Question

The point of intersection of the lines x-2y = 3 and 4x+3y = 1 is *​

Answer 1

POINT OF INTERSECTION

Answer (1, -1)

Given linear equation in two variables are,

⇒ x - 2y = 3 ...(1)

⇒ 4x + 3y = 1 ...(2)

The point at which the two lines will be intersected is the solution of these two linear equations.

First, Let us make sure that these two lines will be intersected at a unique point.

We know, If the lines are intersected at a unique point then the given condition evaluates true.

⇒ a1 / a2 ≠ b1 / b2

⇒ 1 / 4 ≠ -2/3

Which is true.

Here,

• a1 = Coefficient of x of (1)
• a2 = Coefficient of x of (2)
• b1 = Coefficient of y of (1)
• b2 = Coefficient of y of (2)

Multiply (1) by 4 , to make the coefficient of x make same in both the equations.

⇒ 4(x - 2y) = 4×3

⇒ 4x - 8y = 12 ...(3)

Now, Subtract (2) from (3), we get

⇒ 4x - 8y - (4x + 3y) = 12 - 1

⇒ 4x - 8y - 4x - 3y = 11

⇒ -11y = 11

⇒ y = -1

We got the ordinate of the intersection point. Substitute [y = -1] in (1) to get the Abscissa as well.

⇒ x - 2(-1) = 3

⇒ x + 2 = 3

⇒ x = 1

Therefore, The point at which these two lines would be intersected is (1, -1).

Some Information :-

☛ Two lines will be conincide when the given condition evaluates true.

a1 / a2 = b1 / b2 = c1 / c2

☛ Two lines will be parallel to each other when the given condition evaluates true.

a1 / a2 = b1 / b2 ≠ c1 / c2

Where,

• a1 , a2 are the coefficients of x of both the equations respectively.
• b1 , b2 are the coefficients of y of both the equations respectively.

Answer 2

Given ,

The two equations of the straight lines are

• x - 2y - 3 = 0 --- (i)
• 4x + 3y - 1 = 0 --- (ii)

Multiply eq (i) by 4 , we get

4x - 8y - 12 = 0 --- (iii)

Subtract eq (ii) from eq (iii) , we get

4x - 8y - (4x + 3y) = 12 - 1

-11y = 11

y = -11/11

y = -1

Put the value of y = -1 in eq (i) , we get

x - 2(-1) - 3 = 0

x + 2 = 3

x = 1

$\sf \therefore \underline{{The \: point \: of \: intersection \: of \: two \: given \: lines \: is \: (1,-1)}}$