Question

Find the equation of the straight line passing through the points (1, -1) and (-2, 5).​

Answer 1

Answer:

Equation of the straight line passing through the points A and B is x – intercept = 3, y – intercept = 3/2 Area of ∆OAB = 1/2|(x – intercept) ( y – intercept)| = 1/2|(3)(3/2) = 9/4 sq.units.Read more on Sarthaks.com - https://www.sarthaks.com/532891/find-the-equation-of-the-straight-line-passing-through-the-points-1-2-and-5-1-and-also-find

Answer 2

Given the two points are (1,-1). and (-2,5)

we have to find :

The equation of the straight line passing through those points.

Solution:

${\blue{ \boxed{ \boxed{ \begin{array}{cc} \bf \to \: we \: know \: that : \\ \\ \sf \: the \: equation \: of \: the \: straight \: line \\ \sf \: passing \: through \: the \: points \: \\ \rm \: (x_1,y_1) \: and \: \: (x_2,y_2) \: \: is \: : \\ \\ \red{ \boxed{ \blue{ \rm \: \frac{x - x_1}{x_1 - x_2} = \frac{y - y_1}{y_1 - y_2}}} } \\ \\ \end{array}}}}}$

According to the question,

$\rm \: x_1 =1 \\ \rm \: x_2 = - 2 \\ \rm \: y_1 = - 1 \\ \rm \: y_2 = 5$

So,

${\orange{ \boxed{ \boxed{ \begin{array}{cc} \rm \:the \: equation \: of \: the \: straight \: line \\ \rm \: passing \: through \: the \: points \: \\ \rm \: (1, - 1) \: and \: ( - 2,5) \: is : \\ \\ \\ \rm \: \frac{x - 1}{1 - ( - 2)} = \frac{y - ( - 1)}{ - 1 - 5} \\ \\ \rm \implies \: \frac{x - 1}{1 + 2} = \frac{y + 1}{ - 6} \\ \\ \rm \implies \: \frac{x - 1}{ \cancel3} = \frac{y + 1}{ - \cancel6_ {\: 2}} \\ \\ \rm \implies \:x - 1 = \frac{y + 1}{ - 2} \\ \\ \rm \implies \: - 2(x - 1) = y + 1 \\ \\ \rm \implies \: - 2x + 2 = y + 1 \\ \\ \rm \implies \:y + 2x - 1 = 0 \\ \\ \underbrace{\boxed{ \pink{\therefore \: \rm \: y + 2x - 1 = 0}}}_{ \red{\sf \: answer}}\\ \\ \end{array}}}}}$