Question

100 points! Help Alicia! :)

The equation of two sides AB and AC of a
triangle ABC are respectively 3x + 4y +9-0 and
4x - 3y +16--0. If the third side BC passes through
D (5,2) and BDD: DC 4:5.Find the equation of
the third side. The given answer is X -5
Please answer this quickly. Explain your answer
nicely.

Answer 1

here many concepts will use
so , I am solving this one stepwise .

step 1 :- here given equations of two sides AB ; 3x +4y + 9 = 0
AC ; 4x -3y + 16 =0

both are perpendicular lines .
becoz slope of AB× slope of AC = -1
so, angle A = 90°

step 2 :- let slope of line BC = m
then,
concept of angle between two lines ,
e.g
tan∅ = | m2 - m1 |/| 1 + m1 × m2 |

now , for angle ABC = ∅ ( let )

tan∅ = | m +3/4 |/| 1-3m/4 | -------(1)

for angle ACB = 90-∅

tan(90-∅) = cot∅ =| m-4/3|/| 1+ 4m/3 | ------(2)

step 3 :- multiply (1) and (2)

tan∅.cot∅ = |m +3/4| ×|m-4/3| /| 1-3m/4 | × |1+4m/3 |

|4m +3 | × | 3m -4 | = | 4 -3m | × | 3 + 4m |

( 4m +3)(3m -4) =± (4 -3m)( 3 +4m)

case 1 :-
(4m +3)(3m-4) = (4-3m)(3+4m)

2(4m +3)( 3m -4) = 0
m = -3/4 and 4/3

case 2 :-
(4m+3)(3m-4) = -(4-3m)(3+4m)

here no value of m found ,

if in co-ordinate for value of m doesn't found any results then , we let slope = ∞

so, m = ∞

now, you see here
m = ∞ , 4/3 and -3/4.

these are slopes of all lines of ∆ABC.

we know, m =4/3 is slope of AC , m = -3/4 is slope of AB .

then, rest m = ∞
is slope of BC.

now, equation of line BC is
(y -2) =∞ (x -5)

(y -2) = 1/0(x -5)

x = 5

hence, x = 5 is equation of lines .


[note :- you can check this by using questions data ,D is the point which divide the line BC in 4: 5 ]

Answer 2

Answer is in attachment

thank you