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−axis


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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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