Consider a triangle ABC with AB=3units,AC=7units,B=(0,0)and C=(3,4).the possible equation of the altitude of triangle drawn from A is

A) 6x-8y+15=0

B)6x+8y-15=0

C)6x+8y+15=0

D)8x+6y+15=0

AB = 3 unit , AC = 7 unit

BC = √(3² + 4²) = 5 unit ( by distance formula)

Let coordinates of A be ( x,y )

At first calculate the value of x & y

AC = √(x-3)² + (y-4)² = 7 …….(1)

AB = √(x² + y²)=3 ……. (2)

=> AC² = x²+9 - 6x + y² +16 - 8y = 49

=> x² + y² -6x - 8y = 24 ……….. (1)’

& AB² = x² + y² = 9 …….. (2)’

So, Eq (1)’ is

9–6x -8y = 24

=> 6x + 8y = -15 ……….. (3)

Now, area (triangle ABC) is calculated by Heron's formula

= √{7.5 * 0.5 * 4.5 * 2.5}

= (15√3)/4 cm² ………. (4)

Now area( tri ABC) is calculated by coordinates

=0.5*[x1 ( y2-y3) + x2( y3-y1) + x3( y1-y2) ]

= 0.5* [ -4x + 3y ] cm² ……….(5)

Since eq(4) = eq(5)

=> -4x +3y = (15√3/4) * 2

=> -8x + 6y = 15√3 ……… (6)

Now, by solving eq (3) & eq (6)

We get, x= -3 & y = 0.4

Now, To Find : the equation of altitude AD

Slope of BC = (4–0)/(3–0) = 4/3

,& slope of perpendicular AD = -1 ÷ (4/3) = -3/4

Now, using the coordinates of A & the slope of AD , we can find the equation of altitude AD

Since, (Y - b) = m( X -a) , where, b= y & a= x

=> Y-0.4 = -3/4 { X-(-3) }

=> Y - 0.4 = -3X/4 - 9/4

=> Y + 3X/4 = - 9/4 + 0.4

=> 4Y + 3X = - 7.5

So altitude AD is 3X+ 4Y = – 7.5

6x + 8y + 15 = 0