Question

two sides of a rhombus ABCD are parallel to the line x-y=5and 7x-y=3.the diagonal intersect at (2,1) then the equation of the diagonal are
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Answer 1

Given : two sides of a rhombus ABCD are parallel to the line x-y=5and 7x-y=3.the diagonal intersect at (2,1)

To find : the equation of the diagonal

Solution:

two sides of a rhombus ABCD are parallel to the

line  x-y=5  and   7x-y= 3.

slope of lines =  1    &  7

Diagonal of rhombus is angle bisector

=> Let say Slope of Diagonal  = m

now using   tan α = | (m₁ - m₂ ) /( 1 + m₁m₂) |

|  (m - 7) / ( 1 + 7m) |  = | (m - 1) / ( 1  + m ) |

Case 1 :    (m - 7) / ( 1 + 7m )   = (m- 1) / ( 1  + m )

=>  m² - 6m - 7 = 7m² - 6m - 1

=> 6m² = -6

=> m² = - 1  

not possible

Case 2 :

(m - 7) / ( 1 + 7m )   = -(m- 1) / ( 1  + m )

=>  m² - 6m - 7 = -7m² + 6m + 1

=>  8m² - 12m - 8 = 0

=> 2m² - 3m - 2 = 0

=> 2m² - 4m + m - 2 = 0

=> 2m(m - 2) + 1(m - 2) = 0

=> (2m + 1)(m - 2) =0

=> m = -1/2  & m = 2

(-1/2) 2 = - 1

Hence these two slopes are perpendicular to each other

=> these two are slopes of two Diagonal of Rhombus

Equation of diagonal becomes

y  = 2x  + c     &  y = -x/2  + c

passes through ( 2, 1)

=> 1 = 2(2) + c => c = - 3  hence y = 2x - 3   => 2x - y = 3

=> 1 = -2/2 + c => c = 2  hence y = -x/2  + 2   => x + 2y  = 4

2x - y = 3  and  x + 2y  = 4 are the equation of diagonals

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