If the equations of three sides of a triangle | vedclass

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

Question

(A) The equations of the three sides are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.$1$. Intersection of $L_1$ and $L_2$: $x = 2, y = -1 \implie

Vedclass · Accepted Answer

$(4, 0)$

Vedclass · Answer

(A) The equations of the three sides are $L_1: x = 2$,$L_2: y = -1$,and $L_3: x + 2y = 4$.$1$. Intersection of $L_1$ and $L_2$: $x = 2, y = -1 \implies A(2, -1)$.$2$. Intersection of $L_1$ and $L_3$: $x = 2, 2 + 2y = 4 \implies 2y = 2 \implies y = 1 \implies B(2, 1)$.$3$. Intersection of $L_2$ and $L_3$: $y = -1, x + 2(-1) = 4 \implies x - 2 = 4 \implies x = 6 \implies C(6, -1)$.Let the circumcentre be $O(x, y)$. The distance $OA^2 = OB^2 = OC^2$.$OA^2 = (x - 2)^2 + (y + 1)^2$$OB^2 = (x - 2)^2 + (y - 1)^2$$OC^2 = (x - 6)^2 + (y + 1)^2$Equating $OA^2 = OB^2$: $(x - 2)^2 + (y + 1)^2 = (x - 2)^2 + (y - 1)^2 \implies (y + 1)^2 = (y - 1)^2 \implies y^2 + 2y + 1 = y^2 - 2y + 1 \implies 4y = 0 \implies y = 0$.Equating $OA^2 = OC^2$ with $y = 0$: $(x - 2)^2 + (0 + 1)^2 = (x - 6)^2 + (0 + 1)^2 \implies (x - 2)^2 = (x - 6)^2 \implies x^2 - 4x + 4 = x^2 - 12x + 36 \implies 8x = 32 \implies x = 4$.Thus,the circumcentre is $(4, 0)$.

If the equations of three sides of a triangle are $x = 2$,$y + 1 = 0$,and $x + 2y = 4$,then the coordinates of the circumcentre of this triangle are:

Similar Questions

The area of the quadrilateral formed by the lines $x+2y+3=0$,$2x+4y+9=0$,$x-2y+3=0$,and $3x-6y+11=0$ is

The incentre of the triangle formed by the lines $x+y=1$,$x=1$,and $y=1$ is

Let one of the sides of a triangle be $17 \text{ cm}$ and the sum of all the sides of the triangle be $40 \text{ cm}$. If the sum of two adjacent sides is $35 \text{ cm}$, then the area (in $\text{cm}^2$) of the triangle is (in $\sqrt{2}$)

Let $a, b, c$ be the side-lengths of a triangle and $l, m, n$ be the lengths of its medians. Put $K = \frac{l+m+n}{a+b+c}$. Then,as $a, b, c$ vary,$K$ can assume every value in the interval

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

Vedclass Test Series

Exam Paper Generator

Online Exam Module