Question

Find the equation of the line passing through the point (1,-2),(-2,3)

Answer 1

Answer:

$5x+3y=-1$ is the required equation of the line.

Step-by-step explanation:

• Given two points $(1,2) and (-2,3)$ through which a line passes.

Equation of line passing through two points $(x_{1}, y_{1} )and( x_{2} ,y_{2} )$ is given by,

$\frac{y_{} -y_{1} }{x_{}- x_{1} } = m$    $\frac{y_{} -y_{2} }{x_{}- x_{2} } = m$

m is the slope of the line and m is given by ,

        $m = \frac{y_{2} -y_{1} }{x_{2}- x_{1} }$

Here $x_{1}= 1, y_{1} = -2 and x_{2} = -2,y_{2}= 3$

• so the slope of the required line is given by,

                   $\frac{y_{2} -y_{1} }{x_{2}- x_{1} } = \frac{3-(-2)}{-2-1} =\frac{5}{-3}$

                        $m =- \frac{5}{3}$

• Equation of a required line can be found using the formula,

                         $\frac{y_{} -y_{2} }{x_{}- x_{2} } = m$

                         $\frac{y-3}{x+2} = \frac{-5}{3}$

upon cross-multiplication, we get

          $3(y-3) = -5(x+2)$

               $3y-9 = -5x-10$

                  $3y+5x = -1$

Therefore, the required equation of the line is given by $3y+5x = -1$ .

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Answer 2

Answer:

The required equation of the line is $5x+3y+1=0$.

Step-by-step explanation:

Concept:-

• An equation of a straight line is obtained when two points $(x_1,y_1)$ and $(x_2,y_2)$ are connected with minimum distance between them.
• The general form of the equation in terms of $x$ and $y$ is:-

        $y-y_1=m(x-x_1)$, where $m$ = slope of the line.

• Slope of line:-

        $m=\frac{y_2-y_1}{x_2-x_1}$

Step 1 of 1

Consider the points as follows:

$(x_1,y_1)=(1,-2)$

$(x_2,y_2)=(-2,3)$

Calculate the required slope of the line as follows:

⇒ $m=\frac{3-(-2)}{-2-1}$

⇒ $m=\frac{3+2}{-3}$

⇒ $m=-\frac{5}{3}$

Thus, the slope of the line is -5/3.

So, the equation of the line is,

$y-y_1=m(x-x_1)$

Substitute the values of $x_1,y_1$ and $m$ as follows:

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

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

Further, simplify as follows:

$3(y+2)=-5(x-1)$

$3y+6=-5x+5$

$5x+3y+1=0$

Final Answer: The equation of the line is $5x+3y+1=0$.

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