Please solve q 101?

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1 Answer
May 23, 2018

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As the type of triangle is not mentioned in the question, I would take a right angled isosceles triangle right angled at B with A(0,12), B(0,0) and C(12,0) .

Now , the point D divides AB in the ratio 1:3,

So, D(x,y)=((m_1x_2+m_2x_1)/(m_1+m_2), (m_1y_2+m_2y_1)/(m_1+m_2))

=((1*0+3*0)/(1+3),(1*0+3*12)/(1+3))=(0,9)

Similarly, E(x,y)=((m_1x_2+m_2x_1)/(m_1+m_2), (m_1y_2+m_2y_1)/(m_1+m_2))

=((1*12+3*0)/(1+3),(1*0+3*0)/(1+3))=(9,0)

Equation of line passing through A(0,12) and E(3,0) is

rarry-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)

rarry-12=(0-12)/(3-0)(x-0)

rarr4x+y-12=0.....[1]

Similarly, Equation of line passing through C(12,0) and E(0,9) is

rarry-y_1=(y_2-y_1)/(x_2-x_1)(x-x_1)

rarry-0=(9-0)/(0-12)(x-12)

rarr3x+4y-36=0.....[2]

Solving [1] and [2] by the rule of cross multiplication, we get,

rarrx/(4xx(-2)-(-36)xx1)=y/(-3xx(-12)+4xx(-36)=)=1/(3-4*4)

rarrx=12/12 and y=108/13

So, the co-ordinates of F is (12/13,108/13).

Now, (CF)^2/(FD)^2=((12/13-12)^2+(108/13-0)^2)/((0-12/13)^2+(9-108/13)^2)=(144^2+108^2)/(12^2+9^2)=144=12^2

So, (CF)/(FD)=12