ex. 12a q7

Asked by nilesh.dhote74 | 7th May, 2020, 10:35: PM

Expert Answer:

TO find the slopes of (i) BC  (ii) altitude AD  (iii) median AM
Here, A(7, 3), B(4, -1) and C(6, 5)
 
left parenthesis i right parenthesis space
S l o p e space o f space B C equals fraction numerator y subscript 2 minus y subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction equals fraction numerator 5 minus left parenthesis negative 1 right parenthesis over denominator 6 minus 4 end fraction equals 6 over 2 equals 3
left parenthesis i i right parenthesis
A l t i t u d e space A D space i s space p e r p e n d i c u l a r space t o space s i d e space B C.
therefore space S l o p e space o f space A D equals negative fraction numerator 1 over denominator S l o p e space o f space B C end fraction
therefore space S l o p e space o f space A D space equals space minus 1 third
left parenthesis i i i right parenthesis
M space i s space t h e space m i d p o i n t space o f space s i d e space B C
rightwards double arrow C o o r d i n a t e s space o f space M space w i l l space b e space open parentheses fraction numerator 4 plus 6 over denominator 2 end fraction comma fraction numerator negative 1 plus 5 over denominator 2 end fraction close parentheses
rightwards double arrow M identical to open parentheses 5 comma space 2 close parentheses
rightwards double arrow S l o p e space o f space A M equals fraction numerator y subscript 2 minus y subscript 1 over denominator x subscript 2 minus x subscript 1 end fraction equals fraction numerator 2 minus 3 over denominator 5 minus 7 end fraction equals 1 half

Answered by Renu Varma | 8th May, 2020, 11:40: AM