Slope of a line m joining (x1,y1) and (x2,y2) = y2- y1 / x2 - x1 The coordinates of the vertices of triangle XYZ are X(-5,5), Y(-3,-2) and Z(4,0) The coordinates of the vertices of quadrilateral DEFG are D(-2,5), E(2,4), F(0,0) and G(-4,1) The coordinates of the vertices of quadrilateral JKLM are J(-4,1), K(2,3), L(5.-3) and M(0,-5) The coordinates of the vertices of quadrilateral PQRS are P(-4,2), Q(3,4), R(5,0) and S(-3,-2) Slope of RS m1 = (4-1) /[-1-(-3)] = 3/2 Slope of RT m2 = (1-1)/[3-(-3)] = 0/6 =0 Slope of ST m3 = (1-4) /[3-(-1)] = -3/4 As, no two of above slopes are in negative (opposite) reciprocals or no two of these slopes have a product of -1, non of two lines are perpendicular to each other. So, Triangle RST is not a right triangle because no two sides are perpendicular. The slope of XZ is -5/9, the slope of YZ is 2/7 and the slope of XY is -7/2. Triangle XYZ is a right triangle because two of these slopes have a product of -1. **( YZ x XY = 2/7 x -7/2 = -1) Slope of DE is 4-5/2-(-2) = -1/4, Slope of EF is 0-4/0-2 = 2 Slope of FG is 1-0/-4-0 = -1/4 Slope of GD is 5-1/-2-(-4) = 2 AS the slopes of DE=FG = -1/4, and the slopes of EF=GD = 2 ...(equal) DE parallel to FG and EF is parallel to GD. DE = √ [ 4^2 + (-1)^2] = 3 EF = √ [ 1^2 + 2^2] = √5 FG = √[ 4^2 + (-1)^2] = 3 GD = √ [ 1^2 + 2^2] = √5 DE = FG and EF=GD So, quadrilateral DEFG is not a rhombus because opposite sides are parallel but four sides do not all have the same length. The slope of JK is 1/3, the slope of LK is -2, the slope of ML is 2/5 and the slope of MJ is -3/2 Quadrilateral JKLM is not a parallelgram because neither opposite sides are parallel. **( Parallel lines are having equal slope) Slope of PQ is 4-2/3-(-4) = 2/7, Slope of QR is 0-4/5-3 = -2 Slope of RS is -2-0/-3-5 = 1/4 Slope of SP is 2-(-2)/-4-(-3) = -4 Only RSxSP = -1, and non of products of two consecutive lines is -1. Only RS is perpendicular to SP. So quadrilateral PQRS is not a rectangle because it has only one right angle.