If two equal sides of an isosceles triangle a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The third side of an isosceles triangle is perpendicular to the angle bisector of the two equal sides. However,in this specific geometry,the third

Vedclass · Accepted Answer

$x-3y=17$

Vedclass · Answer

(A) The third side of an isosceles triangle is perpendicular to the angle bisector of the two equal sides. However,in this specific geometry,the third side is parallel to the angle bisector of the given lines $7x-y+3=0$ and $x+y-3=0$ if the vertex is the intersection point. First,find the intersection point of the two lines: $7x-y+3=0$ and $x+y-3=0$. Adding them gives $8x=0$,so $x=0$,which implies $y=3$. The vertex is $(0,3)$. The angle bisectors are given by $\frac{7x-y+3}{\sqrt{7^2+(-1)^2}} = \pm \frac{x+y-3}{\sqrt{1^2+1^2}}$. $\frac{7x-y+3}{\sqrt{50}} = \pm \frac{x+y-3}{\sqrt{2}} \Rightarrow 7x-y+3 = \pm 5(x+y-3)$. Case $1$: $7x-y+3 = 5x+5y-15$ $\Rightarrow 2x-6y+18=0$ $\Rightarrow x-3y+9=0$. Case $2$: $7x-y+3 = -5x-5y+15$ $\Rightarrow 12x+4y-12=0$ $\Rightarrow 3x+y-3=0$. The third side is perpendicular to these bisectors. For $x-3y+9=0$,the perpendicular line is $3x+y+k=0$. Passing through $(2,-5)$: $3(2)+(-5)+k=0$ $\Rightarrow 6-5+k=0$ $\Rightarrow k=-1$. So,$3x+y-1=0$. For $3x+y-3=0$,the perpendicular line is $x-3y+k=0$. Passing through $(2,-5)$: $2-3(-5)+k=0$ $\Rightarrow 2+15+k=0$ $\Rightarrow k=-17$. So,$x-3y=17$. Comparing with options,$x-3y=17$ is present.

If two equal sides of an isosceles triangle are given by the equations $7x-y+3=0$ and $x+y-3=0$,then the equation of its third side passing through the point $(2,-5)$ is

Similar Questions

Find the equation of the locus of points equidistant from the lines $3x + 4y - 11 = 0$ and $12x + 5y + 2 = 0$.

The equation of the angle bisectors between the lines $3x + 4y - 7 = 0$ and $12x + 5y + 17 = 0$ is:

The straight line $x+y+1=0$ bisects an angle between the pair of lines of which one is $2x+3y-4=0$. Then,the equation of the other line is

The equation of the line which bisects the obtuse angle between the lines $x - 2y + 4 = 0$ and $4x - 3y + 2 = 0$ is:

The equation of the bisector of the acute angle between the lines $2x - y + 4 = 0$ and $x - 2y - 1 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module