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find the equation of the straight line passing through the point (0,-1) which is perpendicular to the line y= -3/4 x-3

Sagot :

VirαtKσhli

Answer:

[tex]4x - 3y -3=0[/tex]

Step-by-step explanation:

Given equation ,

[tex]\implies y =\dfrac{-3}{4}x - 3 [/tex]

Compare it to slope intercept form to find the slope which is [tex]y=mx+c[/tex] , we have ,

  • [tex] m =\dfrac{-3}{4}[/tex]

Now we know that the product of slope of two perpendicular lines is-1 . Hence the slope of the perpendicular line is ,

  • [tex]m_{perp}= \dfrac{4}{3}[/tex]

The given point is (0,-1) , on using Point slope form of the line we have,

[tex]\implies y-y_1 = m(x - x_1) \\\\\implies y -(-1) = \dfrac{4}{3}(x -0 ) \\\\\implies y + 1 =\dfrac{4}{3}x \\\\\implies \underline{\boxed{\red{ 4x - 3y -3 = 0 }}}[/tex]

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