Three vertices of triangle ABC are A(-1 ,1 ), B (-9,-8) and C (15,-2) then the equation of angle bisector of angle A is?

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Three vertices of triangle ABC are A(-1 ,1 ), B (-9,-8) and C (15,-2) then the equation of angle bisector of angle A is?

Ans : 4x + y = 7

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Answer 1 · 2018-01-10T10:49:17

$\frac{9 x - 8 y + 17}{\sqrt{29}} = \pm \frac{3 x + 16 y - 13}{\sqrt{53}}$ $(9x-8y+17)/sqrt29=+-(3x+16y-13)/sqrt53$ .

Explanation:

Let $P (x, y)$ $P(x,y)$ be any point on the $∠ - bisector$ $/_-"bisector"$ of $∠ A$ $/_A$ .

Then, from Geometry, we know that, $P$ $P$ is equidistant

from the sides $A B and A C$ $AB and AC$ .

$A (- 1, 1), & B (- 9, - 8) \Rightarrow t h e e q n . o f$ $A(-1,1), & B(-9,-8) rArr the eqn. of$ AB# is given by,

$A B : y - 1 = {\frac{1 - (- 8)}{- 1 - (- 9)}} (x - (- 1)), i . e .,$ $AB : y-1={(1-(-8))/(-1-(-9))}(x-(-1)), i.e.,$

$A B : 8 (y - 1) = 9 (x + 1), or, 9 x - 8 y + 17 = 0$ $AB : 8(y-1)=9(x+1), or, 9x-8y+17=0$ .

Similarly, $A C : 3 x + 16 y - 13 = 0$ $AC : 3x+16y-13=0$ .

Now, the $⊥ - distance d_{1}$ $bot-"distance "d_1$ from $P to A B$ $P" to "AB$ is given by,

$d_{1} = \frac{| 9 x - 8 y + 17 |}{\sqrt{9^{2} + {(- 8)}^{2}}} = \frac{| 9 x - 8 y + 17 |}{\sqrt{145}}$ $d_1=|9x-8y+17|/sqrt(9^2+(-8)^2)=|9x-8y+17|/sqrt145$ .

The $⊥ - distance d_{2}$ $bot-"distance "d_2$ from $P to A C$ $P" to "AC$ is,

$d_{2} = \frac{| 3 x + 16 y - 13 |}{\sqrt{3^{2} + 16^{2}}} = \frac{| 3 x + 16 y - 13 |}{\sqrt{265}}$ $d_2=|3x+16y-13|/sqrt(3^2+16^2)=|3x+16y-13|/sqrt265$ .

But, $d_{1} = d_{2}$ $d_1=d_2$ .

$:. |9x-8y+17|/sqrt145=|3x+16y-13|/sqrt265$ .

$:. (9x-8y+17)/sqrt29=+-(3x+16y-13)/sqrt53$ , are the desired

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Three vertices of triangle ABC are A(-1 ,1 ), B (-9,-8) and C (15,-2) then the equation of angle bisector of angle A is?

Ans : 4x + y = 7

1 Answer

Explanation:

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