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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
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 isA) 6x-8y+15=0B)6x+8y-15=0C)6x+8y+15=0D)8x+6y+15=0

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2 years ago

```							AB = 3 unit , AC = 7 unitBC = √(3² + 4²) = 5 unit ( by distance formula)Let coordinates of A be ( x,y )At first calculate the value of x & yAC = √(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)’ is9–6x -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 = 0.4Now, To Find : the equation of altitude ADSlope of BC = (4–0)/(3–0) = 4/3,& slope of perpendicular AD = -1 ÷ (4/3) = -3/4Now, using the coordinates of A & the slope of AD , we can find the equation of altitude ADSince, (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 = - 7.5So altitude AD is 3X+ 4Y = – 7.5 6x + 8y + 15 = 0
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2 years ago
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