Question

7. A triangle is formed by the lines 3x + 4y = 10, 4x - 3y = 5 and 7x + y + 10 = 0. Find the
internal bisector of the angle opposite to the side 7x + y + 10 = 0.​

Answer 1

Answer:

Step-by-step explanation:

3x+4y=10  Eq1

4x-3y=5    Eq2

7x+y+10=0   Eq3

Lets first find vertex of triangle formed by given lines

3*eq1 + 4 * eq2

=> 25x = 50

=> x = 2  , y = 1

(2 , 1)

Eq1 - 4*eq3

=> -25x - 40 = 10

=> x = - 2  , y  = 4

(-2 , 4)

Eq2 + 3 * eq3

=> 25x + 30 = 5

=> x = - 1 , y = -3

(-1 , -3)

Area of Δ (2 , 1) , (-2 , 4) (-1 , -3)

= (1/2) |  2 ( 4 + 3)  -2(-3 - 1)  - 1(1 - 4)|

= (1/2) | 14 + 8 + 3|

= 25/2

if Origin (0 , 0) lies with in triangle

Then Sum of Triangle of (2 , 1) , (-2 , 4) , (0 , 0)   &  (-1 , -3)  (2 , 1) , (0, 0)  & (-2 , 4) (-1 , -3) , (0 , 0)  = 25/2

(1/2)| 2(4 -0) - 2(0 - 1) + 0(1 - 4) |  + (1/2) | -1(1 - 0) + 2(0 + 3) + 0(-3 - 1)|  +  (1/2) | -2(-3 - 0) - 1(0 - 4) + 0(4 + 3)|

= (1/2)| 8 + 2 + 0 |  + (1/2) | -1 + 6 + 0|  + (1/2) | 6 + 4 + 0|

= (1/2) | 10 |+ (1/2) | 5|  + (1/2) | 10 |

= 25/2

Hence Origin lies with in triangle

Answer 2

Given:

The triangle lines $3x + 4y = 10, 4x - 3y = 5$ and $7x+y+10=0$.

To find:

The angle opposite to the side $7x+y+10=0$.

Step-by-step explanation:

Equation of angle for is,

$\frac{A_{1} x+B_{1} y+C_{1} }{\sqrt{A_{1} ^{2} +B_{1} ^{2} } }$=± $\frac{A_{2}x+B_{2}y+C_{2}}{\sqrt{Ac^{2}+Bc^{2}}}$

$\frac{3_{x}+4_{y}-10}{\sqrt{9+16} } =\frac{4_{x}-3_{y}-5}{\sqrt{9+16} }$

$3x+4y-10=4x-3y-5$

$4x-3x-3y-4y-5+10=0$

$x-7y+5=0$

as,$7x+y-15=0$ is $7x+y+10=0$ are parallel.

so,$x-7y+5=0$

Answer:

  The angle opposite to the side ​$x-7y+5=0$.