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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
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