Q14 of 112 Page 156

The vertices of a ΔABC are A (–5, 7), B (–4, –5) and C (4, 5). Find the slopes of the altitudes of the triangle.

Let AD, BE and CF be the altitudes of a ΔABC.



Since, the altitude AD is perpendicular to BC,


Slope of BC




(slope of BC) × (Slope of AD) = –1 (∵ m1m2 = –1)


Let slope of AD be m1.


⇒


⇒


Since, the altitude BE is perpendicular to AC,


Slope of AC




(slope of AC) × (Slope of BE) = –1 (∵ m1m2 = –1)


Let slope of BE be m2.


⇒


⇒


Since, the altitude CF is perpendicular to AB,


Slope of AB




(slope of AB) × (Slope of CF) = –1 (∵ m1m2 = –1)


Let slope of CF be m3.


⇒


⇒


More from this chapter

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12

The vertices of ΔABC are A (1, 8), B (–2, 4), C (8, –5). If M and N are the midpoints of AB and AC respectively, find the slope of MN and hence verify that MN is parallel to BC.

13

A triangle has vertices at (6, 7), (2, –9) and (–4, 1). Find the slopes of its medians.

 15

Using the concept of slope, show that the vertices (1, 2), (–2, 2), (–4, –3) and (–1, –3) taken in order form a parallelogram.

 16

Show that the opposite sides of a quadrilateral with vertices A (–2, –4), B (5, –1), C (6, 4) and D (–1, 1) taken in order are parallel.