If the vertices of Delta LMN are L(1,4), M(4, | vedclass

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(B) To determine the type of $\Delta LMN$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$

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an isosceles right angled

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(B) To determine the type of $\Delta LMN$,we calculate the lengths of its sides using the distance formula: $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$.$1$. Length of $LM = \sqrt{(4 - 1)^2 + (1 - 4)^2} = \sqrt{3^2 + (-3)^2} = \sqrt{9 + 9} = \sqrt{18} = 3\sqrt{2}$.$2$. Length of $MN = \sqrt{(4 - 4)^2 + (4 - 1)^2} = \sqrt{0^2 + 3^2} = \sqrt{9} = 3$.$3$. Length of $NL = \sqrt{(4 - 1)^2 + (4 - 4)^2} = \sqrt{3^2 + 0^2} = \sqrt{9} = 3$.Since $MN = NL = 3$,the triangle is isosceles.Also,check for a right angle using the Pythagorean theorem: $MN^2 + NL^2 = 3^2 + 3^2 = 9 + 9 = 18$,and $LM^2 = (3\sqrt{2})^2 = 18$.Since $MN^2 + NL^2 = LM^2$,the triangle is a right-angled triangle.Therefore,$\Delta LMN$ is an isosceles right-angled triangle.

If the vertices of $\Delta LMN$ are $L(1,4), M(4,1)$ and $N(4,4),$ then $\Delta LMN$ is.......

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