Question

The points A(-4, 0), B(4, 0) and C(0, 3) are the vertices of a triangle, which is

(a) isosceles (b) equilateral (c) scalene (d) right angled

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),
$AB=\sqrt{(4-(-4){)}^2+(0-0{)}^2}$
$=\sqrt{(4+4{)}^2}=\surd 64=8$
$DistancebetweenB(4,0)andC(0,3),$
$BC=\sqrt{(0-4{)}^2+(3-0{)}^2}=\surd (16+9)=\surd 25=5$
Distance between A(-4,0) and C(0,3)
$AC=\sqrt{\left(\right[0-(-4){]}^2+(3-0{)}^2)}=\sqrt{(16+9)}=\surd 25=5$
$BC=AC$
Hence, ΔABC is an isosceles triangle because an isosceles triangle has two sides equal.