Question

Find the equation of the line joining the points whose x coordinates on the curve y = 2x^2 - 3 are -1 and 1.​

Answer 1

Answer:

The equation of the line joining the points (-1,-1), (1,-1) is

y +1 =0              

Step-by-step explanation:

Step(i):-

Given curve y = 2 x² -3  , x = -1 and x=1

substitute x =-1 in given curve y = 2 x² -3

                                              y  = 2 (-1 )² - 3

                                             y = 2 -3

                                            y = - 1

First point A( -1 ,-1)

substitute x =1 in given curve y = 2 x² -3

                                                y = 2 (1)² - 3

                                             y = -1

Second point B( 1 , -1 )

Step(ii):-

A( -1, -1 ) and B( 1,-1)

slope of the line

                $m = \frac{y_{2}-y_{1} }{x_{2} -x_{1} } = \frac{-1-(-1)}{1-(-1)} =0$

The equation of the straight line passing through the point ( -1,-1 ) having slope m =0

              y - y₁ = m ( x - x₁ )

              y - (-1) = 0( x-(-1)

              y +1 =0

The equation of the line joining the points (-1,-1), (1,-1) is  y +1 =0