Solution :-

GIVEN that, the vertices of the triangle are :-

• A = (1, 3)
• B = (2, -1)
• C = (0, -3)

Let us assume that,

• AD ⟂ BC .
• CF ⟂ AB .
• BE ⟂ AC .

so,

→ Slope of AB = (y2 - y1) / (X2 - x1) = (-1 - 3) / (2 - 1) = (-4)/1 = (-4)

then,

→ Slope of CF = Perpendicular slope of AB

→ Slope of CF = (-1) / Slope of AB

→ Slope of CF = (-1) / (-4)

→ Slope of CF = (1/4) .

now, Equation of CF :-

→ y - y1 = m(x - x1)

→ y + 3 = (1/4)(x - 0)

→ 4y + 12 = x

→ x - 4y = 12 ----------------- EQN.(1)

Similarly,

→ Slope of BC = (y2 - y1) / (x2 - x1) = (-3 + 1) / (0 - 2) = (-2)/(-2) = 1

then,

→ Slope of AD = Perpendicular slope of BC

→ Slope of AD = (-1) / Slope of BC

→ Slope of AD = (-1) / 1

→ Slope of AD = (-1) .

now, Equation of AD :-

→ y - y1 = m(x - x1)

→ y - 3 = (-1)(x - 1)

→ y - 3 = -x + 1

→ x + y = 4 ----------------- Eqn.(2)

Subtracting Eqn.(2) from Eqn.(1) we GET,

→ (x - 4y) - (x + y) = 12 - 4

→ x - x - 4y - y = 8

→ (-5y) = 8

→ y = (-8/5)

putting value of y in Eqn.(2) ,

→ x - (8/5) = 4

→ x = 4 + (8/5)

→ x = (28/5)

therefore, the orthocentre of the triangle are (28/5 , -8/5) .

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