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Asked by araima2001 | 15th Feb, 2017, 01:01: PM

Expert Answer:

Q. The equation of the sides of the triangle having (3,-1) as a vertex and x-4y+10=0 and 6x-10y-59=0
as angle bisector and as median respectively drawn from different vertices are 
(A) 6x - 7y- 13 = 0            (B) 2x + 9y- 65 = 0
(C) 18x - 13y- 41 = 0         (D) 6x - 7y- 25 = 0
 
Dear student, in this question, the equation 6x-10y-59=0 should be 6x+10y-59=0, otherwise we will not get the correct answer which matches the options. 
Solution:
 
begin mathsize 16px style Since space open parentheses straight x subscript 1 comma space straight y subscript 1 close parentheses space space lies space on space straight x minus 4 straight y plus 10 equals 0 comma space so space it space satisfies space the space equation
rightwards double arrow space straight x subscript 1 minus 4 straight y subscript 1 plus 10 equals 0 space space space.... left parenthesis straight i right parenthesis
open parentheses fraction numerator straight x subscript 1 plus 3 over denominator 2 end fraction comma space fraction numerator straight y subscript 1 minus 1 over denominator 2 end fraction comma close parentheses space rightwards double arrow space lies space on space 6 straight x plus 10 straight y minus 59 equals 0
rightwards double arrow 6 open parentheses fraction numerator straight x subscript 1 plus 3 over denominator 2 end fraction close parentheses bold plus 10 open parentheses fraction numerator straight y subscript 1 minus 1 over denominator 2 end fraction close parentheses minus 59 equals 0
rightwards double arrow 3 open parentheses straight x subscript 1 plus 3 close parentheses plus 5 open parentheses straight y subscript 1 minus 1 close parentheses minus 59 equals 0
rightwards double arrow 3 straight x subscript 1 plus 9 plus 5 straight y subscript 1 minus 5 minus 59 equals 0
rightwards double arrow 3 straight x subscript 1 plus 5 straight y subscript 1 minus 55 equals 0 space..... left parenthesis ii right parenthesis
Solving space equations space left parenthesis straight i right parenthesis space and space left parenthesis ii right parenthesis space simultaneously comma space we space get
straight y subscript 1 equals 5 space and space straight x subscript 1 equals 10
open parentheses straight x subscript 2 comma space straight y subscript 2 close parentheses space space lies space on space space 6 straight x bold plus 10 straight y minus 59 equals 0 comma space so space it space satisfies space the space equation
rightwards double arrow 6 straight x subscript 2 plus 10 straight y subscript 2 minus 59 equals 0 space space.... left parenthesis iii right parenthesis space
Now comma space
tan space angle CBE equals tan angle EBA space..... left parenthesis Since space BE space is space the space bisector right parenthesis
rightwards double arrow fraction numerator begin display style fraction numerator straight y subscript 2 minus 5 over denominator straight x subscript 2 minus 10 end fraction minus 1 fourth end style over denominator 1 plus 1 fourth open parentheses fraction numerator straight y subscript 2 minus 5 over denominator straight x subscript 2 minus 10 end fraction close parentheses end fraction equals fraction numerator 1 fourth minus 6 over 7 over denominator 1 plus 1 fourth 6 over 7 end fraction space space space..... open square brackets Since space the space angle space space between space straight space lines space of space known space gradients space is space given space by space tanφ equals fraction numerator straight m subscript 2 minus straight m subscript 1 over denominator 1 plus straight m subscript 1 straight m subscript 2 end fraction close square brackets
rightwards double arrow fraction numerator begin display style fraction numerator 4 straight y subscript 2 minus 20 minus straight x subscript 2 plus 10 over denominator straight x subscript 2 minus 10 end fraction end style over denominator fraction numerator 4 straight x subscript 2 minus 40 plus straight y subscript 2 minus 5 over denominator 4 straight x subscript 2 minus 40 end fraction end fraction equals fraction numerator negative 17 over 28 over denominator 34 over 28 end fraction
rightwards double arrow fraction numerator begin display style fraction numerator 4 straight y subscript 2 minus 10 minus straight x subscript 2 over denominator straight x subscript 2 minus 10 end fraction end style over denominator fraction numerator 4 straight x subscript 2 minus 45 plus straight y subscript 2 over denominator 4 straight x subscript 2 minus 40 end fraction end fraction equals negative 1 half
On space simplifying space we space get comma
rightwards double arrow 2 straight x subscript 2 plus 9 straight y subscript 2 minus 65 equals 0.... left parenthesis iv right parenthesis

Solving space left parenthesis iii right parenthesis space and space left parenthesis iv right parenthesis comma space we space get
straight x subscript 2 equals fraction numerator negative 7 over denominator 2 end fraction space and space straight y subscript 2 equals 8
Now space using space the space two space point space form space of space AB comma space BC space and space CA comma space we space can space get space the space required space equations
which space are space 6 straight x minus 7 straight y minus 25 equals 0 comma space 2 straight x plus 9 straight y minus 65 equals 0 space and space 18 straight x minus 13 straight y minus 41 equals 0
Hence comma space options space left parenthesis straight B right parenthesis comma space left parenthesis straight C right parenthesis space and space left parenthesis straight D right parenthesis space are space correct.

Two space point space form colon
straight y minus straight y subscript 1 equals fraction numerator straight y subscript 2 minus straight y subscript 1 over denominator straight x subscript 2 minus straight x subscript 1 end fraction left parenthesis straight x minus straight x subscript 1 right parenthesis end style

Answered by Rebecca Fernandes | 16th Feb, 2017, 08:35: AM