Question

Q.6) The diagram shows (triangle)ABC with vertices A(-2, 1),
B(1,1) and C(3, 4). Given that K is the point (t, 4) and the area of
(triangle)BCK is 12 units², find the possible values of t.​

Answer 1

Question: The area of ΔBCK is 12 square units when three points B(1, 1), C(3, 4), and K(t, 4) are given. What are the values of t?

The area of a triangle is equal to $\sf{\dfrac{1}{2} \times height\times base}$.

The base $\sf{\overline{BC}}$ is equal to $\sf{\sqrt{(3-1)^2+(4-1)^2} =\sqrt{13} }$ units.

So the triangle area is equal to $\sf{\dfrac{1}{2} \times height\times \sqrt{13}}$ which is given as 12 units².

→ $\sf{\dfrac{1}{2} \times height\times \sqrt{13}=12}$

→ $\sf{height=12\times\dfrac{2}{\sqrt{13} } }$

$\sf{\therefore height=\dfrac{24}{\sqrt{13} }\;units}$

The height is the shortest distance from a point to a line.

When a line is given in the form of $\sf{ax+by+c=0}$, and a point is $\sf{P(x_1,\;y_1)}$ the distance is $\sf{\dfrac{|ax_1+by_1+c|}{\sqrt{a^2+b^2} } }$.

→ $\sf{\overleftrightarrow{BC}:\;y=\dfrac{3}{2}x-\dfrac{1}{2}}$

→ $\sf{\overleftrightarrow{BC}:\;2y=3x-1}$

→ $\sf{\overleftrightarrow{BC}:\;3x-2y-1=0}$ and a given point is K(t, 4).

According to the form, the distance is equal to $\sf{\dfrac{|3\times t-2\times4-1|}{\sqrt{(3)^2+(-2)^2} } }$.

→ $\sf{\dfrac{|3t-9|}{\sqrt{13} } =\dfrac{24}{\sqrt{13} } }$

→ $\sf{|3t-9|=24}$

→ $\sf{3|t-3|=24}$

→ $\sf{|t-3|=8}$

→ $\sf{t-3=\pm8}$

$\sf{\therefore t=11\;or\;t=-5}$