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Question

# The vertices of a triangle are A(10,4), B(-4,9), and (-2,-1). Find the equation of its altitude which passes through (10,4)

A

x - 5y + 10 = 0

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B

5x - y + 20 = 0

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C

5x - y + 10 = 0

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D

x - 5y + 20 = 0

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Solution

## The correct option is A x - 5y + 10 = 0 Attitude which passes through A is perpendicular to the line BC. Slope of the line segment BC= =y2−y1x2−x1=−1−9−2+4=−102=−5 Let slope of the attitude be m Since, attitude and line segment BC is perpendicular m1,m2=−1 (−5),m=−1 m=15 Equation of attitude which passes through (10,4) is y−y1=m(x−x1) y−4=15(x−10) 5y-20=x-10 x-5y+10=0

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