These equations can be written as x + y + 0z = 5 0x + y + z = 3 x + 0y + z = 4 D=110011101=1(1 0) 1(0 1)+0(0 1)=1(1) 1( 1)+0=2D1=510311401=5(1 0) 1(3 4)+0(0 4) ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Bisectors of Angle between Two Lines

x+y=5y+z=3x+z...

Question

x+ y = 5
y + z = 3
x + z = 4

Open in App

Solution

These equations can be written as
x + y + 0z = 5
0x + y + z = 3
x + 0y + z = 4

$D = |\begin{matrix} 1 & 1 & 0 \\ 0 & 1 & 1 \\ 1 & 0 & 1 \end{matrix}| = 1 (1 - 0) - 1 (0 - 1) + 0 (0 - 1) = 1 (1) - 1 (- 1) + 0 = 2 D_{1} = |\begin{matrix} 5 & 1 & 0 \\ 3 & 1 & 1 \\ 4 & 0 & 1 \end{matrix}| = 5 (1 - 0) - 1 (3 - 4) + 0 (0 - 4) = 5 (1) - 1 (- 1) = 6 D_{2} = |\begin{matrix} 1 & 5 & 0 \\ 0 & 3 & 1 \\ 1 & 4 & 1 \end{matrix}| = 1 (3 - 4) - 5 (0 - 1) + 0 (0 - 4) = 1 (- 1) - 5 (- 1) = 4 D_{3} = |\begin{matrix} 1 & 1 & 5 \\ 0 & 1 & 3 \\ 1 & 0 & 4 \end{matrix}| = 1 (4 - 0) - 1 (0 - 3) + 5 (0 - 1) = 1 (4) - 1 (- 3) + 5 (- 1) = 2 Now, x = \frac{D_{1}}{D} = \frac{6}{2} = 3 y = \frac{D_{2}}{D} = \frac{4}{2} = 2 z = \frac{D_{3}}{D} = \frac{2}{2} = 1 ∴ x = 3, y = 2 and z = 1$

Suggest Corrections

Similar questions

Q. If

\frac{log 3}{x - y} = \frac{log 5}{y - z} = \frac{log 7}{z - x}

the

3^{x + y} \cdot 5^{y + z} \cdot 7^{z + x}

is equal to

Q. Solve the following equations

x + y + z = 9

3 x + 5 y + 7 z = 52

2 x + y - z = 0

Factorise:(3x- 5y)³+ (5y - 2z)³+ (2z - 3x)³

Q. Maximize

z = 3 x + 5 y

, Subject to

x + 4 y \leq 24

3 x + y \leq 21, x + y \leq 9, x \geq 0

. Also find maximum value of

z

Q. Factorise each of the following expression:
(i)

a^{3} x + a^{2} (x - y) - a (y + z) - z

(ii)

(x^{2} + 3 x)^{2} - 5 (x^{2} + 3 x) - y (x^{2} + 3 x) + 5 y

Join BYJU'S Learning Program

x+ y = 5 y + z = 3 x + z = 4

x+ y = 5
y + z = 3
x + z = 4