Let the three sides of a triangle are on the lines 4 x − 7 y + 10 = 0 , x + y = 5 and 7 x + 4 y = 15 . Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines x =

Question

Let the three sides of a triangle are on the lines $4 x - 7 y + 10 = 0, x + y = 5$ and $7 x + 4 y = 15$ . Then the distance of its orthocentre from the orthocentre of the triangle formed by the lines $x = 0, y = 0$ and $x + y = 1$ is

Question diagram: Let the three sides of a triangle are on the lines 4 x − 7 y

Options:

Answer:

B

Solution:

$7 x + 4 y = 15 7 x + 7 y + 10 = 0$
distance between P and $B = 5$

TestHub · Accepted Answer

7 x + 4 y = 15 7 x + 7 y + 10 = 0 ​ distance between P and B = 5 ​

Mathematics - Point Question with Solution | TestHub

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