if the line 3x+2y−1=0 transformed by matrix A= (((1 a)),((b 2)) ) such that the image is the line 2x+8y+c=0 find the value of a×b×c


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Question Number 95260 by i jagooll last updated on 24/May/20

$if the line 3 x + 2 y - 1 = 0 transformed$ $by matrix A = (\begin{matrix} 1 a \\ b 2 \end{matrix}) such that$ $the image is the line 2 x + 8 y + c = 0$ $find the value of a \times b \times c$
Commented by mr W last updated on 24/May/20

$(x, y) \to (u, v) w i t h A = (\begin{matrix} 1 a \\ b 2 \end{matrix})$ $u = x + a y$ $v = b x + 2 y$ $2 (x + a y) + 8 (b x + 2 y) + c = 0$ $2 (1 + 4 b) x + 2 (a + 8) y + c = 0$ $\equiv 3 x + 2 y - 1 = 0$ $2 (1 + 4 b) = 3 \Rightarrow b = \frac{1}{8}$ $2 (a + 8) = 2 \Rightarrow a = - 7$ $c = - 1$ $\Rightarrow a \times b \times c = \frac{7}{8}$
Commented by mr W last updated on 24/May/20

$i a m n o t s u r e i f i u n d e r s t a n d t h e$ $q u e s t i o n c o r r e c t l y . p l e a s e c h e c k!$
Commented by i jagooll last updated on 24/May/20

$thank you sir .$ $you and sir john got the same result$
	Answered by john santu last updated on 24/May/20

	$(\begin{matrix} x^{'} \\ y^{^{'}} \end{matrix}) = (\begin{matrix} 1 a \\ b 2 \end{matrix}) (\begin{matrix} x \\ y \end{matrix}) = (\begin{matrix} x + ay \\ bx + 2 y \end{matrix})$ $image of the line {2 x}^{'} + {8 y}^{'} + c = 0$ $\Rightarrow 2 (x + ay) + 8 (bx + 2 y) + c = 0$ $it similar to line 3 x + 2 y - 1 = 0$ $\Rightarrow (2 + 8 b) x + (2 a + 16) y + c = 0$ $we get 2 + 8 b = 3 \Rightarrow b = \frac{1}{8}$ $2 a + 16 = 2 \Rightarrow a = - 7$ $and c = - 1 so a \times b \times c = \frac{7}{8} .$
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