The straight lines x+y=0,5x+y=4,and x+5y=4 fo | vedclass

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

Question

(A) The given lines are $L_1: x+y=0$,$L_2: 5x+y=4$,and $L_3: x+5y=4$.Solving $L_1$ and $L_2$: $x+y=0 \implies y=-x$. Substituting into $L_2$: $5x-x=4

Vedclass · Accepted Answer

an isosceles triangle

Vedclass · Answer

(A) The given lines are $L_1: x+y=0$,$L_2: 5x+y=4$,and $L_3: x+5y=4$.Solving $L_1$ and $L_2$: $x+y=0 \implies y=-x$. Substituting into $L_2$: $5x-x=4 \implies 4x=4 \implies x=1, y=-1$. So,$A = (1, -1)$.Solving $L_1$ and $L_3$: $x+y=0 \implies x=-y$. Substituting into $L_3$: $-y+5y=4 \implies 4y=4 \implies y=1, x=-1$. So,$C = (-1, 1)$.Solving $L_2$ and $L_3$: $5x+y=4$ and $x+5y=4$. Subtracting: $4x-4y=0 \implies x=y$. Substituting into $L_2$: $5x+x=4 \implies 6x=4 \implies x=2/3, y=2/3$. So,$B = (2/3, 2/3)$.Calculating side lengths:$AB = \sqrt{(2/3 - 1)^2 + (2/3 + 1)^2} = \sqrt{(-1/3)^2 + (5/3)^2} = \sqrt{1/9 + 25/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.$BC = \sqrt{(-1 - 2/3)^2 + (1 - 2/3)^2} = \sqrt{(-5/3)^2 + (1/3)^2} = \sqrt{25/9 + 1/9} = \sqrt{26/9} = \frac{\sqrt{26}}{3}$.$CA = \sqrt{(-1 - 1)^2 + (1 - (-1))^2} = \sqrt{(-2)^2 + 2^2} = \sqrt{4+4} = \sqrt{8} = 2\sqrt{2}$.Since $AB = BC$,the triangle is an isosceles triangle.

The straight lines $x+y=0$,$5x+y=4$,and $x+5y=4$ form

Similar Questions

If the points $(k, 3), (2, k),$ and $(-k, 3)$ are collinear,then the values of $k$ are

The opposite angular points of a square are $(3, 4)$ and $(1, -1)$. Then the coordinates of the other two points are

If the equation of the base of an equilateral triangle is $x+y=6$ and the opposite vertex is the point $(-1, -1)$,then the area of the triangle is equal to $k$ sq. units. Find the value of $k$.

Let $ABCD$ be a square of side length $1$. Let $P, Q, R, S$ be points in the interiors of the sides $AD, BC, AB, CD$ respectively,such that $PQ$ and $RS$ intersect at right angles. If $PQ = \frac{3\sqrt{3}}{4}$,then $RS$ equals

Let the equations of two sides of a triangle be $3x - 2y + 6 = 0$ and $4x + 5y - 20 = 0$. If the orthocentre of this triangle is at $(1, 1)$,then the equation of its third side is

Vedclass Test Series

Exam Paper Generator

Online Exam Module