Basic straight line problems

Geometry Level 3

If a triangle is created by A(3,4); B(2t,5); C(6,t) points and its area is 19.5 square unit,then what are the values of t?

-3,8 − 3 2 \frac{-3}{2} ,5 5,-1 -2, 7 1 2 7\frac{1}{2}

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Mohammad Khaza by Mohammad Khaza , 20, Bangladesh

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1 solution

Mohammad Khaza
Jan 25, 2018

as the three points of the triangle is A(3,4); B(2t,5);C(6,t), \text{as the three points of the triangle is A(3,4); B(2t,5);C(6,t),}

so, ∆ABC= \text{∆ABC=} 1 2 \frac{1}{2} [ 3 4 1 2 t 5 1 6 t 1 ] \begin{bmatrix}3&4 &1\\2t&5&1\\6&t&1\end{bmatrix}

= 1 2 ( 15 − 3 t − 8 t + 24 + 2 t 2 − 30 ) =\frac{1}{2}(15-3t-8t+24+2t^2-30)

= 1 2 ( 2 t 2 − 11 t + 9 ) =\frac{1}{2}(2t^2-11t+9)

now, \text{now,}

∆ A B C = 19.5 ∆ABC=19.5

or, 1 2 ( 2 t 2 − 11 t + 9 ) = 19.5 \frac{1}{2}(2t^2-11t+9)=19.5

or, 2 t 2 − 11 t + 9 = 39 2t^2-11t+9=39

or, 2 t 2 − 11 t − 30 = 0 2t^2-11t-30=0

or, ( 2 t − 15 ) ( t + 2 ) = 0 (2t-15)(t+2)=0

so, t=-2 or, t=7.5 \text{t=-2 or, t=7.5}

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