Question

Find the equation of line passing through the point (2,1) and parallel to the line 2x+3y=4.​

Answer 1

Answer:

Step-by-step explanation:

x=1 and y=2

is answer

Answer 2

$\sf\large\red{\underline{\underline{Question}}}:$

• Find the equation of line passing through the point (2,1) and parallel to the line 2x+3y=4.​

$\sf\large\red{\underline{\underline{Solution}}}:$

First of all we will find the slope of line 2x + 3y = 4 by placing it in slope intercept form.

Slope intercept form :

$\red\bigstar\sf\boxed{\bf y = mx+c}$

Where,

• m is slope of the line .

$\longmapsto \sf 2x+3y=4$

$\dashrightarrow \sf3y=4-2x\\ \\\dashrightarrow \sf 3y=-2x+4\\\\\dashrightarrow \sf y=\dfrac{-2}{3}x+4$

So, here we got the slope of line 2x+3y=4x is -2/3.

Since we know that both lines are parallel to each other, there slope will be same . So, slope of the line passing through point (2,1) is -2/3.

Now, we know that equation of a line passing through point is : y -y₁ = m(x-x₁)

here, (x₁,y₁) is point form which line is passing.

So, Equation of line passing through the point (2,1) is :

$\longmapsto \sf y-1=\dfrac{-2}{3}(x-2)$

$\dashrightarrow\sf 3y-3=-2x+4 \\\\\dashrightarrow \sf 2x+3y-7=0$

Therefore,

• Equation of line passing through the point (2,1) and parallel to the line 2x+3y=4 is 2x+3y-7=0 .