100)
2. The three lines 4x-7y+10=0,x+y=5 and 7x+4y=15 form the sides of a triangle.
Then the point (1,2) is it
(1.2)
(A) controid
(5) Incontre
(C) orthocante
(D) circumference
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Answer:
Given lines
4x−7y+10=0−−−−(1)
x+y−5=0−−−−(2)
7x+4y−15=0−−−−(3)
On solving eq (1) and (3)
7x+4(
7
4x+10
)−15=0
49x+16x+40−105=0
65x=65⇒x=1
From eq (1)
4−7y=−10
−7y=−14
y=2
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