110 likes | 334 Views
Altitudes and Medians. Median of A meets the mid point of BC. A. B. ( , ). x 1 +x 2 y 1 +y 2. Median. C. 2 2. m 1 m 2 = -1. Altitude. A. B. C. Altitude of A meets BC at right angles. A. Altitude from B. B. median from C. C. Altitudes and Medians.
E N D
Median of A meets the mid point of BC A B ( , ) x1+x2 y1+y2 Median C 2 2
m1m2 = -1 Altitude A B C Altitude of A meets BC at right angles
A Altitude from B B median from C C
Altitudes and Medians Altitude of A meets BC at right angles A m1m2 = -1 B ( , ) x1+x2y1+y2 2 2 C Median of A meets the mid point of BC
A(2,1) B(5,8) C(9,6) are vertices of ABC Find equation of Median through A √ Need m and (a,b) ? find mid point B(5,8) √ √ T (7 , 7) T = (7 , 7) find mAT C(9,6) either point will do A(2,1) use y – b = m(x – a)
A(2,1) B(5,8) C(9,6) are vertices of ABC Find equation of Altitude through B Need m and (a,b) ? perpendicular to AC √ B(5,8) √ mAC =6 – 1 = 5/7 9 – 2 C(9,6) maltB = - 7/5 A(2,1) use y – b = m(x – a)
Perpendicular Bisector These cut line in half at right angles Find equation of perpendicular bisector to AB mAB =7 – 1 = -3 = 6/-2 1 – 3 A (1 , 7) (m1m2 = -1) mbisector = 1/3 m = 1/3, (a , b) = (2 , 4) use y – b = m(x – a) (2 , 4) y – 4 = 1/3(x – 2 ) 3y – 12 = 1(x – 2) B (3 , 1) 3y = x + 10
Key Question Find equation of perpendicular bisector to AB for points A(-1 , 17) B (3 , 1) mAB =1 – 17 = -4 = -16/4 3 + 1 A (-1 , 17) (m1m2 = -1) mbisector = 1/4 m = 1/4, (a , b) = (1 , 9) use y – b = m(x – a) (1 , 9) y – 9 = 1/4(x – 1 ) 4y – 36 = 1(x – 1) B (3 , 1) 4y = x + 35
and Finally Perpendicular Bisector cut line in half at right angles x1+x2y1+y2 2 2 m1m2 = -1
Key Question A(-1,3) B(4,2) C(-2,-4) are vertices of ABC Find equation of Median through A Need m and (a,b) find mid point A(-1,3) ? √ √ G = (1 , -1) 3 + 1 B(4,2) mAG = 4/-2 -2 = = -1 – 1 G m = -2 , (a , b) = (1 , -1) (1 , -1) C(-2,-4) use y – b = m(x – a) y + 1 = -2(x – 1) y + 1 = -2x + 2 y = -2x + 1