Correct Answer - B
Let equation of AB bd 4x-5y-20=0 and AC be 3x-3y+6=0
Clearly, slope of `AC=3/2`
[`because` slope of ax + by +c = 0 is `-a/b`]
`:.` Slope of altitude BH, which is perpendicular to `AC=-2/3" "(becausem_("BH")=-1/m_("AC"))`
Equation of BH is given by `y-y_(1)=m(x-x_(1))`
Here, `m=-2/3,x_(1)=1" and "y_(1)=1`
`:." "y-1=-2/3(x-1)`
`rArr 2x+3y-5=0`
Now, equation of AB is 4x+5y-20=0 and equation of BH is 2x+3y-5=0
Sloving these, we get point of intersection (i.e. coordinates of B)
`{:(4x+5y-20,=,0),(4x+6y-10,=,0):}}rArry=-10`
On substituting y = -10 in 2x + 3y - 5 = 0, we get `x=(35)/2)`
`:. " "B((35)/2,-10)`
Solving 4x+5y-20=0 and 3x-2y+6=0, we get coordinate of A.
`{:(12x+15y-60,=,0),(12x-8y+24,=,0):}}rArr23y=84`
`rArr" "y=(84)/(23)rArrx=(10)/(23)`
`:." "A(10/23,84/23)`
Now, slpe of `AH=((y_(2)-y_(1))/(x_(2)-x_(1)))=((84/23-1)/(10/23-1))=61/(-13)`
`because` BC is perpendicular to AH.
`:.`Slope of BC is `13/61" "(becausem_("BC")=-1/m_("AH"))`
Now, equation of line BC is given by `y-y_(1)=m_(1)(x-x_(1)),` Where `(x_(1),y_(1))` are coordinates of B.
`:." "y-(-10)=13/61(x-35/2)`
`rArr" "y+10=13/(61xx2)(2x-35)`
`rArr" "122y+1220=26x-455`
`rArr" "26x-122y--1675=0`
