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-

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} .