Question

Question 8
The points (-4,0), (4,0) and (0,3) are the vertices of a
(A) Right angled triangle
(B) Isosceles triangle
(C) Equilateral triangle
(D) Scalene triangle

Answer 1

Let A(-4,0), B(4,0) and C(0,3) are the given vertices.
Now, distance between A(-4,0) and B(4,0),
$\because$ Distance between two points $(x_1,y_1)$ and $(x_2,y_2)$
$D=\sqrt{(x_2-x_1{)}^2+(y_2-y_1{)}^2}$
$AB=\sqrt{[4-(-4){]}^2+(0-0{)}^2]}$
$=\sqrt{(4+4{)}^2}=\sqrt{8^2}=8$
Distance between B(4,0) and C(0,3),
BC=$\sqrt{(0-4{)}^2+(3-0{)}^2}=\sqrt{16+9}=\sqrt{25}=5$
Distance between A(-4,0) and C(0,3),
AC=$\sqrt{[0-(-4){]}^2+(3-0{)}^2}=\sqrt{16+9}=\sqrt{25}=5$
$\therefore$ BC=AC
Hence, $\Delta ABC$ is an isosceles triangle because an isosceles triangle has two sides equal.