Question

The bisector of acute angle between the lines x+y-3=0 7x-y+5=0

Answer 1

Answer:

x-3y+10=0

Step-by-step explanation:

Equation of acute angle bisector of ,

a1x+b1y+c1=0 and a2x+b2y+c2=0 is

(a1x+b1y+c1)/(a1^2+b1^2)^(1/2)==(a2x+b2y+c2)/(a2^2+b2^2)^(1/2)

x+y-3/5(2)^(1/2)=7x-y+5/5(2)^(1/2)

==> x-3y+10=0

Answer 2

$12\sqrt2x + (5\sqrt2+1)y - 20\sqrt2 = 0$ is required equation

Step-by-step explanation:

One of the given lines, x + y - 3= 0 & 7x - y + 5 = 0

Thus it is in the form of $a_1x + b_1y + c_1 = 0$

$a_1 = 1, b_1 = 1\ and\ c_1 = -3$  """(1)

Other line,

7x - y + 5 = 0  """(2)

$a_2 = 7, b_2 = -1\ and\ c_2 = 1$

Now, $a_1a_2 + b_1b_2 = 1\times 7 + 1 \times -1$

= 7 - 1 = 6

And 6 > 0

So, Equation of bisector of acute angle is -

$\frac {a_1x + b_1y + c_1}{\sqrt{a_1^2 + b_1^2}} = \frac {a_2x + b_2y + c_2}{\sqrt{a_2^2 + b_2^2}}$

$\frac {x+y-3}{\sqrt{1^2+1^2}} = -\frac {7x-y+5}{\sqrt{7^2+ (-1)^2}}$

$\frac {x+y-3}{\sqrt2} = \frac {7x-y+5}{5\sqrt2}$

$(x+y-3)5\sqrt2 = -\sqrt2(7x-y+5)$

$5\sqrt2x + 5\sqrt2y -15\sqrt2 = -7\sqrt2x + y\sqrt2 -5\sqrt2$

$(5+7)\sqrt2x + (5\sqrt2+1)y - (15+5)\sqrt2 = 0$

$12\sqrt2x + (5\sqrt2+1)y - 20\sqrt2 = 0$ Answer !