The vertices of a △ABC are A(1,2),B(−4,5) and C(0,1). Find the slopes of the altitudes of the triangle.
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Solution
Let AD,BE and CF be the altitudes of a △ABC.
Slope of BC=1−50+4=−1
Since the altitude AD is a perpendicular to BC,
Slope of AD=1∵m1m2=−1
Slope of AC=1−20−1=1
Thus, Slope of BE=−1∵BE⊥AC
Also, Slope of AB=5−2−4−1=−35∴ Slope of CF=53∵CF⊥AB