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Find the equations of the diagonals of a rectangle whose sides are x + 1 = 0, x - 4 = 0, y + 1 = 0 and y - 2 = 0.

Straight Line Eq

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From graph,

Find the equations of the diagonals of a rectangle whose sides are x + 1 = 0, x - 4 = 0, y + 1 = 0 and y - 2 = 0. Mixed Practice, Concise Mathematics Solutions ICSE Class 10.

The lines intersect at point I, J, K and L.

By two point formula,

Equation of line : y - y1 = y2−y1x2−x1(x−x1)\dfrac{y2- y1}{x2 - x1}(x - x_1)

Equation of diagonal IK is

⇒y−2=−1−24−(−1)[x−(−1)]⇒y−2=−35[x+1]⇒5(y−2)=−3[x+1]⇒5y−10=−3x−3⇒5y+3x−10+3=0⇒5y+3x−7=0.\Rightarrow y - 2 = \dfrac{-1 - 2}{4 - (-1)}[x - (-1)] \\[1em] \Rightarrow y - 2 = -\dfrac{3}{5}[x + 1] \\[1em] \Rightarrow 5(y - 2) = -3[x + 1] \\[1em] \Rightarrow 5y - 10 = -3x - 3 \\[1em] \Rightarrow 5y + 3x - 10 + 3 = 0 \\[1em] \Rightarrow 5y + 3x - 7 = 0.

Equation of diagonal LJ is

⇒y−(−1)=2−(−1)4−(−1)[x−(−1)]⇒y+1=35[x+1]⇒5(y+1)=3[x+1]⇒5y+5=3x+3⇒5y−3x+5−3=0⇒5y−3x+2=0.\Rightarrow y - (-1) = \dfrac{2 - (-1)}{4 - (-1)}[x - (-1)] \\[1em] \Rightarrow y + 1 = \dfrac{3}{5}[x + 1] \\[1em] \Rightarrow 5(y + 1) = 3[x + 1] \\[1em] \Rightarrow 5y + 5 = 3x + 3 \\[1em] \Rightarrow 5y - 3x + 5 - 3 = 0 \\[1em] \Rightarrow 5y - 3x + 2 = 0.

Hence, equation of diagonals are 5y + 3x - 7 = 0 and 5y - 3x + 2 = 0

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