\frac{1(-2) - (-1)(-3)}{2(-1) - 3(1)}$ = $\frac{-5}{-5} = 1$

and y = $\frac{3(-3) - 2(-2)}{2 (-1) - 3 (1)}$ = $\frac{-5}{-5} = 1$

(x , y ) = (1,1)

Final answer:

Hence , (1,1) is the point of intersection of the line 2x +y -3 = 0 and

3x -y -2 = 0 .

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