x+8y-22=0,5x+2y-34=0,and 2x-3y+13=0 are the t | vedclass

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

Question

(B) Given the three sides of a triangle are:$x+8y-22=0$ $(i)$$5x+2y-34=0$ $(ii)$$2x-3y+13=0$ $(iii)$Solving $(i)$ and $(ii)$:$x=6, y=2$. Vertex $A = (

Vedclass · Accepted Answer

$19$ sq units

Vedclass · Answer

(B) Given the three sides of a triangle are:$x+8y-22=0$ $(i)$$5x+2y-34=0$ $(ii)$$2x-3y+13=0$ $(iii)$Solving $(i)$ and $(ii)$:$x=6, y=2$. Vertex $A = (6, 2)$.Solving $(ii)$ and $(iii)$:$x=4, y=7$. Vertex $B = (4, 7)$.Solving $(i)$ and $(iii)$:$x=-2, y=3$. Vertex $C = (-2, 3)$.The area of the triangle with vertices $(x_1, y_1), (x_2, y_2), (x_3, y_3)$ is given by:$	ext{Area} = \frac{1}{2} |x_1(y_2-y_3) + x_2(y_3-y_1) + x_3(y_1-y_2)|$$	ext{Area} = \frac{1}{2} |6(7-3) + 4(3-2) + (-2)(2-7)|$$	ext{Area} = \frac{1}{2} |6(4) + 4(1) - 2(-5)|$$	ext{Area} = \frac{1}{2} |24 + 4 + 10|$$	ext{Area} = \frac{1}{2} |38| = 19$ sq units.

$x+8y-22=0$,$5x+2y-34=0$,and $2x-3y+13=0$ are the three sides of a triangle. The area of the triangle is

Similar Questions

The equations of the sides of a triangle are $x - 3y = 0$,$4x + 3y = 5$,and $3x + y = 0$. The line $3x - 4y = 0$ passes through:

The points $(0, 0)$,$(a, 0)$,and $\left( \frac{a}{2}, \frac{a\sqrt{3}}{2} \right)$ are vertices of:

Let $P$ be an interior point of a convex quadrilateral $ABCD$ and $K, L, M, N$ be the mid-points of $AB, BC, CD, DA$ respectively. If $\text{Area}(PKAN) = 25$,$\text{Area}(PLBK) = 36$,and $\text{Area}(PMDN) = 41$,then $\text{Area}(PLCM)$ is

If the points $(a, 0)$,$(0, b)$,and $(1, 1)$ are collinear,then:

$A$ straight line $4x+y-1=0$ passing through the point $A(2,-7)$ meets the line $BC$ whose equation is $3x-4y+1=0$ at the point $B$. Then the equation of the line $AC$ such that $AB=AC$,is

Vedclass Test Series

Exam Paper Generator

Online Exam Module