The points (-4,0), (4,0), (0,3) are the verti | vedclass

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

Question

(C) Let the vertices be $A(-4,0), B(4,0),$ and $C(0,3)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(4 - (-4))^2

Vedclass · Accepted Answer

isosceles triangle

Vedclass · Answer

(C) Let the vertices be $A(-4,0), B(4,0),$ and $C(0,3)$.Using the distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$:$AB = \sqrt{(4 - (-4))^2 + (0 - 0)^2} = \sqrt{8^2 + 0^2} = 8$$BC = \sqrt{(0 - 4)^2 + (3 - 0)^2} = \sqrt{(-4)^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$$AC = \sqrt{(0 - (-4))^2 + (3 - 0)^2} = \sqrt{4^2 + 3^2} = \sqrt{16 + 9} = \sqrt{25} = 5$Since $BC = AC = 5$,two sides of the triangle are equal.Therefore,$	riangle ABC$ is an isosceles triangle.

The points $(-4,0), (4,0), (0,3)$ are the vertices of a

Similar Questions

If the distance between $A(-4, -3)$ and $B(6, a)$ is $10$,then $a = \ldots \ldots \ldots . . .$

If one of the end points of the diameter of a circle is $(0,0)$ and the centre of the circle is $(-2,4),$ then the other end point of the diameter is.........

The centre of a circle is $(2a, a-7)$. Find the values of $a$ if the circle passes through the point $(11, -9)$ and has a diameter of $10\sqrt{2}$ units.

Using the distance formula,show that the points $(0, -3), (1, -1),$ and $(2, 1)$ are collinear.

Two vertices of the triangle are $(3, 1)$ and $(1, 5)$. If the centroid of the triangle is the origin,then the third vertex is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module