Given-the-lines-l-1-3mx-3y-9-and-l-2-y-mx-c-find-the-value-of-m-and-c-if-the-point-1-2-lie-on-both-lines-hence-the-tangent-of-the-curve-y-mx-c-2-when-it-moves-across-the-x-

Question Number 39591 by Rio Mike last updated on 08/Jul/18

G i v e n t h e l i n e s

l_{1} : - 3 m x + 3 y = 9

a n d l_{2} : y = m x + c

f i n d t h e v a l u e o f m a n d c i f

t h e p o i n t (1, 2) l i e o n b o t h l i n e s .

h e n c e t h e t a n g e n t o f t h e

c u r v e y = {(m x + c)}^{2}

w h e n i t m o v e s a c r o s s t h e x - a x i s

Answered by MJS last updated on 08/Jul/18

(\begin{matrix} 1 \\ 2 \end{matrix}) on l_{1} : - 3 m + 6 = 9 \Rightarrow m = - 1

(\begin{matrix} 1 \\ 2 \end{matrix}) on l_{2} : 2 = - 1 \times 1 + c \Rightarrow c = 3

y = {(3 - x)}^{2}

{(3 - x)}^{2} = 0 \Rightarrow x = 3

curve touches x - axis \Rightarrow tangent = x - axis

in point (\begin{matrix} 3 \\ 0 \end{matrix}) \Rightarrow t : y = 0

if you mean curve moves across y - axis :

y = {(3 - 0)}^{2} = 9

tangent in (\begin{matrix} 0 \\ 9 \end{matrix}) :

y = k x + d

d = 9

f (x) = {(3 - x)}^{2}

f^{'} (x) = 2 x - 6

k = f^{'} (0) = - 6

t : y = - 6 x + 9

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