he orthocentre of triangle formed by lines 4x−7y+10=0, x+y=54𝑥-7𝑦+10=0, 𝑥+𝑦=5 and 7x+4y=157𝑥+4𝑦=15 is
Question
Solution 1
The orthocentre of a triangle is the point of intersection of the altitudes of the triangle. The altitudes of a triangle are the perpendiculars drawn from the vertices of the triangle to the opposite sides.
The given lines are:
- 4x - 7y + 10 = 0
- x + y = 5
- 7x + 4y = 15
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