Question

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
Mark​

Answer 1

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